Structural Engineering & Statics
where the load goes
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The statics of things that do not move — page 3

Essays 49 to 72 of 97 on this thread, in the same order.
A buckling load with no compression anywhere in the member. A tube of radius 81.6 mm and wall 5 mm under pure torque — no axial load at all — buckles into a helix at T = 2πEI/L, which for the section drawn is 1877 kNm. Two things are absent from that expression and both are surprising: the length appears to the first power rather than the second, and the shear modulus does not appear at all, so the torsional stiffness of the member has nothing to do with the torque at which it buckles in torsion. The comparison is the torque that yields the same tube, 42.9 kNm — a factor of 43.8 below it. The mode arrives first only past a length of 263 m, which is πE/τ_y = 3219 radii and contains no thickness whatever. Stability

The buckling load with no compression in it

Twist a straight bar hard enough and it snaps into a helix, with no axial load on it anywhere. The load at which that happens is 2πEI/L — first power of the length, and no shear modulus in it at all, so how stiff the bar is in torsion has nothing to do with the torque that buckles it in torsion.

 The circle is searched for, and the first guess is 39 per cent optimistic. The same slope with 81 trial circles evaluated, each one through the toe and each one giving its own factor of safety. There is no equation whose solution is the answer: the slip surface is a shape the ground chooses, so the calculation is a search over shapes and the answer is the smallest number found — 1.191 against 1.650 for the circle a first guess puts through the toe from above the middle of the slope, which is 39 per cent optimistic. A slope analysis that reports one circle has reported nothing. Equilibrium

The surface that has to be searched for

Every other check in this collection is made at a section somebody drew. A slope has no section — the failure surface is a shape the ground chooses, so the calculation is a search over shapes, and the answer is the smallest number found rather than the solution of anything.

 Every pressure points at the pin, so the water lifts nothing. A radial gate of radius 8 m holding 6 m of water, with its pivot 6 m above the sill. The pressure on a curved surface cannot be obtained by multiplying anything by anything, so it is integrated round the arc: the horizontal component comes to 176.6 kN/m and the vertical to 110.5. Both are recoverable without any integral at all — the horizontal is the pressure force on the surface's own vertical projection, γH²/2 = 176.6, and the vertical is the weight of the water standing above it, 110.5. They agree to 0.000 per cent. And because every pressure is normal to a circle, every one of them passes through the centre: the moment of the whole 208 kN/m about the pivot is -3.4e-15 kNm, against 353 for a flat gate on the same hinge. Equilibrium

Every pressure points at the pin

Pressure acts normal to a surface, so on a curved one every element pushes in a different direction and no multiplication gives the resultant. Two free bodies recover it without an integral — and on a circular surface a third observation makes the whole force disappear from the equation a hoist has to satisfy.

 The ductility number depends on the ruler. Elongation after fracture against the gauge length it was measured over, in units of √S₀. A tensile specimen extends uniformly until the ultimate load and then localises into a neck, so the total is a strain (16 per cent here) plus a length (7.2 mm), and dividing a length by the gauge length is what makes the curve fall. At the two standard gauges the same steel reports 27.5 per cent over 5.65√S₀ and 21.8 over 11.3√S₀ — a ratio of 1.264, and 42 per cent of the first number is a property of the specimen rather than of the material. Materials

The ductility that depends on the ruler

Percentage elongation after fracture is the most quoted ductility measure in the subject and one of the least well defined. A specimen stretches uniformly until the ultimate load and then localises, so the number is a strain plus a length — and dividing a length by the gauge length makes the answer a property of the specimen.

 The reaction lies inside the cone, so the block stands. A block of 48 on a plane at 22°, against a coefficient of friction of 0.6. Resolving across and along the plane gives a normal force of 44.5 and a friction demand of 18.0, against a capacity of μN = 26.7 — a ratio of 0.67. Added together the two make one contact reaction leaning 22.0° from the normal, and the admissible reactions fill a cone of half-angle arctan μ = 31.0°. Equilibrium is possible exactly when the demanded reaction lies inside that cone, which here it does. The weight enters neither the cone nor the lean: a block of any weight on this slope leans its reaction by the same 22.0°, which is why the angle of repose is a material property and the size of a heap of sand is not. Equilibrium

The area that is not in the equation

Friction is proportional to the force pressing two surfaces together and independent of how large they are, which sounds like an approximation and is not. The area is absent because the contact that carries the load is a tiny fraction of the contact that is drawn, and that fraction grows in exact proportion to the load.

 Weight is the only thing holding it down. A body 1.6 m wide and 4.5 m tall weighing 22 kN, under a wind pressure of 1 kN/m². The wind delivers 2 kN and an overturning moment of 4 kNm about the leeward toe; the weight restores 18 kNm, a factor of 4.35. The resultant lands 0.18 m from the centre against a middle third of ±0.27 m, so the base is still wholly in bearing. Equilibrium

Whether it tips or slides

A free body pushed sideways has two ways of leaving, and which one it takes is decided before any load is known. The condition is a width divided by a height set against a coefficient of friction, and the weight, the wind pressure and the depth of the body all cancel out of it.

 A preloaded joint, before and after it slips. Eight preloaded bolts at 100 kN each, on two friction faces at μ = 0.35. The joint carries 560 kN by friction with the bolts in tension and not in shear at all; past that it slips into bearing and carries 900 kN with the bolts now in shear. Two different mechanisms, one joint. Equilibrium

The force that is capped on purpose

Everywhere else in this collection friction is a nuisance whose value nobody controls, checked with a coefficient known to one figure. In a friction damper the inequality is the design intent — the device is specified so that a member behind it can never be asked for more than a stated force.

 One pulse, two blocks of the same shape. Rotation as a fraction of the toppling angle, under a single 0.8 s sine pulse of 1.00 g, for two blocks of identical proportion whose sizes differ by a factor of 3. Both lift off at the same instant, because uplift depends on the shape alone. The small one reaches the toppling angle and goes over; the large one reaches 46%. The kinks are impacts: there is no dashpot anywhere in this model, and the only energy the block loses is lost when it lands on its other corner, at a velocity ratio of 0.926 per landing — 14% of the energy each time, decided by the block's shape and by nothing else. Dynamics

The only damping is the landing

A rocking block has no dashpot in it. The only energy it ever loses is lost at the instant it lands on its other corner, and how much is a property of the block's proportions — 14 per cent for a slender one and 38 for a stocky one. That single number decides whether it settles or goes over, and a real base does not deliver the value the theory computes.

 Two load sets with the same resultant and different work. The two ways of putting a uniform load of 10 kN/m onto a beam element 6.00 m long. Both put 30.0 kN at each node, so both have the same resultant and the same moment about any point — they are equivalent for a rigid body. The consistent set adds a couple of 30.0 kN·m at each end, in opposite senses, which is what makes it do the same virtual work over the element's shape functions as the real load does. The couples cancel in the resultant, which is exactly why the resultant cannot see them, and they are the whole difference between an exact answer and one that is a third out. Equilibrium

Equivalent in work, not in resultant

Two force systems with the same resultant and the same moment about every point are interchangeable — for a rigid body. A finite element is not a rigid body, and substituting one for the other on a beam element leaves the tip of a cantilever a third too low with no warning of any kind.

 Every number in the table is a rotation, and none of them is a moment. The rotation contributions of a three-span beam of 8, 10, 8 m under 24 kN/m, sweep by sweep. There are six of them, one per member end, and not one is a bending moment: the moment is assembled at the end from M = FEM + 2m′ + m′ of the far end, and until that is done the table holds quantities that mean nothing on their own. That is the trade. A moment distribution stopped after two cycles hands over moments that are wrong by a known amount and can be used; this table stopped after two sweeps hands over nothing that can be read at all — and it gets there in four sweeps against Cross's own count on the same beam, writing six numbers a sweep rather than one per distributed member end plus a carry-over. Deflection

The table that cannot be read halfway

Kani's method converges at exactly the rate moment distribution does, sweep for sweep and digit for digit, because it is the same iteration. What it changes is what is written in the boxes — rotations rather than moments — and that buys a shorter table that repairs its own mistakes and cannot be stopped early.

 The pier grips, gives, and grips again. A sliding bearing carrying 3000 kN on a pier head of 20 kN/mm, dragged by a deck expanding at 1.7 mm an hour, with a static coefficient of 0.05 and a kinetic one of 0.03. The force in the pier climbs while the bearing grips, reaches 150 kN, and falls in a fraction of a second to 30 kN: the pier springs back under only the kinetic friction, overshoots the 90 kN that friction would hold it at, and grips again. The swing is 120 kN — 2.00 times the 60 kN between the two coefficients — and the pier head jumps 6.00 mm each time, three times in 12 hours. Equilibrium

The pier that moves in jumps

A sliding bearing whose static friction is larger than its kinetic friction does not release a slow thermal movement as a drift. It grips, gives and grips again, and each time the force in the pier swings by twice the difference between the two coefficients — whatever the pier is made of.

 The same pulse on a wall that is designed to rock. Rotation as a fraction of the toppling angle under one 0.8 s sine pulse of 1.00 g, for the same 2.0 × 8.0 m wall of 400 kN three ways. Bare, it lifts at 0.250 g; with a 600 kN tendon it lifts at 0.625 g. The bare wall reaches 79 per cent of its toppling angle with one landing; with the tendon it reaches 27 per cent of its toppling angle with eight landings; with tendon and bars it reaches 12 per cent of its toppling angle with 43 landings. With its bars it comes to rest upright. The landings are where the bare wall loses energy; the bars add a loss that does not wait for a landing. Dynamics

A wall that is allowed to lift

A block that rocks inherits everything that decides whether it survives — a restoring moment set by its weight and shape that falls as it leans, and a loss of energy set by its proportions at each landing. Put a tendon through a wall and yielding bars across its base and both become design quantities, and the ratio between them decides whether the wall comes home.

 After a yield, the prestress left does not remember the prestress put in. The prestress left in the tendon of a 2.0 × 8.0 m rocking wall of 400 kN once it has rocked to each rotation and come back upright, for a tendon yielding at 1,370 kN and stiffening by 20 kN a millimetre as the base opens, prestressed to 300 kN, 600 kN and 900 kN. Each keeps all its prestress until it yields — at 53.5 mrad, 38.5 mrad and 23.5 mrad — and then loses 20 kN for every further milliradian, so past 53.5 mrad the three lines are one: the yield force less the stiffness times the stretch. Every one of them is slack at upright past 68.5 mrad. With bars of 400 kN the wall re-centres only while 400 kN remains, dashed, and it keeps that only up to 48.5 mrad, whatever it was prestressed to. Dynamics

The tendon that forgets its prestress

A rocking wall comes home because a tendon pulls it, and the tendon must not yield — yet the rotations that test the wall are exactly the ones that stretch it. Past its yield, the force a tendon keeps is its yield force less its stiffness times the stretch, whatever it was prestressed to, so the guarantee a design needs is a limit on rotation, and more prestress only reaches that limit sooner.

 Three moment diagrams for one pair of beams. Two 15 m spans carrying 12.0 kN/m, drawn sagging downward. As built they are simple spans: 337.5 kN·m at each midspan and nothing over the middle support. Built monolithic they would carry 337.5 kN·m of hogging over the support and 168.8 at midspan. Loaded at 28 days and made continuous at 60, creep takes them 46 per cent of the way from the first to the second: 155.2 kN·m over the support and 259.9 at midspan, after twenty years in which nothing about the loading changed. The support had no moment on the day it was cast and no drawing of the finished structure shows why it has one now. Materials

The support that had no moment when it was cast

Two beams are set on their bearings, carry their own weight for a month, and are then stitched together over the middle support. Nothing about the loading changes afterwards. Twenty years later the stitch is carrying 155 kilonewton-metres, because the concrete went on creeping and the joint would not let it — and how much arrives is decided by a crane schedule.

 The bolt force is the larger of two lines. What an M20 bolt in a 25 mm tee flange actually carries, against the tension applied to the flange. Preloaded to 171 kN it starts there and climbs at Φ = 0.185 — the bolt's own stiffness over the bolt's plus the clamped plates', 857 against 3781 kN/mm — so 18 per cent of every kilonewton applied reaches it and the rest is unloading the contact. At 138 kN the contact runs out and the line joins the one an ordinary bolt has followed from the start, climbing at 2.12. The two lines meet, so the strength is the same either way; what differs is the slope, by a factor of 11.5. The flange's own mechanism is at 172 kN, comfortably past the crossing. Connections

The bolt that was already stretched

Prying is a lever that needs the flange to lift before it can act, and a preloaded bolt does not let it. The bolt carries a fifth of every kilonewton applied until the plates part, and everything after that is the ordinary calculation — which is why preload changes the fatigue answer by a factor of a thousand and the strength answer by nothing at all.

 A unit load carried by the prop taken away. A beam fixed at its left end and propped at its right, under 4 kN/m over 8 m, asked how far it moves at 4 m. The real moment is the propped cantilever's own, with -32.0 kNm at the wall. The unit load is carried by the prop taken away, whose moment diagram peaks at 4.00. Their product has 98.92 of area on one side and -13.33 on the other, and the net, divided by EI, is 85.33 — the propped cantilever's closed-form deflection, 85.33, which no part of this calculation was given. Deflection

Any structure will carry the unit load

Virtual work has two readings and each is free exactly where the other is bound. A unit load needs only something to stand on in equilibrium, so the deflection of a beam statics cannot solve comes out of a cantilever statics can. A virtual displacement needs only to fit together, so a reaction comes out of pushing a mechanism — and on a redundant beam the unknown cancels out of the equation and nothing is found at all.

 The damping that stops helping. The storey drift's white-noise root-mean-square as the bearings' damping is raised from 2 to 50 per cent, each divided by the exact value at 2 per cent. The classical analysis promises that every increment helps, and at 50 per cent predicts 0.20 of the lightly damped drift. The exact analysis flattens: 0.34 at 20 per cent and 0.29 at 50, because the damping force at the bearings is transmitted into the superstructure's own mode, which the classical analysis has decoupled from it. Dynamics

The damping that belongs to no mode

Give every mode its own damping ratio and throw the rest of the damping matrix away, and an isolated building's periods and damping come out right to within a fifth of a per cent. Its storey drift at the superstructure's frequency comes out six times too small. The bearings' dashpot pushes on both modes at once, and the more of it there is the less extra damping buys: the classical analysis promises the drift keeps falling, and it stops.

 Three routes to one deflection, and the one that is wrong. Two 2500 mm bars of 250 MPa proof stress meeting at a loaded apex, the drop of the apex against the load. The line is geometry: each bar's extension from its own stress, divided by the sine of its slope. The dots are the derivative of the total complementary energy with respect to the load, and lie on it. The dashed line is the derivative of the strain energy — Castigliano's theorem applied to a material that is not linear — which leaves the truth by ten per cent at 99 kN and at 170 kN gives 308 mm for a deflection of 46.0. Deflection

The other area under the curve

Castigliano's theorem says a deflection is the derivative of the strain energy with respect to the load, and it is true only while the material is linear. Past that, the right energy is the area on the other side of the stress–strain curve. On two aluminium bars at their proof stress the strain energy gives a deflection four times too large, and on a redundant truss minimising it picks a set of forces in perfect equilibrium that no deformed shape can produce.

 The same cycles, and two different lives. Two sequences made of exactly the same cycles — blocks of 6.00M at 40 N/mm² alternating with blocks of 0.22M at 90, on a detail starting with a 0.50 mm flaw. Taking the small cycles first, the crack reaches its critical 87.7 mm after 12.36M cycles; taking the large ones first, after 6.36M. Miner's sum at the moment of failure is 0.99 for the first and 0.68 for the second, so a rule that predicts failure at a sum of one is right to within a per cent about the first and 48 per cent unconservative about the second. The mechanism is on the axes: ΔK rises with the crack, so a large block met late finds a longer crack and does more with it. Materials

The record played backwards

Miner's rule adds damage, and a sum has no order. A crack does, twice over: a large block met late finds a longer crack and does more with it, and an overload leaves a plastic zone that slows everything after it. The same cycles rearranged fail at 12.4 million or at 6.4, and a Miner sum that is right to one per cent about the first is out by half about the second. One cycle in four million can add fifty-four per cent to a life.

 Bearing damping that shakes the roof harder. The roof's white-noise root-mean-square acceleration as the bearings' damping rises from 3 per cent to 58 per cent, exact and classical, both divided by the exact value at 3 per cent, for the six-storey building. The classical analysis falls all the way, to 0.33. The exact one falls to a minimum of 0.44 at about 28 per cent and then rises, to 0.50 at 58 per cent: past the minimum, every extra per cent of damping at the bearings shakes the top floor harder, while the classical analysis says it is still helping. Dynamics

The top floor the bearings shake

On a building of several storeys, the damping at the isolation bearings is tied less tightly to each higher structural mode than to the one below. It still does more harm in each, because the ground barely reaches those modes on its own. The error the usual analysis throws away gathers in the upper storeys, and past about a third of critical damping at the bearings the roof shakes harder while the analysis says it is still helping. How much damping is best depends on how tall the building is.

 A mode that points wherever the asymmetry points. A square floor 24 m on a side, equally stiff in both directions, with a period of 0.80 s. Four copies are drawn on top of each other, each made stiffer by one part in 1,000,000 along a different direction: 0°, 20°, 45°, 70°. The first mode each one returns lies along 90°, 110°, 135°, 160° — at right angles to its stiffening, whatever the size of it — and the two periods differ by one part in 1,000,000. Four structures no instrument could tell apart have first modes pointing four different ways. The floor with no asymmetry at all has no first mode: every direction is one. Dynamics

Two modes that are really a plane

A building equally stiff in both directions has two translational modes with one period, and they are not a pair of shapes but a whole plane of them. The pair an analysis returns is chosen by asymmetries of a millionth, so any result that depends on the pair — a square-root combination, a comparison with measured modes — inherits a choice the building never made. Damping then decides whether the difference can be seen at all.

 The same mode's curvature, which can. The first mode's drift in each storey — the difference between the floors above and below it, which for a shear building plays the part a beam's curvature plays — intact, solid, and with storey 3 10 per cent less stiff, dashed, each scaled to the roof. The damaged storey's drift rises by 9.6 per cent; the storeys above it change by at most 1.8 and below it by at most 1.5. The largest change is in storey 3: the drift finds the damage and says where it is, which the frequency's fall of 0.91 per cent cannot. Dynamics

What a mode shape notices that a frequency does not

Take a tenth of the stiffness out of one storey of a ten-storey building and its first frequency falls by a per cent at most, and by almost nothing if the storey is near the top. The mode shape looks unchanged, its match to the old one 0.99997. But the mode's storey drift — its curvature — rises by about a tenth in the damaged storey and hardly anywhere else, wherever that storey is. The instrument that sees the damage is the one almost nobody installs.

 The modes of five masses joined by springs and held by nothing. Five equal masses joined in a line by equal springs, with nothing holding them to the ground. The first four modes, at zero, 0.62, 1.18, 1.62 times the frequency of one mass on one spring. The first is every mass moving together with no spring stretched at all: a mode at exactly zero frequency, a real solution of the eigenvalue problem, which carries all of the mass and none of the strain. Every other mode has the ends moving against each other, and carries none of the mass under a uniform acceleration. Dynamics

The modes at zero frequency

A structure held by nothing — a span being launched, a segment on a crane, a pontoon — has a mode in which it moves as one body and stretches nothing, at a frequency of exactly zero. That is a real mode, not a glitch in the stiffness matrix. It carries every kilogram of the structure under a uniform acceleration and leaves the flexible modes none at all. A load that is not uniform is a different matter: pushed suddenly from one end, the structure has no static answer to give, only an acceleration with a vibration riding on it.

 The vertical response on layers of different depth. The vertical amplitude of a 150 tonne block 5.0 m across on soil with a shear wave speed of 178 m/s, under a harmonic force of fixed size, as a multiple of its static deflection on a half-space, against frequency. On a half-space the mode is at 12.5 Hz and the peak is 1.10. On 10.0 m the mode is at 14.3 Hz against a cut-off of 7.4 and the peak is 1.03. On 6.0 m the mode is at 15.5 Hz against a cut-off of 12.3 and the peak is 1.74. On 4.5 m the mode is at 16.3 Hz against a cut-off of 16.4 and the peak is 5.84. The layer stiffens the footing and lifts the frequency a little; what changes the picture is the cut-off, which rises as the layer thins and, once it passes the mode, takes the radiation damping away and leaves only the soil's own 5 per cent. Radiation below the cut-off is taken as nothing and its rise above it as linear, which is the usual idealisation of the exact layered solution. Dynamics

The rock that sends the waves back

A machine block on the ground is damped by the waves it launches, and a half-space lets them all escape. Put rock four and a half metres down and none of them can: below the soil layer's own lowest frequency there is no wave that travels, and a vertical mode damped at 47 per cent of critical keeps the soil's own 5. Its resonance grows fivefold, and the heavy block the textbook rule recommends is the one that loses its damping on the deepest ground.

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