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Which failure arrives first — page 16

Essays 361 to 376 of 376 on this thread, in the same order.

The limit moment goes as the square root of the ring's stiffness, and pressure is part of it. Brazier's limit moment for a steel tube of radius 600 mm and wall 2.5 mm (r/t = 240), as a multiple of its value with no pressure, against the pressure as a multiple of the ring's own buckling pressure, pc = 3D/r³ = 4.07 kPa; positive inside. The dots are the peaks of the moment–curvature path found by minimising the energy at each curvature; the line is √(1 + p/pc). Half pc outside leaves 0.71 of the bending capacity, and the capacity is gone at pc; pc inside gives 1.41, three times pc 2.00. The pressure enters only by adding to the ring's stiffness against the oval, (3D/r³ + p) where it had 3D/r³.

Three kilopascals that halve a tube in bending

A long thin tube in bending flattens until its section can carry no more, at an oval of exactly two ninths of its radius. Put a pressure on it and nothing about that shape changes. What changes is how much bending it takes to get there — as the square root of one plus the pressure over the ring's own buckling pressure — and for a steel tube 1.2 m across with a 2.5 mm wall that unit is 4 kPa. Three kilopascals of suction halve its bending capacity; a fifth of an atmosphere inside doubles it and hands the tube to another failure entirely.

How long the pumps must run is set by how fast the water comes back. The floors that must be cast before the pumps stop under a 20 × 30 m basement dug 6 m below a water table at the surface, its 0.9 m slab and walls weighing 18.7 MN against 35.3 MN of uplift when the water is back, so that the weight stays ahead of the water at every instant, against the water's recovery time on a logarithmic scale, for a floor every 5, 7 and 10 days. If the water returns at once, five floors — the building has to be nearly heavy enough on its own. With a floor every 5 days: 3 at a 3-day recovery, 2 at 7, 0 at 14; a floor every 7 days: 4 at a 3-day recovery, 3 at 7, 1 at 14; a floor every 10 days: 4 at a 3-day recovery, 3 at 7, 2 at 14. Dashed, the closed form for weight added continuously at a floor a week, which runs about one floor below the steps: a floor arrives at the end of its cycle, not through it.

Equilibrium

The water that comes back while the floors go up

A basement below the water table is built dry, inside an excavation kept pumped, and the building that will one day hold it down is not there yet. When the pumps stop, the water comes back over days and the frame goes up a floor a week, and the basement is safe only if the weight stays ahead of the water at every instant. How long the pumps must run is not set by how heavy the building is. It is set by one ratio — how much weight the frame adds while the ground refills — and for a basement that needs five floors on it to hold itself down, a seven-day recovery asks for three and a fourteen-day one for one.

Connections

The column on the other side of the joint

A partial-strength joint is sold as a fuse: it caps the moment a beam can put into its support at the joint's own resistance. At an interior column that is true. At the edge of a frame the column is not a support that stays still; it is a spring in series with the joint, so the joint delivers less than its resistance in service and almost exactly what a rigid joint would. The column then carries three times the moment the simple-construction rule gave it, and at the top storey, where it stands alone and is lightest, it is weaker than the joint it was meant to be protected by.

Past a wall ratio of 2.22, a thicker tube gains nothing more. The most autofrettage can raise a thick cylinder's elastic pressure, as a multiple of its first-yield pressure, against its wall ratio b/a. Yielding the whole wall would buy σy ln k ÷ py (faint), which keeps rising; the release caps it at 1 + β, where β is the reverse yield strength over the forward one. With β = 1.0 the two meet at b/a = 2.22, and the gain is 2.00 for every thicker wall; with β = 0.7 the two meet at b/a = 1.80, and the gain is 1.70 for every thicker wall; with β = 0.5 the two meet at b/a = 1.55, and the gain is 1.50 for every thicker wall. Thinner walls than that can be yielded right through and gain less.

Materials

The overstrain the release gives back

A thick tube is pressurised past yield once, at the factory, so that releasing the pressure leaves its bore in compression and the working pressure must overcome that before it does any harm. The obvious rule is to overstrain as far as possible — right through the wall. For a tube whose outside is more than 2.22 times its bore that is wrong: the release itself reverses the bore past yield, and every newton of overstrain beyond twice the first-yield pressure is given back on the way down. The best overstrain stops part-way through the wall, and a steel that yields early in reverse stops it sooner.

Six millimetres of slip draw the Eurocode's line. The least degree of shear connection at which a 457 × 191 × 67 UB in S355 under a 130 mm slab of C30/37 2.5 m wide, with 19 mm studs of 80 kN can reach 95 per cent of its plastic moment without its end studs slipping more than 6 mm, against the span (solid), and EN 1994-1-1's minimum for ductile studs in a beam with equal steel flanges, 1 − (355/fy)(0.75 − 0.03L), not less than 0.4 (dashed). 6 m: 0.24 against 0.43; 8 m: 0.51 against 0.49; 10 m: 0.61 against 0.55; 12 m: 0.69 against 0.61; 14 m: 0.75 against 0.67; 16 m: 0.77 against 0.73; 18 m: 0.80 against 0.79; 20 m: 0.88 against 0.85. From 8 m up the two differ by at most 0.08: a stud model and a slip capacity reproduce the rule's slope as well as its level. At 6 m the slip alone would allow 0.24, and the rule asks for 0.43.

Internal forces

The slip the plastic moment asks of the studs

A composite beam with half the studs that full interaction needs still has 85 per cent of the full plastic moment, because the slab's compression block simply gets thinner and the steel finds its own neutral axis. That is the stress-block answer, and it assumes every stud slips as far as the beam asks. Follow the slip along the beam and the end studs of a 12 m beam need 6.6 mm to reach nine tenths of that moment and 9 mm to reach 95 per cent — past the 6 mm a stud is trusted to give. Require 6 mm and the Eurocode's minimum degree of shear connection comes out of the arithmetic, slope and all.

The force each section may carry rises from the end, and the strands must climb under it. The prestress each section of a pretensioned beam 300 × 700 mm spanning 12 m, with eight strands of 125 kN at 220 mm below its centroid may carry at transfer without its top fibre passing −2.50 N/mm² or its bottom 18 (thick), against the distance from the end: 739 kN at 0.5 m, 1,101 at 2 m, 1,507 at mid-span, where the self-weight moment is largest. All eight strands bonded from the end give 1,000 kN once their transmission length is past (dashed), which the beam can carry only near mid-span. Debonded — six bonded from the end, one debonded for 0.50 m, one debonded for 1.00 m — their force (solid) climbs in steps that stay under the line everywhere.

Sections and stress

The strands that are sleeved at the ends

A post-tensioned tendon can be draped up towards the centroid near the supports, where the self-weight moment that made its eccentricity safe has gone. A pretensioned strand cannot: it runs straight between the abutments of the casting bed. Choose eight strands at 220 mm below the centroid of a 12 m beam and mid-span passes every limit at transfer, while a metre in from each end the top fibre cracks. The cure is to stop some strands gripping the concrete near the ends, and how many is a trade: every strand saved at mid-span by lowering the strands is a strand to sleeve at the ends.

A secondary holds more than it twists once its connection passes a few tens of kilonewton-metres per radian. The mid-span twist of the 8 m primary beam of 12 kN/m at 75 mm off its shear centre, with a 6 m 356 × 171 × 51 UB framing in at mid-span whose 60 kN reaction acts 60 mm off the shear centre, against the rotational stiffness of the secondary's connection on a logarithmic scale, in series with the secondary's own 3EI/L of 14,453 kN·m/rad. Dashed, the 6.25° with no secondary. On the deck's side the secondary is worse than none below 33 kN·m/rad and better above it: 9.12° at 10, 3.28 at 100, 0.47 at 1,000. On the far side its torque opposes the deck's (0.89° at 10); as a pair, the torques cancel (1.05° at 100). EN 1993-1-8 calls this secondary's connection nominally pinned below 2,409 kN·m/rad (dotted).

Deflection

The secondary beam that twists what it holds

A secondary beam framing into a twisting primary at mid-span is the best restraint a primary can have — concentrated exactly where the twist is. It is also a load: its reaction arrives on the primary's web, off the shear centre, as a torque at exactly the same place. Which wins is decided by the connection, and the threshold is low: a fin plate stiffer than about 33 kN·m per radian, a seventieth of what still counts as a pin, makes the secondary hold far more than it twists. The bolts' free play is another matter, because the torque acts from the first millimetre and the restraint only once the play is taken up.

Dynamics

Half a spring is neither tuning nor freedom

An actuator pushing on a tuned mass is safe if its law cancels the mass's spring, and it then needs half the applied force to do much better than the passive mass. The obvious compromise is to cancel only part of the spring, keeping some of the tuning in case the law is not enough. Search every law in the family and the compromise turns out to be the worst choice available: each share of spring kept costs performance and makes more of the laws unstable, and from about 85 per cent kept upward no stable law under the same limits does as well as the mass with no law at all.

A counterweight that follows the jib leaves only the payload, halved. The moment in the mast of a crane whose 50 m jib weighs 300 kN and luffs between 15° and 85°, lifting 60 kN at its tip, against the luffing angle, with a counterweight linked to the luff so that it balances the jib and half the payload at every angle: 1,449 kN·m loaded and −1,449 empty at 15°, 131 and −131 at 85°. What is left is the payload's moment less the half the counterweight was set for, plus or minus 1,449 kN·m at most, against 4,744 with a fixed counterweight.

Equilibrium

The counterweight that follows the jib

A counterweight that stays put balances one position and leaves the mast a moment at every other. Make it move and it can balance anything that moves slowly enough to be followed. On a luffing crane the thing that moves most is not the payload but the jib itself, whose own moment about the mast changes more than twice as much through its luff as the payload's ever is — and a counterweight linked to the luff cancels that change exactly. What it cannot cancel is the payload, which leaves the ground in a second. The best any counterweight can do with that is wait halfway, and the mast is left with half the payload's moment at the longest radius, whatever else the counterweight does.

The linear answer has the ground holding the ends down. The contact pressure along a strip 16 m long, EI 5.40 × 10⁵ kN·m², on a bed of 50 × 10³ kN/m² per metre of its length, under a column of 1,000 kN. With springs that pull as readily as they push (dashed) the bed presses 197 kN/m under the column and pulls down on the ends at up to 34 kN/m — tension the ground cannot supply. With the pulling springs released (solid), the strip touches the ground only from 3.97 m to 12.03 m, 8.05 m in all, and presses 213 kN/m under the column, 8 per cent more; beyond that it carries nothing.

Internal forces

The ends the ground was holding down

Winkler's springs pull as readily as they push, and on a long footing strip the linear answer quietly uses that: it has the ground holding the strip's ends down. Take the pull away and the correction under the column is modest — eight or nine per cent on the pressure and the moment — but the shape of the answer changes completely. A weightless strip longer than about eight metres keeps exactly eight metres of itself on the ground, whatever length was poured, and the rest rises off as straight cantilevers carrying nothing. The middle-third rule turns out to be a property of rigid footings: an eight-metre strip lifts at a fiftieth of the eccentricity the kern allows.

Below its ceiling, a row of braces is only its total. The critical moment of a beam 24 m long whose unbraced critical moment in uniform bending is 837 kN·m, held by one, two, three or five torsional braces equally spaced along it, against the braces' total rotational stiffness. At 4,336 kN·m/rad in all, the rows of two, three and five give 3,815, 3,726, 3,578 kN·m: how the stiffness is divided hardly matters, while one has already stopped. Each row then stops at its own ceiling, where the braces are stiff enough to be nodes and the beam buckles between them in one more half-wave than it has braces: 2,637 kN·m for one, 5,585 kN·m for two, 9,704 kN·m for three, 21,466 kN·m for five. The dashed line is 5,627 kN·m, the moment the beam is asked to carry; one and two cannot reach it however stiff they are.

Stability

The stiffness a row of braces shares out

One torsional brace at midspan has a ceiling, the moment at which the beam gives up twisting it and buckles in two half-waves instead. A row of braces raises the ceiling with every brace added, and the stiffness each must have to reach its ceiling rises with the count — seventy-four times as much for eight braces as for one. That is the right answer to the wrong question. Asked for the moment the beam has to carry rather than for its ceiling, a row of braces needs very nearly the same total stiffness however many it is divided between, so each brace gets softer as the count goes up. The count has one job, which is to put the ceiling above the moment; the stiffness has the other.

The preload lifts the young slab and lands on the one below it. The load on one slab of a frame of 250 mm slabs on a 7.5 m grid cast a floor every 7 days at 20 °C, on one level of shores and one of backprops, as a multiple of its own weight, from the day it is struck at 7 days to 28 days, with creep. With its backprops put in snug (dashed) it takes 1.52 w at the cast above it, creeps down to 1.29 by the end of the week, takes 1.48 w when the slab above it is struck and cast on in turn, and creeps up to 1.71. With every backprop jacked to 0.30 w (solid) the young slab starts its week at 1.22 w and ends it at 1.14; a week later the preload of the backprops above lands on it, and it takes 1.78 w, creeping to 1.86.

Materials

The backprop jacked tight lands on the floor below

A backprop put in snug shares only the load that arrives after it, and creep then pushes the young slab above onto it anyway. Jack it to a preload instead and the young slab is lifted at once — but the preload survives the week, so the lift lands on the slab below the moment the next level is struck, and about half of it is load creep would have moved within the week regardless. The best preload balances the week-old slab against the fortnight-old one, and with creep it is small: a fourteenth of a slab's weight, about six hundred newtons a prop. Anything over a fifth leaves the frame worse than snug. What a small preload does buy completely is immunity to slack.

The tub girder's shear centre at three stages of its building. A steel trough 1,500 mm deep, 3,000 mm between its web tops and 2,000 mm across its 12 mm bottom flange, with 12 mm webs and a 400 × 20 mm flange on each web, drawn to scale. Open, as it is lifted, its shear centre lies 706 mm below the bottom flange, outside the steel entirely. Closed across the top by a bracing truss equivalent to a plate 0.50 mm thick, which carries shear and no bending stress, it is still 538 mm below. With a 250 mm concrete deck cast and hardened across the web tops it is 1,127 mm above the bottom flange, inside the box. The steel's centroid, 673 mm up, does not move until the deck is cast.

Sections and stress

The box that is a trough until its deck is cast

A composite tub girder is a box only once its concrete deck has hardened. Before that it is an open steel trough — two leaning webs and a bottom flange — whose shear centre lies seven hundred millimetres below its bottom flange, in the air, and whose resistance to twisting is the thickness of its plates cubed. A light truss across the top is what holds it together while the deck is poured, and it is a plate for shear and nothing else: an angle of ordinary size is worth half a millimetre of steel. That half-millimetre multiplies the trough's torsion constant two thousandfold, and leaves its shear centre almost exactly where it was. The bracing closes the box for twisting; it does not close it for the shear centre.

A crack at the cope's corner, and the hole its tip needs. A 457 × 190 beam end coped 150 mm long and 80 mm deep, reacting 350 kN of which 70 per cent comes and goes with traffic, to scale. A fatigue crack has grown 30 mm down the web from the corner where the cope's two cuts meet, the most stressed point of the coped section. At that depth the live load's range of stress intensity at its tip is 15.6 MPa√m, and a hole drilled with the tip inside it stops the crack only if its radius is at least (ΔK/10.5√fy)², 6.2 mm: a 12 mm hole. The crack and the hole together leave 341 mm of the 377 mm coped section below them.

Connections

The hole at the tip of the crack

The commonest reason to repair a coped beam end is a fatigue crack at the corner of the cope, and the commonest repair needs no plate at all: a hole drilled so that the crack's tip is inside it, turning a crack into a smooth notch that the traffic can no longer drive. The hole has to be big enough, and how big is set by the crack — its radius grows in step with the crack's length. A 30 mm crack at a stringer's cope needs a 12 mm hole, a 60 mm crack a 24 mm one. The repair stops working at about 100 mm, where the hole outgrows a site drill and the section left beside it runs out of capacity at nearly the same depth. Most of the crack's life is spent far shorter than that, which makes the repair a matter of how early the crack is found.

A cambered girder is an arch on its slings. A 40 m precast girder of 11 kN/m, lateral stiffness 2.55 × 10⁵ kN·m², hung 0.90 m below its roll axis, picked 2 per cent of its length in from each end, with 100 mm of camber at mid-length, drawn in elevation with the camber exaggerated. The girder's centre of gravity is two-thirds of the camber above the line through its ends, 67 mm; its lifting points, on the arc 2 per cent in from each end, are 8 mm above it. It rolls about the line through the picks (dashed), so the camber has lifted its centre of gravity 59 mm towards the roll axis: the girder hangs as though its hook were that much lower.

Deflection

The camber that lowers the hook

A camber is built into a girder so that it ends up level in the finished structure, and nothing about it matters until then — except on the day it is lifted. A cambered girder hanging from its ends is a shallow arch, its centre of gravity two-thirds of the camber above the line through its lifting points, and it rolls about that line as though its hook were that much lower. On a long precast girder already close to its lifting limit, a hundred millimetres of camber takes the factor of safety against roll from 2.3 to 1.4. The camber grows while the girder waits in the yard, so the same girder becomes harder to lift every week it is stored. Pick it a fifth of its length in from each end and the camber drops out of the problem altogether.

A filler that knows the speed meets the collision hardest at its start. The contact force against the depth crushed in the hardest collision, 500 t and 300 t closing at 2.85 m/s with 761 kJ, on a 40 mm filler between 500 t and 300 t, crushable over 24 mm: uniform (dashed), with 50 per cent of its strength at that speed depending on the speed, and with all of it. Each is as strong as it must be to stop the collision in 24 mm. Uniform: 30.5 MN at first contact; 50 per cent on speed: 38.4 MN at first contact; all on speed: 63.4 MN at first contact. The crushing speed is highest at the first instant and falls to nothing, so a filler whose strength follows it spends its force at the start; the same area under a falling line needs a higher peak than under a level one, and a purely viscous filler's is twice the uniform's.

Dynamics

The filler that knows how fast it is hit

A filler graded through its depth meets the small collisions softly and the hard one with a strong layer that has to be very strong, because every kilojoule the soft layer misses must be absorbed in less depth. A filler whose strength rises with the speed it is crushed at ranks collisions by speed instead, with no layers at all, and treats the middling collisions better than grading does. But it meets every collision hardest at its first instant, when the crushing speed is greatest, and a force that falls as the collision proceeds needs a higher peak to absorb the same energy in the same depth. A purely viscous filler that stops the hardest collision in 24 mm hits it with exactly twice the force of a uniform one.

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