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Geometry beats material — page 6

Essays 121 to 144 of 329 on this thread, in the same order.

Two drawings of one deck, and they are not the same structure. A 112 m viaduct on five supports, articulated two ways. Above, the fixed point is at the left abutment: the far end has to be given 45 mm of movement, and the friction of every sliding bearing runs one way, so the fixed support takes 660 kN before any wind or braking is applied. Below, the fixed point is at the middle pier: the largest joint halves to 22 mm and the friction now cancels across the fixed point, leaving 0 kN. The movement arrows are drawn at 900 times the scale of the deck, because a 45 mm movement on a 112 m span is thinner than the line the deck is drawn with. Nothing about the deck, the loads or the ground has changed between the two.

Where the structure is allowed to move

One drawing decides how big every movement joint on a bridge is and where every horizontal force goes, it takes an afternoon, and it appears on no calculation sheet. Move the fixed point from an abutment to the middle pier and the largest joint halves and the horizontal force on that support drops from the whole of the friction to none of it.

A transverse load with nothing applied. Web slenderness against web thickness, with the limit the flange's own curvature sets. A flange carrying 6213 kN and curved to a radius of 592 m needs 10.5 N per millimetre of radial force to stay on its curve, and the only thing available to supply it is the web. Nothing has been applied to the girder: the load comes from the deflected shape, which is why a straight beam has none of it and a beam at a plastic hinge has a great deal. Setting the radial force against the web's own plate-buckling resistance gives, in four lines, h_w/t_w ≤ k·(E/f_yf)·√(A_w/A_fc) — the form the codes use, arrived at without them. The constants differ: an elastic flange strain gives k = 1.34 and the rule uses 0.3, a factor of 4.5, and the gap is the curvature assumed. k goes as the inverse square root of the flange strain, so 0.3 is a flange strained to 3.4% — which is what a plastic hinge does to it. The rule is not conservative; it is written about a different beam.

Stability

The web that is crushed from inside

A plate girder's compression flange is curved by the beam's own deflection, and a curved force needs a transverse load to stay on its curve. The only thing available to supply it is the web. So a deep girder can buckle its web vertically with nothing applied to it at all, and the rule that prevents it is the only clause in the codes about a load no load case contains.

$Ten decades of time for forty per cent of the strength. Strength as a fraction of the five-minute test value, against the length of time the load is held, over 12 decades of seconds. The relation is a straight line on this axis, which is the reason a single duration factor works at all: every decade of time costs about the same amount of strength. A load held for 60 years leaves 59% of the short-term strength, and the member need not have moved — this is not creep, and it is not fatigue, because nothing about the load varies. It is that a static stress slowly breaks fibres. The curve has an asymptote at 18% that nobody quotes: below about a fifth of its short-term strength a member has no time to failure at all, so there is a stress under which duration stops being a question rather than merely becoming a small one.$ Materials

The load that was left on too long

A timber beam that carries a load for fifty years fails at about fifty-nine per cent of the stress the same beam carries for five minutes in a testing machine. Nothing about the load varies, the member need not have deflected, and the relation between the two is a straight line on a logarithmic time axis over ten decades.

The strength was bought in a furnace and the welder gives it back. Proof stress as delivered and beside a weld, for four aluminium alloys. The heat-treated alloys lose half of it: the strength of a 6xxx extrusion is in precipitates formed by an ageing treatment, and the arc dissolves them for 32 mm either side of the weld, permanently. The work-hardened tempers lose nearly as much, because the heat undoes exactly the work. The annealed ones lose nothing at all, because there is nothing left in them to anneal. The consequence is the crossover: 5083-H22 is 1.04 times 6061-T6 as delivered and 0.92 times it once welded, so the stronger alloy is the weaker member. The 240 mm member drawn, with two longitudinal welds, keeps 87% of its parent capacity — a weld along a member softens a strip and leaves a section, and the same weld across it softens the whole of one.

Materials

The strength the welder gives back

A 6082-T6 extrusion is twice as strong as a 5083-H111 plate and, welded across, the two are within a few per cent of each other. The heat of the arc anneals the metal for thirty millimetres either side, permanently, and the strength that was bought in a furnace is given back at the first joint.

Connections

The detail decides and the steel does not

A fatigue check contains no material strength anywhere. The same detail in a steel twice as strong lies on exactly the same line, because a fatigue life is decided by the geometry of a weld and by the stress range it sees — and two per cent of the traffic does most of the damage, because life goes as the inverse cube of the range.

A restoring moment that gets smaller the further it leans. Restoring moment against rotation for a block 0.90 m wide and 4.20 m tall, both normalised — the moment by its value at first uplift, the rotation by the angle α = 12.1° at which the block topples. It is mgR·sin(α − θ), which is a weight times a geometry with no material property in it at all, and it has two features an elastic system does not. There is a jump at the origin: the moment is whatever the ground demands until uplift and then it is mgR sinα, so the law is discontinuous where a spring's is steepest. And the slope is negative — the further it leans the less it pushes back — so the equilibrium at θ = 0 is stable only because of that jump, and there is no stiffness to divide into a mass. The straight line is what a spring of the same first-uplift strength would have done.

Dynamics

The block that is safer for being bigger

A block resting on the ground lifts off at an acceleration that depends only on its shape and not at all on its size, and then falls over at one that depends strongly on its size. Two objects of identical proportion begin rocking at the same instant and only the smaller one topples — which is why the slender water towers stood in Chile and the squat tanks beside them did not.

The edge, and the length over which it is forgotten. A cylinder of radius 4.00 m and wall 12 mm under 0.6 N/mm² of internal pressure, held at its base. Away from the base the wall carries the pressure as pure hoop tension and bends nowhere, which is why a pressure vessel is a cylinder. At the base the hoop force is zero, because the wall cannot grow there, and the difference is made up by a boundary layer of bending that dies out inward. The length it dies out over is 1/β = 170 mm — 0.778√(Rt), a geometric mean of the radius and the thickness — and the moment is under a twentieth of its edge value by 3.07 of them. Nothing in that length is the load. The base moment is p/2β², and the bending stress it produces is 1.82 times the membrane hoop stress the whole design is about, at every pressure, every radius and every thickness: the ratio is √3/√(1 − ν²) and contains none of them. The hoop force overshoots by 4.3% at 3.2 lengths in, which is the wall springing back past where it was going.

Internal forces

The length a structure was never given

A disturbance applied at one place dies out over a distance, and the distance is not something anybody chose. A beam forgets a badly applied load over its own depth. A beam on the ground forgets a point load over the fourth root of its stiffness against the soil's. A shell forgets a held edge over the square root of the radius times the thickness — a geometric mean of two lengths three orders of magnitude apart, which is neither of them and is not near either.

The axis moves when the section yields. Six sections, each drawn to its own scale, with their elastic neutral axis — the centroid, dashed — and their plastic neutral axis, the equal-area axis, solid. For the symmetric ones the two lines are the same line and the distinction never arises, which is why it is so easily missed. For the tee they are 23% of the depth apart, because the axis that makes the first moment of area vanish is not the axis that makes the two areas equal. The shape factors run from 1.144 to 1.800 across these six, and they are ratios of moduli taken about two DIFFERENT axes — which is also why an asymmetric section has two elastic section moduli, one to each extreme fibre, and only one plastic modulus. The tee's two elastic moduli differ by a factor of 2.78; a fully plastic section does not care which fibre reached yield first, so it has nothing to be two of.

Sections and stress

The axis that moves when the section yields

An elastic section bends about its centroid. A fully plastic one bends about the axis that halves its area, and for anything symmetric those are the same line — which is why the distinction is almost never met. For a tee they are a fifth of the depth apart, and three things follow that the elastic calculation gives no warning of.

$The lining that carries less for being weaker. Bending moment and hoop thrust in a circular lining, against the lining's own bending stiffness, both as fractions of the free-ring values. The ground arrives already stressed — 500 kPa vertically and 300 horizontally at K₀ = 0.6 — and the difference between them tries to squash the hole into an ellipse. A lining stiff enough to refuse absolutely collects the whole distortion pressure, p₂R²/3 = 300 kNm/m; one flexible enough to go with the ground collects nothing, because there is no curvature change left to resist. The thrust is the flat line: it comes from the mean stress rather than the difference, so it does not move at all. Putting 8 joints in this ring drops the moment to 34% of the solid one and leaves the thrust exactly where it was, which is why a segmental lining is jointed and why the intuition carried over from a beam is inverted here.$ Structural form

The lining that is stronger for being weaker

A tunnel lining is not loaded. The ground arrives already stressed and the hole wants to squash into an ellipse; the lining's only job is to refuse, and how much moment it collects depends entirely on how hard it refuses. Make it stiffer and it takes more. Make it flexible — put joints in it, make it thin — and it takes almost none, while the hoop thrust it carries does not move at all.

The average is not the answer, and it is unsafe. Critical load of a pinned column whose middle third has been given a different stiffness, against the whole-column Euler load, with the two numbers a hand check reaches for beside it. The eigenvalue is taken from K − P·Kg over 24 elements, so nothing here is a formula for a stepped column — it is the same computation the uniform case gets. At a middle third of 0.50 times the rest the true load is 0.612 of Euler's, the arithmetic average says 0.832 and the weakest segment says 0.496. The average is high by 36% and it is high on the unsafe side, because the third of the column it is averaging over is the third where the mode has all its curvature. The weakest-segment answer is safe everywhere and wasteful by about as much.

Stability

An average stiffness is not a safe stiffness

Euler's load belongs to a column of one EI. Give the same column two, and the temptation is to average them — which is wrong, and wrong in the unsafe direction by a quarter. Buckling weights stiffness by the square of the curvature of the mode, so the middle of a pinned column decides everything and the ends decide almost nothing.

Two identical pipes, and one carries three times the other. Load per metre on a buried conduit against the depth of cover, in trench widths, with the weight of the prism of soil directly above it drawn between them. A conduit laid in a narrow trench is stiffer than nothing and softer than the sides: the backfill settles relative to the undisturbed ground, the friction on the trench walls acts upward, and the conduit gets 64% of the prism. Lay the same conduit on the ground and build an embankment over it and it is now stiffer than the fill beside it, the interior prism settles less, the friction acts downward, and it gets 172% — a factor of 2.71 between two pipes with nothing different but which way the ground moved. The equation is Janssen's, the same one a silo wall obeys, with a trench for a silo; both curves start on the prism line, because with no depth there is no shear to redistribute anything. This is why a flexible pipe is buried rather than a rigid one: making the conduit weaker moves it down the page.

Equilibrium

The pipe decides what the soil weighs

A buried conduit is not loaded by the soil above it. It is loaded by whatever share of that soil the relative movement leaves it — and which way the shear on the sides of the prism acts depends on whether the conduit settles more or less than the ground beside it. Two identical pipes under identical fill, one carrying two thirds of the prism and one carrying nearly twice it.

The same curve, computed twice and from opposite ends. A cantilever of 6 m, with its deflected shape drawn from a double integration of M/EI and the bending moment of its conjugate beam drawn on top of it. The conjugate is the same span carrying M/EI as a load, with its supports transformed — a real fixed end becomes a free end and a real free end becomes a fixed one, because a fixed end has no slope and no deflection and the conjugate therefore needs no shear and no moment there. The two curves agree to 4.7e-6 of the largest deflection, which is the trapezium rule and not the method. The largest deflection is 36.00 mm against the closed form's 36.00. The reaction of the conjugate beam is -9.000 milliradians, which is the real beam's rotation at that support — so the whole of a slope calculation is one reaction.

Deflection

The beam whose moment is a deflection

A beam's bending moment is the second integral of its load. Its deflection is the second integral of M/EI. They are the same problem, so a deflection can be found by loading a fictitious beam with M/EI and asking a statics question — and the only thing to remember is the supports, which are not remembered but derived, one boundary condition at a time.

$The strength that does not keep pace. Concrete's mean tensile strength against its characteristic compressive strength, with the ratio of the two on the same axis, scaled. The tensile strength goes as f_ck^⅔, so it rises from 1.57 to 5.04 N/mm² over a sixfold rise in the compressive strength, and the ratio between them falls from 11.1% at C20 to 6.0% at C80 — a factor of 1.83. Nothing in a bending or a column calculation ever uses the lower curve, and everything that decides a transition does: when the section cracks, how much minimum steel it needs, how far a bar has to be lapped, what a member without links carries. So a stronger concrete needs more minimum reinforcement, longer laps and a bigger crack-control check, in a member whose ultimate capacity has barely moved. Its real scale is a length: E·G_F/f_ct² is 299 mm here, which is the size at which a member stops behaving plastically and starts behaving like a fracture problem.$ Materials

The strength that is never used

Concrete's tensile strength appears in no bending calculation, no column calculation and no shear calculation with links in it. The whole design philosophy is that it cracks and the steel takes over. And it decides where nearly every transition in the subject sits — when a section cracks, how much minimum steel it needs, how far a bar has to be lapped, what a member without links can carry, and how wide a crack opens.

Prestress buys shear as a square root, not as a sum. The shear stress an uncracked web can take before the principal tension reaches the concrete's tensile strength, against the axial compression the prestress put there. With no prestress it is 1.35 N/mm², the tensile strength itself, because pure shear has a principal tension of exactly its own magnitude at forty-five degrees. Adding compression gives √(f_ct² + σ_cp·f_ct), which is a square root and therefore flattens: the first newton of prestress is worth far more than the last. At the 7.14 N/mm² drawn the limit is 3.39 N/mm², a gain of 2.51, and doubling the prestress from there takes it only to 4.59. The straight line is what a rule that simply added the two strengths would have promised.

Internal forces

The crack that never reached forty-five degrees

Every shear expression for reinforced concrete is a curve fitted to tests, because a cracked section has no free body worth drawing. An uncracked prestressed web has one — a single point, a Mohr's circle and a principal tension — and it is the only shear check in the subject that is derived rather than measured.

A soap film over a hole, and its volume is the torsion constant. The stress function for Saint-Venant torsion of a square of 100 mm, relaxed on a 97 by 97 grid until it stopped moving — 385 sweeps. Its contours are the lines the shear stress runs along, its slope is the magnitude of that stress, and twice its volume is the torsion constant: 1.4053e+7 mm⁴ against a closed form of 1.4058e+7, an error of -0.035 per cent. The steepest slope is 67.515 at the boundary, at the point on it nearest the centre — which is why the peak shear in a solid section is at the middle of the longest side and never at a corner, where the film comes down to zero from two directions and its slope vanishes.

Sections and stress

Two volumes, and both of them are torques

Torsion of a solid section that is not a circle has no elementary answer, and for thirty years after Saint-Venant posed it the only way to get one was to blow a soap film over a hole cut in a plate and measure it. The film is not an illustration of the solution. It is the solution, and so is a heap of sand poured on the same hole.

The web down the middle carries no torsion at all. A 2-cell box 3.0 m wide and 1.5 m deep under 4000 kNm of torque. Statics gives one equation, T = 2 Σ q_i A_i, and there are 2 unknown flows — so the section is torsionally redundant and the missing statements are that every cell twists by the same amount. Solving that system gives 444.4 and 444.4 N/mm in the cells, so the internal web carries 0.00 N/mm — 0.00 per cent of the outer wall's. J is 2.3679e+12 mm⁴ against 2.3679e+12 for the same outline with no internal web at all, a ratio of 1.000000. The same walls slit open would give 7.413e+10, so closure is worth 32 times and the internal web is worth what the ratio says.

Sections and stress

Two cells, one equation, and a web with nothing in it

Bredt's formula answers a single closed cell because a single closed cell has one unknown and one equation. Put a web down the middle and there are two unknowns and still one equation — and the answer, when the missing statement is supplied, is that the new web carries exactly nothing.

The abutment force is the sag turned upside down. A 100 m ribbon carrying 35 kN/m at a sag of 2.0 per cent of its span. H = wL²/8f, so the horizontal force at each abutment is 21875 kN — 6.25 times the entire weight of the deck, and five times what a suspension bridge of the same span and weight at a tenth would have needed. The curve is a reciprocal and it has no flat part: halving the sag doubles the force, at any sag. What stops a designer flattening it further is not the ribbon, which is in tension and cannot buckle. It is what the ground at each end will take, and at 6.25 deck-weights that is usually rock or a very large anchor block.

Structural form

The deck that is its own cable

Every other cable structure hangs something from the cable. A stressed ribbon hangs nothing — the walking surface is the catenary, laid at a fiftieth of the span rather than a tenth, because a footbridge has to be walkable. That one decision hands the abutments six and a quarter times the entire weight of the bridge.

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.

A stiffener is a boundary condition, and it is bought at a threshold. The buckling stress of a 2400 × 12 mm plate with one longitudinal stiffener, against how rigid that stiffener is. Below γ the stiffener rides on the buckle and the plate takes the whole-width mode; at γ the stiffener stays straight and the plate buckles between stiffeners at 74 N/mm², 4.0 times the bare plate's 18.5. Above γ nothing further happens at all, because the sub-panel mode does not know the stiffener is there. The curve is a ramp and then a horizontal line, so a stiffener at twice γ is exactly as good as one at γ. Here γ = 31.5, which asks for an outstand of 144 mm; the 150 mm one drawn gives γ = 35.5, a margin of 1.13.

Stability

The rib that is a boundary condition

A rib on a plate is not a member carrying load. It is a line the buckle is not allowed to cross — and it becomes one at a threshold. Below the required rigidity it rides on the buckle and buys a fraction; at the threshold it stays straight and the plate buckles between stiffeners; above it, nothing further happens at all.

The tube flattens because of the bending, and then cannot carry it. Moment against curvature for a long tube of radius 300 mm and wall 4 mm. Compression on one face and tension on the other are both directed along a curved line, so each produces an inward transverse pressure and the circle is squashed into an oval by the bending it is carrying. That reduces the second moment, so the curve bends over and reaches a limit point — no bifurcation, no imperfection, nothing to be sensitive to. It arrives at an ovalisation of exactly 2/9 for every tube of every size in every material, at 1018 kNm, where the secant stiffness has fallen to 67 per cent of the undeformed value and the tangent stiffness is zero. The relaxed path reproduces the closed form to 0.004 per cent.

Stability

The tube that flattens itself

Bend a tube and the compression on one face and the tension on the other are both running along a curve, so both push inward. The circle becomes an oval, the second moment falls, and the moment–curvature curve turns over at a limit point that needs no imperfection, no bifurcation and nothing to be sensitive to.

The deformation with no limit against it. A 8 m open section carrying 12 kN/m at an eccentricity of 75 mm from its shear centre. The torque is small — 900 Nmm per mm — and the twist is not: 2.55° at mid-span with the ends restrained against warping, against 9.09° if they are not, a factor of 3.57. What that angle does is move the flange tip sideways by 8.9 mm — 62 per cent of the member's own vertical deflection, and 28 per cent of the span/250 that vertical deflection is checked against. The same load on a closed section of the same depth moves it 0.15 mm, 60 times less. No code gives a limit for this quantity, so it is the one movement in the collection that is computed only after somebody has complained about it.

Deflection

The movement with no limit against it

Every code in the world gives a deflection limit. None gives a twist limit — and a beam loaded off its shear centre twists. On an open section a modest eccentricity moves the flange tip sideways by three fifths of the sag that does get checked, and nothing anywhere says whether that is acceptable.

Materials

The coincidence reinforced concrete stands on

Steel expands at twelve microstrain per degree and concrete at ten. Nobody chose either number, they are not equal, and the seventeen per cent between them is the smallest mismatch of any pair of materials engineering bonds together — which is the reason the most-used structural material on earth does not tear itself apart every summer.

The section that is checked is not the section that was chosen. A 457 mm beam coped 50 mm deep over 120 mm to frame into a girder. What is left is a tee with a section modulus of 3.836e+5 mm³ against the whole section's 1.438e+6 — 27 per cent. The moment at the end of the cope is the reaction on a lever arm of 130 mm: 23.4 kNm, giving 61 N/mm² and a flexural utilisation of 0.17. The web now has a free edge along the cope, so its buckling coefficient collapses from 4 to 0.425 — a factor of 9.4 — and the re-entrant corner has a stress concentration of 5.5 on a 10 mm radius.

Connections

The section that is checked is not the one chosen

A beam framing into a girder has its top flange cut away so the two can sit at the same level. What is left is a tee with a quarter of the section modulus, a web with a free edge, and a re-entrant corner — and the beam was selected on a table entry that describes none of it.

Evenly spaced modes, so one of them is always where the feet are. The first 6 modes of a 120 m stay under 3.50 MN. A taut string's frequencies are an arithmetic progression — every one of them 1.006 Hz above the last — where a beam's go as the square of the mode number and spread out. That difference is the whole of why a cable is a lively member and a beam is not: a beam has a first mode and then a gap, and a cable has a mode every 1.01 Hz for ever. The shaded band is ordinary walking, 1.6 to 2.4 Hz, and mode 2 sits inside it. Nothing about the tension can move a mode out of the band without moving another one in.

Dynamics

The force read off a frequency

Nothing can measure the tension in a stay cable directly — there is no gauge, no accessible end and no place to put a load cell. What there is, is a member whose frequencies are an arithmetic progression whose spacing is the square root of its own tension, so a phone taped to it for thirty seconds returns the force.

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