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Which failure arrives first — page 4

Essays 73 to 96 of 344 on this thread, in the same order.

The corner that moves most

A lateral force is shared out in proportion to stiffness only if it passes through the centre of rigidity, which is not the centre of the plan and not the centre of mass. The distance between the two is a torque, and the wall that pays for it is the one furthest away and carrying least.

The bearing is one length and the web is loaded over another. A load applied over a stiff bearing of 100 mm on the flange of a girder with a 1200 × 8 mm web. The flange bends under it and the yield lines that form spread the load along the web over 559 mm — 5.6 times the bearing, and 82% of the yield resistance is that spread rather than the bearing. The effective length is not a decision anybody made: it is what the flange's own bending stiffness against the web's own strength works out to.

Stability

The load that chooses its own length

Every other load in this collection arrives over a length somebody decided. A wheel on a crane girder does not — the flange bends under it and spreads it along the web, and how far it spreads is an output of the flange's own stiffness against the web's own strength. The effective length is 5.6 times the bearing that produced it.

A built-up column has a second way to bend. A 12 m column of two chords 300 mm apart, joined by double lacing. On the left it buckles the way a solid column does, by bending; on the right the chords stay straight and the lattice racks, which a solid column cannot do at all. Neither happens alone, and the two flexibilities add rather than the two stiffnesses — so the critical load is 1287 kN against an Euler load of 1341 kN, which is 96% of it, and the column behaves as though its slenderness were 80 rather than 79.

Stability

The column made of two columns

A solid column buckles when its bending stiffness runs out. A laced one has a second way to go — the lattice shears, the chords stay straight — and the two flexibilities add rather than the two stiffnesses. A battened column reaches 23% of its own Euler load and behaves as though its slenderness were twice what it is.

How stiff a brace has to be before the frame stops swaying. The effective length factor of a swaying portal against the stiffness of a horizontal spring at its head. The curve starts at k = 1.317, the unbraced value, and falls to 0.774 — the factor for the same frame with its head held — at a brace stiffness of 23.2 EI/L³. Past that point the frame buckles in the non-sway mode, which the brace does not restrain, and further stiffness buys nothing at all. The threshold is worth stating as 1.41 N꜀ᵣ/L, which is the form the number is memorable in: for a storey carrying a thousand kilonewtons over four metres it is about 0.35 kN per millimetre of sway. Against the frame's own lateral stiffness of 12.0 EI/L³ it is a factor of 1.93.

Stability

The most dangerous day is before it is finished

A structure is analysed once, complete, with every restraint present. It spends weeks in states nobody drew — a beam landed with no deck on it holds 17% of the moment its section is worth, a frame not yet braced buckles at a third of the load it will, and a bolt not yet tightened is a pin where the analysis assumed a fixity.

Deflection

The deflection that belongs to the support

A beam calculation answers a question about a beam sitting on things that do not move. Real ones sit on bearings, on other beams and on columns that shorten, and every one of those is a spring in series with the member — so a deflection is the sum of two things and only one of them is a property of the beam.

Equilibrium

A basement is a boat

Every load in this collection presses down and is resisted by strength. Hydrostatic uplift presses up, is resisted by weight, and does not care what is built on it — so the check contains no material property at all. It is a ratio of two weights, and one of them is water.

One of these two curves is a stiffness and the other is a statement of statics. The torque a spandrel beam carries, against how much of its torsional stiffness is left. The rising curve is compatibility torsion — a floor beam framing into the side of the spandrel, which shares its fixed-end moment of 197 kNm between the spandrel's torsional stiffness and its own flexural one. Uncracked, the spandrel takes 51% of it, or 100 kNm; at a quarter of that stiffness it takes 21%, or 41 kNm, and the floor beam picks up what was shed. The flat line is equilibrium torsion — a canopy cantilevering 2.2 m off the same spandrel, whose 116 kNm is fixed by statics and contains no stiffness at all. The first can be designed away by accepting a rotation. The second cannot be designed away by anything.

Internal forces

The torsion that goes away if you let it

A spandrel beam attracts a torque in proportion to its own torsional stiffness. Crack it and the stiffness falls by a factor of four, the torque falls with it, and nothing has failed — because the floor beam it was competing with picks up exactly what was shed. A canopy hung off the same spandrel is a different animal entirely.

The cut that severs the stirrups is the cut that counts them. A cracked web drawn as the truss it has become: a tension chord along the bottom, a compression chord along the top, concrete struts between the cracks and stirrups crossing them. The free body is a cut parallel to the cracks, which over a lever arm of 495 mm severs z·cot θ/s = 8.3 stirrups at cot θ = 2.5. Every one of them is at yield, so the shear is their number times their strength, and flattening the crack raises the count rather than the strength. The same cut passes through the bottom chord, which is where the chord force the moment diagram does not contain comes from.

Internal forces

The beam that becomes a truss

Once a web has cracked in shear there is no shear stress field in it any more. There are concrete struts, two chords and whatever crosses the cracks, and the angle of those cracks is not a property of the material — it is something the designer chooses, and every quantity in the beam moves when it changes.

The strain it wants, the strain it is allowed, and the difference. A bridge deck 1.40 m deep with 18 °C at the top face falling away over 10% of the depth. The left curve is the free thermal strain αT(y); the straight line beside it is what a plane section will actually take, ε₀ + κy with ε₀ = 32.0 microstrain and κ = 0.063 per km. The right-hand block is E times the difference, and it reaches -3.98 N/mm² of compression at the surface and 1.83 of tension 140 mm below it. Its resultant force is 8.3e-14 kN and its resultant moment 3.0e-12 kNm, which is what self-equilibrating means: the field is invisible to every equilibrium check that could be made on the member.

Sections and stress

The stress nobody restrained

A bridge deck lying loose on its bearings, with nothing holding it anywhere, develops four newtons per square millimetre when the sun comes out. The stress is not caused by restraint. It is caused by plane sections, and it is invisible to every equilibrium check that could be made on the member.

$Nothing happens, and then everything happens. The moment capacity left to a section already carrying shear, against the shear as a fraction of what the web can take. The web holds 26.1% of this section's plastic modulus and the flanges hold the rest, and only the web's share is reduced — by the factor √(1 − v²) that von Mises leaves it. So the curve is flat for most of its length: the first per cent of moment is not lost until v = 0.27, half the shear capacity costs 3.5%, and 15% is not reached until v = 0.9. The tangent at v = 1 is vertical, which is why the last tenth of the shear range costs more than the first eight.$ Sections and stress

Both at once, and neither matters until it does

A section carrying shear has less moment capacity, and the reduction is the web's share of the plastic modulus times one minus the root of one minus the shear ratio squared. On a rolled beam that share is a quarter, so half the shear capacity costs three and a half per cent — and then the last tenth costs more than the first eight.

The load did not move; the section did. A lipped channel 200 by 65 mm at 2 mm thick, drawn twice on top of itself: the outline as fabricated, and the part of it still working once the plates have buckled. The web is held on both edges, so it loses its middle; the flanges are held at the web, so an unlipped one would lose its free edge. What survives is not symmetric with what was drawn, so the centroid moves 8.0 mm — and a load applied along the axis it was designed to arrives 8.0 mm off the section that has to carry it. At the 177 kN this section will take, that is 1.42 kNm of bending nobody applied.

Sections and stress

What is left after it ripples

A thin plate that buckles locally has not failed. It has stopped taking load in its middle and gone on taking it near its edges, so the member is now made of a different section from the one that was drawn — and the new one has its centroid somewhere else, which turns a concentric load into an eccentric one.

The arch does not squash; it leans. The first two buckling modes of the same rib, drawn against its undeformed shape at an exaggeration of a few hundred. The first, at wcr L³/EI = 49.7, is antisymmetric: one half rises while the other falls and the crown moves sideways. The second, at 117.1, is symmetric — the whole rib settling. The two differ by a factor of 2.36, which is why an arch is braced against sideways movement of its crown rather than against the load it is carrying, and why a tied arch with a single hanger at midspan is doing nothing for the mode that governs it.

Stability

The arch that leans instead of squashing

A masonry arch is asked whether a line of thrust fits inside it. A steel rib is asked a different question entirely: it is a column carrying an axial force along its whole length, and the mode it buckles in puts one half up and the other half down while the crown moves sideways.

A brace on the wrong flange never gets there, however stiff it is. The critical moment of an 8 m beam against the stiffness of a single midspan brace, drawn three times for the three heights the brace could sit at. On the compression flange it climbs from 143 kNm to the two-half-wave plateau of 447 — the beam braced into two 4.0 m beams — and reaches 99% of it at 447 kN/m. At the shear centre it needs 2252 kN/m, 5.0 times as much. On the tension flange it never arrives at all: at the stiffness that would have done the job on the other flange it has bought a factor of 1.068, and a stiffer brace in the same place buys the same nothing. Past the plateau the beam stops using the brace, which is where the idea of an ideal stiffness comes from.

Stability

The brace on the wrong flange

A brace on a column has one property that matters, and it is stiffness. A brace on a beam has two, and the second decides whether the first is worth anything: put the identical restraint on the tension flange and it does not reach the answer at any stiffness whatever.

A column that has nothing on it but itself. A 30 m column carrying no load except its own weight, with its buckled shape and the axial force that produced it. The force is zero at the top and largest at the base, which is why the answer is not Euler's: the eigenvalue is a load intensity and comes out as q_cr L³/EI = 7.835, a number with no π in it, computed here as the smallest eigenvalue of the same stiffness and geometric-stiffness matrices that give a tip-loaded column its Euler load. It is equivalent to a tip load of 3.18 times as much total weight — a column carries its own weight better than it carries somebody else's, because most of the weight is near the base where the buckle is not. The height limit that follows is a cube root, so a section of radius of gyration 80 mm falls over on its own at 52 m and one of twice that reaches only 82.

Stability

Too tall for nothing but itself

Every critical load on this site so far has been applied at the top of a column. A mast carries a load that is zero at the top and largest at the base, the governing equation stops being harmonic, and the answer comes out as a number with no π in it — along with a maximum height that is almost the same for steel, aluminium and wood.

A tank grows without limit; a silo stops. Vertical pressure against depth in a 8 m silo of a solid weighing 9 kN/m³, beside the straight line a liquid of the same weight would have produced. The free body is a slice: its own weight in, the wall friction out, and the friction is μK times the vertical pressure that generates it. The result saturates at γR/μK = 89 kN/m² and reaches 63% of it at one characteristic depth, R/μK = 9.9 m. At the base the pressure is 85 kN/m² against a liquid's 270 — 69% less — and the wall has taken 69% of the stored weight down with it. The exponent is the capstan's, and for the same reason.

Equilibrium

The pressure that stops growing

A tank of liquid presses harder the deeper it gets, without limit. A silo of grain does not. Wall friction carries part of the weight, the pressure that generates the friction is proportional to the pressure being carried, and the equation that follows is the one that describes a rope round a bollard.

The building does not care how far it went down; it cares how much it tilted. Five footings on soil that is 35% as stiff under one of them, carrying 60 kN/m. They settle between 12 and 54 mm, and the number that matters is neither of those: it is the angular distortion between neighbours, 4.32 per thousand, or one in 231 — against a limit of one in 500 for cracking in finishes, which this does not. A building that went down half a metre uniformly would be undamaged and would need a new front step; this one has moved a twentieth as far and has cracked.

Deflection

The settlement that matters is the difference

A building that goes down half a metre uniformly is undamaged and needs a new front step. One that goes down a twentieth as far, unevenly, has cracked. The superstructure can even the difference out — and the only way it can do so is by carrying the difference itself, as a force.

The same load, two diagrams, both in equilibrium. One span of a pair of 7 m spans under 5 kN/m, drawn twice. The elastic solution puts 31 kNm over the support and 17 in the span. Reducing the support moment by 30% and taking what statics then gives leaves 21 and 21: the section the beam needs falls from 31 kNm to 21, a saving of 30%. Both curves are in equilibrium with the same load — the mid-span ordinate plus half the support moment is the free moment 31 kNm for either — and the second is legitimate for that reason alone. What it costs is 1.0 milliradians of rotation at the support, which the section has to be able to deliver.

Internal forces

The moment that was moved on purpose

The elastic analysis of a continuous beam gives one set of moments. It is not the only set the beam is allowed to have, and taking a smaller one at the support is legal, cheaper, and paid for in a rotation that has to be delivered before the design exists.

The force spreads, and the spreading needs a tie. The end block behind an anchorage of 1200 kN on a 200 mm plate, in a section 700 mm deep. Half the force enters at the quarter point of the plate and leaves at the quarter point of the section, so a strut between the two rises 125 mm and needs a transverse tie to turn it. Placing the tie 0.5 depths from the face makes that tie force 214 kN — and at exactly half a depth this reproduces Guyon's 0.25P(1 − a/h) to the digit, which makes that famous coefficient a lever arm somebody chose rather than a property of concrete. The bearing stress under the plate is 20.0 N/mm² against 5.7 once the force has spread.

Internal forces

The force that splits what it pushes on

A prestressing tendon delivers its whole force through a plate a fraction of the section deep. One depth further along the stress is uniform, and the spreading in between requires a transverse tension nobody applied — the force that splits end blocks, and the only number in the design that no equilibrium equation on the member can see.

Two faces, a couple, and a core that does none of it. A sandwich section 61.4 mm deep: two 0.7 mm faces separated by 60 mm of core. The bending is carried as a couple between the faces — 41 N/mm² of tension in one and compression in the other, over a lever arm of 60.7 mm — and the core carries a shear stress of 0.047 N/mm² and nothing else. The parallel-axis term is 98.8% of the section's second moment; the faces' own bending about their own centroids is 0.004% of it, and the core's is 1.2%. Separated by nothing at all the same two faces would be 2.3e+4 times less stiff.

Sections and stress

Two skins and the space between them

A sandwich panel is a section made of a material that carries the bending and a material that carries none of it. The parallel-axis term is not a correction here — it is 98.8 per cent of the second moment — and the shear deflection is not a correction either.

Every path to the ground goes through the link. A braced bay 8 m by 4 m whose two diagonals stop 800 mm apart instead of meeting. The storey shear reaches the ground through the diagonals, and the vertical components they deliver to the beam have to pass through the segment between them: the link carries 47% of the applied shear as a shear force, at a lever arm short enough that its ends reach 0 kNm while the rest of the beam carries 0. The deflected shape drawn is the solved one, magnified — the real drift under this load is 0.008 mm. Everything outside the link is designed to stay elastic while the link is yielding, which is what makes the mechanism a choice rather than a hope.

Structural form

The part that is meant to be weak

A braced frame is stiff and has nowhere to yield. A moment frame yields everywhere and is soft. Move the two diagonals a metre apart along the beam and the whole storey shear has to pass through the segment between them — which keeps most of the stiffness and puts every yielding in one member the designer chose.

The restraint chooses the buckling length, and it is not the member's. A compression flange 12 m long held sideways not at points but everywhere, by a restraint of 0.35 N/mm per mm of length. Unrestrained it would buckle at 173 kN in a single half-wave, drawn faintly. Restrained it buckles at 1968 kN — 11.4 times as much — in two half-waves, because the sum n²π²EI/L² + kL²/n²π² has its minimum there and every other n is worse. The effective length that answer implies is 3555 mm, which is 0.30 of the member and is a property of the restraint rather than of the span.

Stability

Held everywhere, and it forgets its length

A brace at a point divides a member's buckling length. A restraint spread along the whole member does something else — the member chooses its own number of half-waves, and past a few of them the critical load stops depending on the length at all.

Not where the two loads meet. How much a column loses below the weaker of its two single-mode capacities, against the ratio of its local critical load to its global one. The received claim is that the worst place is where the two coincide; the arithmetic says otherwise. The erosion is largest at a ratio of 0.47 — 23% — sits within a per cent of that for every ratio below about a half, and at exact coincidence is only 2%. What the curve does say is the useful half of the folk claim: once the plates are stocky enough that the local critical load is twice the global one, the interaction is nothing at all, and the section is worth thickening only up to there.

Stability

Two ways of buckling at once

A thin-walled column can bow as a whole or ripple in its plates, and each has its own critical load. The received advice is that the worst arrangement is the one where the two are equal. The arithmetic says the opposite — at coincidence the interaction costs two per cent, and the expensive region is where the plates go first.

The envelope is not a state of the structure. Every arrangement of the imposed load on three spans — 8 of them, since each span is loaded or not — drawn faintly, with the greatest sagging and greatest hogging at each station drawn over them. Each faint curve is a real state of equilibrium and satisfies the free-moment identity exactly: mid-span ordinate minus the mean of the end moments is wL²/8, to 3e-16 of it. The envelope satisfies it nowhere, missing by up to 23% — because it is assembled from different load cases at different stations and no arrangement of load produces it. seven of the 8 arrangements are needed to build it; the rest never govern anywhere.

Equilibrium

The envelope is not a structure

A continuous beam whose imposed load may sit on any span has eight load cases, and every one of them is a genuine state of equilibrium. The curve the design is made against is not one of them — it is assembled from different cases at different stations, and it fails the identity all eight satisfy exactly.

Deflection

The angle nobody limits

Every serviceability rule in this collection limits a displacement. What a bearing, a joint and a cladding gap actually have to accommodate is an angle — and the angle is locked to the displacement by a coefficient that contains no material, no section and no span.

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