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Geometry beats material — page 3

Essays 49 to 72 of 248 on this thread, in the same order.

The hoops change their mind at an angle no proportion chose. The two membrane forces of a spherical dome of radius 30 m under 3 kN/m² of surface, taken from the crown to a base at 60°. The free body for the meridional force is the cap above a cone of half-angle φ, and vertical equilibrium of it gives N_φ = −wR/(1 + cos φ) directly: -45.0 kN/m at the crown falling to -60.0 at the base, compression everywhere. Equilibrium normal to the surface then gives the hoop force, which starts at -45.0 kN/m and reaches 15.0 — it changes sign, and the angle at which it does was found here by bisecting N_θ rather than quoted: 51.827292°. Setting N_θ = 0 gives cos²φ + cos φ − 1 = 0, so cos φ is (√5 − 1)/2, the reciprocal of the golden ratio — an identity this site's solver gate checks against the bisection to nine decimals rather than asserting, because it is too pretty to be believed on sight. Below that parallel the hoops are in tension, which masonry has none of, and that is where every old dome is cracked.

The surface that carries by being curved

A flat plate spends its thickness on a lever arm of a few millimetres. Curve the same sheet and the load is carried in the surface itself, at a thirtieth of the thickness — and the entire bill is presented at the edges.

Four guesses at one buckling mode. A pin-ended column, with four assumed shapes and the load each of them gives. The reference is a ten-term Ritz expansion solved as an eigenvalue problem, at 9.8696 EI/L² — which is π², as it must be. a half sine gives 9.870, its own sag shape gives 9.882, a mid-span sag gives 10.000, a parabola gives 12.000. Every one of them is high and none of them is low, because an assumed shape is a constraint on the column and a constraint can only stiffen it.

Stability

Guessing the shape, and getting the load anyway

A column's buckling load can be had from a shape that is wrong everywhere, because the energy criterion is stationary at the true mode. The error in the load is the square of the error in the shape, and it is always high.

Why the curve sags, and why the two axes are not the same column. The same column curve with the sag computed rather than drawn. A hot-rolled section carries a residual compression of 30% of yield at its flange tips before anything is applied, so the tips yield first and what is left resisting a change of shape is the elastic core. About the major axis the stiffness follows the core's width; about the minor axis it follows its cube. The worst loss is 27% at λ = 74 about the minor axis against 23% about the major, and the whole effect lives between λ = 75 and λ = 89 — outside that band nothing has yielded, or everything has. No imperfection appears anywhere in this figure.

Stability

The column that had yielded before it was loaded

A real column sits below both of the two straight answers over the whole middle of the slenderness range, and the usual explanation — that it was not straight — is only half of it. The other half is that the flange tips had already yielded when it left the rolling mill.

$A buckled panel is a truss that nobody drew. A 1000 × 1000 panel of 6 mm web, at d/t = 167. It buckles in shear at 63.8 N/mm², which is 383 kN — and it then carries 696 kN, 1.82 times as much, because the tension diagonal takes over from the compression one that has gone. The band runs at 22.5° with a membrane stress of 252 N/mm² over a width of 541 mm, and it pulls on the flange at 221.3 N per millimetre of its length. A web that never buckled at all would have reached 953 kN, so the panel ends at 73% of a stocky web's capacity on a fraction of its steel.$ Stability

The panel that carries more after it has failed

Everywhere else in this field a critical load is where the argument ends. A thin web is the exception — it buckles visibly, in waves anybody can see, and then goes on to carry nearly twice as much again by turning itself into a truss nobody drew.

Internal forces

Two beams, or one beam four times as stiff

Stack two planks and they bend as two beams whose faces slide past one another. Bond the faces and the pair has one neutral axis, four times the second moment and half the stress. Nothing was added but a restraint on slip.

The same strain, two moduli, and a width multiplied to say so. A timber section with a steel plate in it, carrying 20.0 kNm. Plane sections stay plane, so the strain at a height is the same in both materials; Hooke's law then puts the stresses in the ratio of the moduli, which here is 19.09. Multiplying the stiffer material's WIDTH by that ratio gives a fictitious section of one material with the same neutral axis and the same forces — 595.2×10⁶ mm⁴ of it, against 351.0 for the same shape with the moduli ignored. The steel plate is 3.8% of the area and carries 43% of the moment, at 96 N/mm² against the timber's 5.0. The transform is not an approximation: it is compatibility and Hooke's law written down.

Sections and stress

A section made of two materials, one of them pretended away

Multiplying a material's width by the ratio of the moduli produces a fictitious section of one material with the right neutral axis and the right forces. It is not a trick — it is compatibility and Hooke's law written down — and it says a stiff material takes what its modulus asks for.

The shear goes round the corner instead of across it. A 6-panel Vierendeel girder, 12 m by 1500 mm, under 100 kN at mid-span. There is no diagonal in it, so each panel's 50 kN of shear is carried as bending in the chords: the curves drawn along them are the chord moments, and every one passes through zero at the middle of its own panel. The local moment is the panel shear times the panel length over four, 25.0 kNm, and it adds to an axial force of 200 kN from the global moment at the same point. The girder deflects 6.20 mm against 3.18 mm for the same members triangulated — 1.95 times — and 68% of that movement is chord bending that a diagonal would have removed entirely.

Structural form

The truss with no diagonals

A rectangle without a diagonal is a mechanism, so a Vierendeel girder makes its corners rigid instead. The shear a diagonal would have carried as axial force now travels as bending in the chords, and bending is a far more expensive way to move a force.

A cable alone goes to a kink, and a kink is not a road. A point load of 1000 at mid-span of a 900 m suspended deck. The upper shape is the cable with no girder at all: two straight lines meeting under the load, because a cable takes the funicular shape of whatever is on it and the funicular of a point load is a kink — 0.0083 radians of it here. The lower shape is the same cable with the girder present, peaking at 1.125 against the bare cable's 1.873. The girder is not carrying the load — it takes only 17% of it — it is spreading it, over a characteristic length of √(EI/H) = 183 m, and what reaches the cable is spread over that length rather than arriving at a point.

Structural form

The deck is not there to carry the load

A cable takes the shape of whatever is on it, which is exactly the problem — under a point load its shape is a kink, and a kink is not a road. The stiffening girder exists to spread the load until what reaches the cable is something the cable's own shape is right for.

The same sheet, twice, and a factor of ten thousand. A 3000 mm developed width of 3 mm sheet, covering 2400 mm in plan — so the legs sit at 36.9° and the fold is 300 mm deep. Flat, its second moment about its own mid-plane is 6750 mm⁴, which spans nothing. Folded, it is 67.50×10⁶ — 10000 times as much, which is exactly the depth in thicknesses squared. The material is identical, the plan cover has fallen by 20%, and the only thing that changed is where the material sits. What limits it is buckling of the leg: at this leg length the flat between the folds goes at 27 N/mm², well below the steel's 275.

Structural form

Folded until it spans

A flat sheet has a second moment of area of B·t³/12 and will not span anything. Folded, the same material has B·t·h²/12, and the gain is exactly the fold depth over the thickness, squared — a ratio with no material in it and no width in it.

The plan a straight beam does not have. A beam of radius 12 m turning through 60°, seen from above, with bending drawn outward from the axis in one colour and torsion in the other. The load is vertical and uniform and nothing is applied off the axis. Bending reaches 446 and torsion 155; the two peaks are in different places, which is why the section has to be chosen for a combination rather than for either.

Internal forces

Bending that arrives as twist

A straight beam under a vertical load carries no torsion unless something applies one. A beam whose axis curves on plan carries torsion everywhere, from the same load, with nothing applied off the axis — and it cannot be simply supported at all.

A section modulus for each face, and only the smaller one is a strength. Four profiles of equal area with the second moment divided by BOTH distances to an extreme fibre rather than by the larger of them. A symmetric section has one section modulus and an asymmetric one has two, differing here by as much as 1.00 to one — so the same member has two bending strengths, and which of them applies is decided by the sign of the moment rather than by anything about the section. The bar is the smaller of the two, which is the one that governs when the moment can go either way.

Sections and stress

Two strengths, depending which way up

A symmetric section has one section modulus. A tee has two, differing by a factor of three, so the same member has two bending strengths and which applies is decided by the sign of the moment. Turn it over and it is a different beam.

The one length a section carries into a column. Four profiles of equal area, with the radius of gyration r = √(I/A) drawn as the distance it is — a pair of lines either side of the centroid, at the depth the whole area would have to sit at to give the section the second moment it has. As a 4 m pin-ended column the same 3000 mm² of material carries between 7 and 3146 kN, in the ratio of the squares of those radii and of nothing else.

Sections and stress

The one length a section takes into a column

A section has an area, a second moment, two section moduli, a shear centre and a torsion constant. A column has heard of exactly one of them, and it is none of those — it is the length √(I/A), which is where the whole area would have to sit to give the section the stiffness it has.

Two curves climbing together, and the one that catches up first. A 6 m member tapering from 200 to 600 mm, with the moment it carries and the moment it can carry drawn on the same scale below it. The demand rises linearly and the capacity as the square of the depth, so the gap between them closes and then opens again. It is narrowest at 3.00 m from the free end, where the member is 400 mm deep and 79% used, against 70% at the root where the moment is largest.

Sections and stress

The section that changes along the span

A prismatic beam is checked where the moment is largest, and everyone knows where that is. A tapered one is not, because the capacity is moving too — and for a cantilever with a load at its tip the governing station is exactly where the depth has doubled, with no length, no load and no material in the answer.

A frame with no plane to be drawn in. The tetrahedron, solved: three equations at every free joint, one axial force in every member. Members drawn heavy carry more; tension and compression are separated by the sign that came back from the solve rather than by inspection. The count is m + r = 12 against 3j = 12, which makes it exactly determinate, and joint equilibrium closes to 1.8e-15.

Structural form

Three equations at every joint

A plane truss is determinate when m + r = 2j. A space frame needs 3j, and that one changed digit is why a cube of twelve bars is six mechanisms short while looking perfectly solid — and why every three-dimensional frame ever built is made of triangles in several planes at once.

Structural form

The floor is a beam lying down

A floor plate spans horizontally between the walls that resist a lateral load, carries a distributed inertia load, and has chords, a web and a span-to-depth ratio like any other beam. Its stiffness decides whether the walls share the load by their stiffness or by the area of floor nearest them — and the familiar tributary answer turns out to be neither limit.

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.

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.

Two null spaces of one matrix, and the count is their difference. Two pin-jointed frames, each with the forces it can carry with nothing applied to it drawn on its bars — tension one colour, compression the other, thickness in proportion. That force set is the null space of the equilibrium matrix; a mechanism is the null space of its transpose; and Maxwell's count b + r − 2j is the difference of their dimensions and knows neither of them separately. A square with both diagonals has s = 1 and m = 0. Two bars in a straight line has s = 1 and m = 1 with a count of 0, so the count is satisfied by a frame that both folds and can be prestressed — and the prestress stiffness is positive, which is why a tensioned pair of collinear bars is stiff at all.

Structural form

The forces that are there with nothing applied

Maxwell's count is the difference between two dimensions, and it knows neither of them separately. A frame can satisfy it exactly and still both fold and be prestressable — and when it does, the second of those is what stops the first.

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