Structural Engineering & Statics
where the load goes
The collection

Every essay — page 7

Essays 145 to 168 of 547, in the same order.
Whether the floor shares the load out by stiffness or by area. The share of a uniform storey force taken by each of three equally stiff walls, against the stiffness of the floor plate that spans between them. A plate far stiffer than the walls translates almost rigidly, every wall deflects the same and the share is the ratio of stiffnesses — 60% to the middle wall here, the same as everyone else. A plate far softer than the walls behaves as a continuous beam over them and the middle wall takes 33%. Neither end is the tributary-area answer of 50%, which assumes a plate that is both soft and discontinuous over the wall, and which nothing here ever reaches. 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.

8 figures · Diaphragm 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.

8 figures · Patch loading 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.

8 figures · Built-up column 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.

8 figures · Erection stability Two differences up the same building, peaking in different places. Differential shortening between a perimeter column and the core of a 40-storey building, plotted up the height. The part driven by load peaks at level 20 — exactly half way up, because a floor near the top has almost nothing built above it to shorten what is beneath, and a floor near the bottom has almost nothing beneath it to shorten. The part driven by shrinkage does not care what is above it at all and accumulates all the way to the roof. Their sum is worst at level 40, at 43 mm, which across a 9 m bay is a floor out of level by one in 208. Deflection

The columns are shorter than the core

Every column in a tall building gets shorter as the building is built on top of it, and the core beside it gets shorter by a different amount. The floors between them tilt by the difference — and the difference is largest exactly half way up, because a floor near the top has almost nothing built above it and a floor near the bottom has almost nothing beneath it.

8 figures · Differential shortening How much of a deflection belongs to the beam. The share of the total deflection that is the beam's own bending, against the stiffness of what it sits on. A 8 m beam on two supports under a uniform load: on rigid supports every millimetre is the beam's, and the share falls away as the supports soften until almost none of it is. The beam drawn beside this figure sits at 51% — so 49% of what it does is happening somewhere a beam calculation never looks. The two flexibilities are in series, which means the softer one governs and stiffening the other buys nothing. 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.

8 figures · Support flexibility A basement is a boat. A 20 by 30 m substructure dug 6 m into ground whose water table stands 2 m down. The head on the underside of the base slab is 4.0 m, so the pressure there is 39.2 kN/m² over the whole plan — 23.5 MN of it, pushing upward. Nothing about the structure changes that number. What resists it is weight: 18.7 MN of concrete and whatever is built above, giving a factor of 0.80. The structure floats if the water reaches 2.82 m below the ground, and a base slab alone would have to be 1.64 m thick to hold it down. 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.

8 figures · Uplift 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.

8 figures · Compatibility torsion A reaction with no load, and the moment it bends the beam with. The prestress moments in a 2-span beam. The primary moment is −P·e, the tendon acting on its own section, and it reaches 540 kNm over the middle support. The secondary moment is what is left when the primary is taken off the total, and it is 306 kNm — 57% of the primary, with the same sign, so it does not cancel anything. It comes from the middle support refusing to let the beam lift: 51.0 kN pressing down there and 25.5 kN lifting at each end, a reaction set that sums to -2e-13 because nothing external was applied. Its diagram is straight between supports to 2.0e-13% of its own peak, which it has to be: reactions are point forces and a point force puts no curvature in a span. Internal forces

The prestress that pushes back

On a simply supported beam a tendon is an internal matter and changes no reaction. Put the same beam on three supports and the tendon lifts it off the middle one, the support refuses, and the force it takes to hold the beam down is a reaction produced with no load applied at all.

8 figures · Secondary prestress 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.

8 figures · Shear truss analogy 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.

8 figures · Thermal gradient 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.

8 figures · Shear moment interaction 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.

8 figures · Effective cross-section 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.

8 figures · Arch buckling 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.

8 figures · Beam bracing 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.

8 figures · Self-weight buckling 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.

8 figures · Self-stress A couple applied to the core, and two columns to make it. A 20-storey core with one outrigger at 59% of its height. The arm is stiff in bending and the perimeter columns are stiff in tension and compression, so between them they resist the core's rotation at that level — a couple of 35283 kNm here, carried as a 294 kN pair in the columns at 120 m centres. The compatibility is one equation: the core's rotation at that level, less what the couple takes back out of it, equals the rotation the arm and its columns allow. The top drift falls from 360 mm to 74, which is 80% of it, and the base moment from 73500 to 38217 kNm. The deflected shape is drawn hugely exaggerated: the real top drift is about one five-hundredth of the height. Structural form

The arm that makes the columns work

The perimeter columns of a tall building are already there, already carrying gravity, and already the furthest thing from the centre. They take almost none of the overturning, because a floor slab transmits shear and not moment — and one storey-deep arm at the right height changes that by nearly a half.

9 figures · Outrigger The force runs along the chord whatever route the member takes. A member pinned at two points 10 m apart, loaded only at those two points, and bowed 1.2 m off the line between them. Equilibrium leaves the two end forces no choice: equal, opposite, and along the chord. So at every section the axial force is P cos α, the shear is P sin α, and the bending moment is the force times the perpendicular offset — 240 kNm at the crown for 200 kN at 1.2 m, which is a multiplication rather than an analysis. Straighten the member and the moment diagram is identically zero; that is the case a truss member is in, and the reason it carries one number. Equilibrium

The member with only one direction

If a body is in equilibrium under forces applied at exactly two points, those forces are equal, opposite and along the line joining the points. It is three conclusions from two equations, it is the shortest real theorem in statics, and nearly everything that follows depends on it without saying so.

8 figures · Two force member 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.

8 figures · Silo pressure A derivative taken with a ruler, and the step that makes it worst. Castigliano's theorem says the deflection is ∂U/∂P, and the derivative here is taken numerically — two solves at ±dQ and a central difference. Against the unit-load answer of 1.720635e-2 it agrees to 1.6e-13, which for a linear structure it must: ∂N/∂P is exactly the force a unit load produces, so the two expressions are the same sum written twice. The error against step size is the classic pair of straight lines — truncation falling as the step shrinks, round-off rising as the difference of two nearly equal energies loses its digits — meeting near dQ = 1.2e+1. For a linear structure the truncation term is exactly zero, so what is drawn here is round-off alone. Deflection

The deflection that is a derivative

A structure's strain energy is one number. Differentiate it with respect to a load and out comes the displacement under that load — and the trick that makes it a method rather than an identity is that the load does not have to be there.

8 figures · Strain energy 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.

8 figures · Differential settlement 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.

8 figures · Moment redistribution Between two beams and one, and much nearer one. How composite a beam is, against the one dimensionless group that decides it: αL, where α² = K·EI∞/(EA*·EI₀). At αL = 0 the layers slide freely and the beam is two beams; past about 20 the connection is stiff enough that the last per cent is unbuyable. The beam drawn sits at αL = 15.8 and is 96% composite, deflecting 36.7 mm against 33.9 for full interaction and 110 for none. The curve is steep where a real design sits, which is why halving the number of studs does not halve anything. Internal forces

Half the studs, and most of the beam

Bonding two layers together quadruples the stiffness of the pair. A real connection is a row of studs that deform, so the layers slide a little and the beam sits between the two answers — but not halfway between, and the number that decides where is a single dimensionless group.

8 figures · Partial interaction