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The load must go somewhere — page 9

Essays 193 to 216 of 450 on this thread, in the same order.

Every section is hogged and sagged before the bridge exists. The bending moment envelope of a launched deck, section by section along its own length, taken over every position of the launch. In service each section has one sign; during the launch 100 per cent of them see both, because each passes over every pier and through every span on its way out. The worst launch moment is 42568 kNm against 40500 in service, and with no launching nose at all it would be 185977. That is why a launched bridge is a constant-depth box with symmetric flanges: the design case is not a load, it is a history.

Every section was somewhere else

A bridge pushed out over its piers subjects each of its cross-sections to a history rather than to a load case. Every one passes over every support and through every span, so the design envelope is the envelope of envelopes — and no in-service condition produces it.

Every level added is a longer span, not a shorter one. Steel per square metre of floor against the number of levels in the hierarchy, for a 12 m bay with a deck that can span 3.0 m. A bending level's weight per unit area is 3ρqrL/8σ — it contains the SPAN and not the spacing — so breaking a floor into more levels cannot make the members lighter by making them closer together. It adds one more system, and the last system always spans the whole bay: a second level costs 52 per cent more steel than one, and a third 107 per cent. The structural zone grows with it, 730 mm to 1346 mm. Hierarchy is not an economy, it is a way of reaching, and it is paid for in both currencies at once.

Structural form

Every level is a longer span

A floor is a hierarchy — deck to joists to beams to girders — and the reason usually given is that breaking a long span into short ones saves material. A bending level's weight per square metre contains its span and not its spacing, so it does not.

The reaction lies inside the cone, so the block stands. A block of 48 on a plane at 22°, against a coefficient of friction of 0.6. Resolving across and along the plane gives a normal force of 44.5 and a friction demand of 18.0, against a capacity of μN = 26.7 — a ratio of 0.67. Added together the two make one contact reaction leaning 22.0° from the normal, and the admissible reactions fill a cone of half-angle arctan μ = 31.0°. Equilibrium is possible exactly when the demanded reaction lies inside that cone, which here it does. The weight enters neither the cone nor the lean: a block of any weight on this slope leans its reaction by the same 22.0°, which is why the angle of repose is a material property and the size of a heap of sand is not.

Equilibrium

The area that is not in the equation

Friction is proportional to the force pressing two surfaces together and independent of how large they are, which sounds like an approximation and is not. The area is absent because the contact that carries the load is a tiny fraction of the contact that is drawn, and that fraction grows in exact proportion to the load.

A preloaded joint, before and after it slips. Eight preloaded bolts at 100 kN each, on two friction faces at μ = 0.35. The joint carries 560 kN by friction with the bolts in tension and not in shear at all; past that it slips into bearing and carries 900 kN with the bolts now in shear. Two different mechanisms, one joint.

Equilibrium

The force that is capped on purpose

Everywhere else in this collection friction is a nuisance whose value nobody controls, checked with a coefficient known to one figure. In a friction damper the inequality is the design intent — the device is specified so that a member behind it can never be asked for more than a stated force.

A closed force polygon. The forces on a joint, laid tip to tail. Equilibrium is the statement that the polygon closes, and the gap when it does not is the out-of-balance force, to scale.

Structural form

One drawing solves the whole truss

The method of joints solves a truss one joint at a time, and each solution is thrown away as soon as the next begins. Drawn instead of computed, the joints share their edges — every member's force appears once in a single figure, and the figure's own closure is the check.

3 continuous spans against 3 simple ones. The bending moment in a continuous beam, solved by the stiffness method, drawn over the moment in the same spans made simply supported. The peak sagging moment falls from 28.1 to 18.0, and a hogging moment of 22.5 appears over the supports where there was none.

Structural form

The chord is a continuous beam

A truss analysis reports one number per member, and for a chord running through eight panels that number is an axial force with no bending in it. The chord is a single piece of steel spanning between its own nodes, and anything landing between them bends it.

Influence line for the shear force at x = 10.5. The shear force at one fixed station, plotted against the position of a unit load walking across the span. The beam was re-solved at 301 load positions. The worst position is x = 10.56, giving 0.560.

Structural form

Two diagonals, one of which is absent

A truss diagonal is sized for the shear in its panel, and near mid-span that shear changes sign depending on where the load stands. A member that can only pull cannot carry the reversed case, so the panel gets a second diagonal — and at any instant one of the pair is not there.

A truss drawn inside a solid, and solved as one. A deep member 5000 mm between bearings and 2500 mm deep, carrying 2400 kN at mid-span. The model is two struts and one tie, on a lever arm of 2000 mm, and it is solved by the truss solver rather than by a formula: the tie comes back at 1500 kN and each strut at 1921 kN, at 38.7° to the horizontal. Spread over a strut width of 1031 mm the compression is 3.7 N/mm² against a limit of 18.1 for concrete cracked across its own strut, and the tie needs 3448 mm² of steel. A beam calculation on the same member would have asked the tie for 1404 kN, which is 7% less than the model does.

Internal forces

The node is the part that is checked

A strut-and-tie model is safe if every part of it can carry what the model asks. The struts are wide and lightly stressed, the ties are steel and easy, and the whole of the difficulty collects at the points where they meet — in a volume of concrete a few hundred millimetres across that no drawing shows.

Flattening the truss saves stirrups and crushes the web. Two capacities against the angle of the cracks, for a web 350 mm wide with a lever arm of 630 mm. The rising line is the stirrups: a cut along the crack severs z·cot θ/s of them, so flattening the crack from 45° to cot θ = 2.5 takes the 315 kN they carry to 787 — 2.5 times as much from the same steel. The falling line is the concrete strut, whose stress is V(cot θ + tan θ)/b_w z and therefore least at 45°. They do not cross in this range, so the stirrups govern throughout and the angle is a free choice.

Internal forces

The angle is a choice, not a property

The truss inside a cracked concrete web has a strut angle, and nothing measures it. The designer picks it, the stirrup requirement falls as it flattens, the web stress rises, and every choice in between is a different structure that carries the same load.

Most of a long pile is doing nothing. Head deflection against embedded length, both measured in the pile's own characteristic length T = 14.67 m. A pile shorter than about two T is a lever with nothing holding its foot and deflects several times as much; past four T the curve is flat to within a per cent, because the ground below that depth is never asked for anything. A lateral check is a check on the top four T of a pile, however deep it goes for its axial load — and adding length to fix a lateral deflection is the one remedy that does not work.

Internal forces

The pile that is too short to bend

A laterally loaded pile is long or short in its own characteristic length, and the two are different structures. The slender pile bends and the ground below four characteristic lengths never hears about it; the monopile is two lengths deep, rotates about a point, and every tabulated coefficient written for the first case is wrong for the second.

Two diagrams for one load, and the second one has no straight-beam ancestor. Bending moment and torsion round a 90° arc of radius 6 m under a uniform load, built in at one end. The bending peaks at 576 and the torsion at 329, 57% of it. Both are zero at the free end and largest at the support, which is where a curved cantilever's bearing has to hold a torque it was probably not asked for.

Internal forces

The torque that has nowhere to go

A curved beam on two supports splits its torsion between them, and the two halves cancel at mid-span. A curved cantilever has one end, so every increment of torque accumulates toward it — and the largest action at the root of a curved balcony is one that a straight beam does not have at all.

More steel across the crack, until the roughness runs out. Shear resistance of the interface against the reinforcement crossing it. The steel clamps rather than carries, so the resistance is the clamping stress times the interlock coefficient and rises in a straight line — until the asperities crush at 5.50 N/mm², which happens at a reinforcement ratio of 0.79%. Past that the line is flat and every further bar is decoration. The dashed line is what the clamping alone would give if the concrete were unbreakable.

Internal forces

Two models of one bracket

A corbel can be designed as a plane that has to be clamped or as a truss that has to be drawn, and the two are not approximations of each other. They describe different failures, they ask for steel in different places, and the honest answer is that both are checked because neither bounds the other.

The same load, two diagrams, both in equilibrium. One span of a pair of 9 m spans under 30 kN/m, drawn twice. The elastic solution puts 304 kNm over the support and 171 in the span. Reducing the support moment by 30% and taking what statics then gives leaves 213 and 207: the section the beam needs falls from 304 kNm to 213, 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 304 kNm for either — and the second is legitimate for that reason alone. What it costs is 13.0 milliradians of rotation at the support, which the section has to be able to deliver.

Internal forces

The moment that was shed has to land

Redistribution takes a moment off a beam's support and pays for it with rotation. On a beam that is the whole story. In a frame the support is a column, the shed moment does not vanish, and it arrives at a member whose section was chosen from the diagram it has just left.

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 1440 kNm over the middle support. The secondary moment is what is left when the primary is taken off the total, and it is 720 kNm — 50% 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: 102.9 kN pressing down there and 51.4 kN lifting at each end, a reaction set that sums to 0e+0 because nothing external was applied. Its diagram is straight between supports to 3.6e-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 tendon that can be moved

Lift a continuous beam's tendon at its interior support without changing its drape and nothing about the beam's total moment changes. The primary falls, the secondary rises by exactly as much, and the pressure line stays where it was — which turns a parasitic effect into a quantity a designer can place.

The studs are evenly spaced and the demand is not. The force per unit length the shear connection carries along half of a 12 m composite beam, from Newmark's solution. It is largest at the support — 282 N/mm — falls to nothing at mid-span, and averages 156: the end studs are asked for 1.81 times the mean. Studs are nevertheless placed at a uniform spacing, and the justification is the one the variable-angle truss uses for its stirrups — a ductile connector sheds what it cannot carry to its neighbours, so the uniform distribution is a plastic redistribution and not a description of the elastic state.

Internal forces

The connection is busiest where the beam is not

A composite beam's studs are spaced evenly along it and the demand on them is not even at all. It peaks at the supports, where the bending stress is nothing, and falls to zero at mid-span, where the section is working hardest — so the connection is designed from a diagram nobody looks at.

The shear centre of a channel. A channel of 100 by 250, with the shear flow in its flanges drawn. Those flows form a couple, so the load has to be applied 38.0 outside the web to leave the section untwisted — a point in the air, outside the material entirely.

Sections and stress

The eccentricity a purlin cannot avoid

A channel's shear centre is outside the material, so a load applied anywhere on the section misses it. The distance is fixed by the proportions rather than by the detailing, it is 38 mm on an ordinary purlin, and the torque it produces is not an error anybody made.

$The lag follows the shear, so it is worst at the supports. Effective width along the span of a simply supported beam under a uniform load, summed over 25 odd harmonics. Each harmonic has its own half-wavelength L/n and its own, smaller, effective width — 0.815 for the first, 0.400 for the third, 0.269 for the fifth — and near a support the short harmonics carry a bigger share of what little moment there is. So the working fraction is 0.603 at the support against 0.839 at mid-span, a difference of 23.6 percentage points on the same flange. The single sinusoid's answer, 0.815, is drawn as the flat line, and it is only right at mid-span. This is the behaviour a code reproduces by shortening L_e near a support, and the reason it does is that shear lag is driven by w‴, which is the shear force.$ Sections and stress

The flange works least where the shear is largest

Shear lag is driven by the shear force rather than by the moment, so the effective width of a wide flange is not a property of the beam. It is a function of position along it, worst at the supports, and a single number quoted for a whole span is right at mid-span and nowhere else.

The load being amplified is not the load doing the amplifying. The sway amplifier 1/(1 − ΣP/P_cr) against the storey's total gravity load, as columns are added that carry load and provide no lateral stiffness. The critical load of the storey is fixed at 8203 kN by the bracing that exists, and every leaning column moves the structure along the axis without changing it. At the storey drawn the amplifier is 1.57, so the second-order sway moment is 57% on top of the first-order one — and none of that 70% of the load which is causing it appears in any stability calculation done column by column. The storey stays stable across the whole of this axis. That is the practical reason a gravity-only column is drawn on the frame model rather than designed on its own: it is not being checked, it is being counted.

Stability

Counted, not checked

A column with pinned ends and no bracing cannot buckle on its own, so nothing about it fails a stability check. It still carries load, and load with no stiffness attached lowers the buckling load of everything around it — which is why a gravity-only column is put on the frame model and never designed by itself.

A base plate, and when the bolts start working. A 550 × 450 mm plate carrying 900 kN and 140 kN·m, so the resultant sits 155.56 mm from the centre against a kern of 91.67 mm. The plate is in partial contact: bearing over 358.33 mm at a peak of 11.16 N/mm², with the holding-down bolts carrying 0 kN. The plate lifts at 82.5 kN·m and crushes at 192.95 kN·m, and the bolts are not needed until 247.5 kN·m.

Stability

The pinned base that is not pinned

A column base drawn as a pin is a plate bearing on grout, and a plate in contact over its whole length resists rotation whether anybody wanted it to or not. The stiffness it delivers depends on the axial load, so the assumption is one a frame can leave and re-enter as its loads change.

Deflection

An influence line is a deflected shape

Finding where a load has to stand to be worst means solving the structure once for every position it could stand in. Reciprocity says the answer is a single deflected shape — release the quantity being asked about, move it by a unit, and the shape the structure takes is the influence line.

The answer arrives in instalments. The hogging moment at support 1 of a three-span beam, cycle by cycle. It starts at the fixed-end moment of 98.0 kNm — the value with every joint clamped — and settles at 156.9 kNm against an exact 156.9. The error falls by about a factor of four per cycle: 21.03, 5.92, 1.54, 0.60 kNm after one, two, three and four. Two cycles is an engineering answer and nobody had to invert anything.

Deflection

Why it converges, and how fast

Moment distribution is an iteration, and iterations do not always converge. This one always does, at a rate the beam's own proportions fix — about a factor of four per cycle on a regular beam and considerably worse on an irregular one, which is where the method's reputation for two cycles being enough comes from and where it stops being true.

$The crack length at which the strength stops mattering. Failure stress against crack length for a toughness of 100 MPa√m, with three steel grades drawn. The falling curve is fracture — Kc divided by Y times the root of pi a — and it does not know what the yield stress is. The horizontal lines are the grades. At 275 N/mm² the two cross at a crack 33.5 mm long; At 355 N/mm² the two cross at a crack 20.1 mm long; At 460 N/mm² the two cross at a crack 12.0 mm long. The stronger grade's transition is the shorter one — raising the yield stress does not raise the strength of a cracked member, it only shortens the crack that takes it away. At a working stress of 120 N/mm² the critical crack is 176.0 mm.$ Materials

Designed to be found in time

A fatigue design that promises a detail will not crack is making a claim about a hundred years of traffic. A damage-tolerant one assumes it will crack, and sets the inspection interval from how long a crack takes to grow from the smallest size anybody can find to the largest the section can survive.

Three specimens cannot see the tail. The factor k applied to the sample's own scatter when a characteristic value is estimated from n specimens. With the scatter known in advance it is z·sqrt(1 + 1/n) and barely moves; with the scatter estimated from the same n results it is the Student t quantile instead, and it runs from 7.73 at two specimens to 1.73 at 30. At n = 4 the characteristic strength comes out at 23.8 N/mm² against 27.9 for a population known exactly — 15% lower, for a material that is identical. A small test programme does not report a worse estimate of the strength; it reports a worse strength.

Materials

The strength that belongs to the test programme

A characteristic strength is a fractile of a distribution, and a distribution estimated from four specimens is not the same object as one known exactly. The same material tested four times reports a strength 15 per cent below what it reports when its scatter is known — and nothing about the material differs.

Throat stress round a fillet weld group. A l shape weld group carrying 150 kN at 200 mm from its centroid. The peak throat stress is 2.17 kN per mm of throat, at (139.53, 100); the worst point at maximum radius from the centroid carries 1.99. Checking by radius is wrong here by 9.17%, and points at identical radius differ by a factor of 1.

Connections

The radius rule, and where it fails

A weld group under an eccentric load is checked at the point furthest from its centroid, on the reasoning that the stress from the twist grows with the radius. That reasoning ignores the direction the two stresses point in, and for one common shape it misses the peak by nine per cent.

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