Structural Engineering & Statics
where the load goes
The collection

Every essay — page 14

Essays 313 to 336 of 728, in the same order.
Where a bar may stop, and how far past there it goes anyway. The tension the bottom steel must carry along a 9 m beam, and the resistance of the bars actually present, drawn as a staircase. The demand is the moment diagram divided by the lever arm and then SHIFTED 270 mm toward midspan, because the truss inside the beam delivers its shear diagonally — the two constructions agree to 0.36 per cent, which is the second-order term and nothing else. Each curtailed layer then runs a further 1150 mm to develop, so the outer layer stops at 245 mm rather than the 1665 mm the moment diagram allows. The tail is 1420 mm at each end — 16 per cent of the span — and it is what turns a 20 per cent saving into 2.2. Sections and stress

Where a bar may stop

The moment diagram falls away from midspan, so the steel midspan needs is not needed everywhere, and curtailing it saves real money. Then two things get in the way, and between them they take nine tenths of what the moment diagram promised.

7 figures · Curtailment One plate, three systems, one stress. The stress at the rib-to-deck weld of an orthotropic deck, split by which system produced it. The main girder contributes 120 N/mm², the crossbeam 34 and the trough 13, and they add to 167 because the deck plate is the top flange of all three at once. A calculation that treats the crossbeam as a support rather than a structure understates the total by 26 per cent. The stress RANGE at that weld is 47 N/mm², which is the number the whole deck is designed by, because the weld is a low fatigue category running the entire length of the bridge. Sections and stress

One plate and three structures

An orthotropic deck is a single steel plate stiffened by troughs, sitting on crossbeams, sitting on main girders. Nothing about that is unusual until it is noticed that the plate is the top flange of all three, so a wheel standing on it loads every one of them at once.

10 figures · Orthotropic deck It is the square of the diagram that destabilises. Four moment diagrams normalised to the same peak, and the buckling factor each one earns. Eliminating the lateral displacement from the coupled buckling equations leaves one functional in the twist, and its destabilising side is ∫M(z)²φ²/EI_z — the SQUARE of the moment, weighted by where the beam wants to twist. A diagram with a peak over a short length has a much smaller weighted square than a flat one of the same maximum, so it buckles at a higher peak: uniform 1.00, uniformly distributed load 1.13, central point load 1.36, cantilever 1.71. The root-mean-square of each diagram, printed beside it, very nearly predicts the order — which is as close to an intuition for C₁ as the subject has. Stability

The shape of the diagram, and not its peak

A beam's lateral-torsional capacity is quoted against uniform moment, which is the one case a beam carrying a load never has. Change the shape of the moment diagram without changing its peak and the buckling moment moves by a factor of nearly three.

7 figures · Moment gradient The best design is where two failures arrive together. A fixed area of steel rolled into tubes of every proportion, with the three things that can end each one. Euler's load goes as r² because I = A r²/2; the local buckling stress goes as 1/r² because the wall thins as the tube grows; squashing does not care. The capacity is the lowest of the three, so it has a maximum — and the maximum is exactly where the two buckling curves cross, at r/t = 129 and 2364 kN, which the closed form r*, the fourth root of αAL² over π³β√3, reproduces to 0.52 per cent. That is the general result and it is not about tubes: the optimum of a minimum of a rising and a falling curve is always their intersection, so optimising a design against two failure modes puts both of them at the design point — which is the one configuration imperfections hurt most. Stability

The best design is the most sensitive one

Take a fixed area of steel and roll it into a tube. Euler's load rises with the radius and local buckling falls with it, so the capacity has a maximum — and the maximum is exactly where the two failure modes arrive together, which is the one configuration imperfections hurt most.

7 figures · Wall optimum A tenth of a per cent of the thrust is all of the moment. The thrust a two-hinged arch loses to its own axial shortening, against rise-to-span. The flexibility equation's denominator has two terms — ∫y²ds/EI for bending and ∫cos²θ ds/EA for shortening — and their ratio is about (15/8)(i/f)², the square of the radius of gyration over the RISE. The loss is a little more than that ratio, because the released rib also shortens under its own shear, and at the 10 per cent rise drawn it is 0.11 per cent of the thrust, which sounds like a rounding error and is not: a parabolic arch under a uniform load is funicular, so the rigid solution has NO crown moment at all, and the 0.11 per cent that the rib shortening removes from the thrust leaves 29 kNm behind. The correction that is a tenth of a per cent of the thrust is a hundred per cent of the bending. At a two per cent rise the loss is 2.6 per cent, because a shallow arch's thrust is enormous and its lever arm is not. Deflection

The arch that gets shorter

A parabolic arch under a uniform load is funicular, so the perfect solution gives it no bending at all. Then the rib shortens under its own thrust by a tenth of a per cent, and every kilonewton-metre of moment the arch will ever carry comes from that.

8 figures · Rib shortening Seventy per cent of the strength and ninety of the stiffness. Strength and modulus against age, each as a fraction of its own twenty-eight-day value. They do not move together: E follows f to the power 0.3, so at seven days the concrete has 78 per cent of its strength and 93 per cent of its stiffness, and at three days 60 and 86. A young structure is much nearer its final deflection than its final capacity. The 20 N/mm² a striking calculation asks for arrives at 2.2 days at 20 °C, 4.6 at five degrees and 1.7 at thirty-five. Materials

The strength it had on the day

Every concrete strength a design quotes is a twenty-eight-day cylinder value, and a structure is loaded long before that — formwork struck at three days, the next storey cast at seven, a prestressing force transferred at two. The number that existed at the moment the load arrived is a different one.

7 figures · Maturity The confinement that slenderness switches off. The capacity of a concrete-filled tube against slenderness, divided by the plain sum of its two materials. Below about λ̄ = 0.5 the concrete is confined and the section is worth more than its parts — up to 29 per cent for a stub. Above it the bonus is gone, because confinement needs the concrete to dilate, dilation needs strain, and a slender column buckles before it gets there. The column drawn is at λ̄ = 0.42 and has 0.3 per cent of a bonus, which is to say none. What does not switch off is the other half: at d/t = 80 an empty tube buckles locally at 286 N/mm², below its own yield of 355, and the filled one reaches 508 because the wall cannot go inward. That is worth more than the confinement ever was, and it applies at every slenderness. Materials

Each one stops the other failing

A concrete cylinder crushes by splitting outward and a thin steel tube fails by rippling inward. Put one inside the other and each material's failure mode requires a movement the other one prevents, which is a much stronger statement than composite action.

7 figures · Filled tube Where the demand crosses the capacity. Base shear against roof displacement for a eight-storey frame pushed with a triangular load pattern. It is elastic to 1994 kN, where the first storey reaches its shear capacity, and flattens as each of the others follows — 7 yield events in the order 3, 4, 2, 1, 5, 6, 7. The demand is a displacement rather than a force: the elastic spectrum at the building's own period of 0.93 s gives 173 mm, and the participation factor of 1.26 makes that 219 mm at roof level. Against an idealised yield of 70 mm that is a ductility demand of 3.10. Dynamics

Pushed over until it will not stand

A structure in an earthquake is asked for a displacement rather than a force. A pushover answers a different question cheaply — how much base shear the frame has against how far its roof moves — and the whole art is in what happens when the two are laid over each other.

7 figures · Pushover The loop a brace has when it cannot buckle. Force against axial deformation for two braces with the same core area, cycled six times at a storey drift of 2 per cent. An ordinary brace yields at 900 kN in tension and buckles at 482 in compression — 54 per cent of it — and the buckled shape leaves a plastic hinge that does not straighten, so the compression side loses capacity every cycle and is at 12 per cent of its first value by the last. A restrained brace has a casing that carries no axial force at all and only holds the core straight, which decouples axial capacity from flexural stiffness — the coupling that makes a strut weaker than a tie — so it yields at the same force both ways and hardens instead. The energy dissipated is 2.07 times as much over the six cycles, and the casing has to satisfy one inequality: π²EI/L² above the fully hardened core force, 2.56 here, which is a buckling check on a member carrying nothing. Dynamics

The brace that yields both ways

An ordinary diagonal yields in tension at its full strength and buckles in compression at half of it, and the buckle leaves a hinge that does not straighten. Stop it buckling with a sleeve that carries no load at all and the loop becomes symmetric.

7 figures · Buckling-restrained The building is a filter with one very sharp tooth. What a component mounted on the roof of a eight-storey building feels, divided by what the same component would have felt on the ground. A rigid item sees the floor's peak acceleration rather than the ground's, which is already 2.29 times as much. An item whose own period matches the building's sees 6.1 times as much, at 0.94 s — a resonance inside a resonance, and the two amplifications multiply rather than add. Far above the building's period the ratio falls back toward one, because a component much softer than its support simply follows the ground. The shape of this curve is a property of the building and not of the earthquake, which is why equipment is qualified against a floor spectrum rather than a ground one. Dynamics

The spectrum a floor hands on

Most of what breaks in an earthquake is not the structure. It is a transformer, a chiller, a rack, a ceiling, a pipe — and every one of them is bolted to a floor rather than to the ground. The motion it feels has already been through a filter with one very sharp tooth in it.

7 figures · Floor spectrum The summer that is worse than the one before it. The earth pressure behind an integral abutment, summer by summer, as a multiple of the at-rest value it started at. A 60 m deck expands by 10.8 mm at each end and pushes the abutment into the backfill. Granular soil under cyclic strain densifies, so the same movement next year needs a higher pressure to achieve, and K climbs from 0.38 toward 0.96 — a factor of 2.49 on the force, reached after about a century. The design load on an integral abutment describes the bridge's whole life rather than a load case, and it is the only load in this collection that gets larger because time has passed rather than because something was added. Structural form

The summer that is worse than the last

An expansion joint is a hole in a deck that leaks salt water onto the bearings underneath it. Remove it and the thermal movement does not go away — it goes into the soil behind the abutment, twice a day for a hundred and twenty years, and granular soil under cyclic strain gets denser.

8 figures · Integral bridge Cross the hangers and the chords stop bending. The same tied arch, the same sixteen hangers, the same load on half the span — hung vertically and hung as a network. Vertical hangers make the two chords a Vierendeel frame, which has no truss action at all, so a partial load is carried by bending: 3316 kNm in the tie and 6234 in the arch. Inclined hangers can carry the shear between the chords axially, and the same load gives 686 and 831 — factors of 4.8 and 7.5. The thrust is identical in both, because that is decided by the span and the rise and nothing else. Structural form

Cross the hangers and the bending goes

A tied arch with vertical hangers is a Vierendeel frame with a curved top chord — it has no truss action at all, so a load on half the span is carried by bending. Incline the hangers so they cross and the same two chords become a truss.

6 figures · Network arch 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. Structural form

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.

7 figures · Launched bridge 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.

4 figures · Hierarchy 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.

7 figures · Friction Weight is the only thing holding it down. A body 1.6 m wide and 4.5 m tall weighing 22 kN, under a wind pressure of 1 kN/m². The wind delivers 2 kN and an overturning moment of 4 kNm about the leeward toe; the weight restores 18 kNm, a factor of 4.35. The resultant lands 0.18 m from the centre against a middle third of ±0.27 m, so the base is still wholly in bearing. Equilibrium

Whether it tips or slides

A free body pushed sideways has two ways of leaving, and which one it takes is decided before any load is known. The condition is a width divided by a height set against a coefficient of friction, and the weight, the wind pressure and the depth of the body all cancel out of it.

7 figures · Overturning 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.

7 figures · Friction 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.

7 figures · Truss A Pratt truss of 8 panels. A Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 14 members came out in tension, 13 in compression and 2 carrying nothing. Structural form

Halving the panel buys a shorter strut

Subdividing a truss into more panels of the same span and depth barely changes the chord forces, because the couple that carries the moment has not moved. What it changes is the length of every compression member, and a buckling capacity goes as the inverse square of a length.

7 figures · Truss 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.

7 figures · Truss 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.

7 figures · Truss 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.

7 figures · Strut-and-tie 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.

7 figures · Concrete shear 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.

7 figures · Lateral pile