Structural Engineering & Statics
where the load goes
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Drawing as calculation — page 4

Essays 73 to 96 of 162 on this thread, in the same order.
The axis moves when the section yields. Six sections, each drawn to its own scale, with their elastic neutral axis — the centroid, dashed — and their plastic neutral axis, the equal-area axis, solid. For the symmetric ones the two lines are the same line and the distinction never arises, which is why it is so easily missed. For the tee they are 23% of the depth apart, because the axis that makes the first moment of area vanish is not the axis that makes the two areas equal. The shape factors run from 1.144 to 1.800 across these six, and they are ratios of moduli taken about two DIFFERENT axes — which is also why an asymmetric section has two elastic section moduli, one to each extreme fibre, and only one plastic modulus. The tee's two elastic moduli differ by a factor of 2.78; a fully plastic section does not care which fibre reached yield first, so it has nothing to be two of. Sections and stress

The axis that moves when the section yields

An elastic section bends about its centroid. A fully plastic one bends about the axis that halves its area, and for anything symmetric those are the same line — which is why the distinction is almost never met. For a tee they are a fifth of the depth apart, and three things follow that the elastic calculation gives no warning of.

 Two frequencies that meet, and a determinant that never moves. The two natural frequencies of Ziegler's two-bar column against the follower load Pℓ/k, with the determinant of its stiffness matrix drawn along the top. The determinant is k² at every load — it varies over this whole axis by 1.1e-16 of itself, which is round-off — so a static buckling analysis of this structure finds no critical load whatever and reports it as stable everywhere. The frequencies say otherwise: they approach, meet at Pℓ/k = 2.0858, the closed form (7 − 2√2)/2, and become a complex pair, which is oscillation that grows. With no damping that merge is also where the column goes unstable. Adding any internal damping at all drops the load at which that happens to 1.4643, which is 41/28 and 30% below the undamped value; the limit of the damped system is not the undamped system, which is the paradox Ziegler found in 1952 and which was taken for an arithmetic error for a decade. Stability

The load it cannot buckle under

Every stability calculation on this site rests on an assumption nobody states: that the load has a potential, so a critical load is where a total potential energy stops being a minimum. A load that turns with the structure it is pushing has no potential, and the static analysis of such a column returns no critical load at all — a determinant that never vanishes, for a column that fails at a perfectly finite one.

 The average is not the answer, and it is unsafe. Critical load of a pinned column whose middle third has been given a different stiffness, against the whole-column Euler load, with the two numbers a hand check reaches for beside it. The eigenvalue is taken from K − P·Kg over 24 elements, so nothing here is a formula for a stepped column — it is the same computation the uniform case gets. At a middle third of 0.50 times the rest the true load is 0.612 of Euler's, the arithmetic average says 0.832 and the weakest segment says 0.496. The average is high by 36% and it is high on the unsafe side, because the third of the column it is averaging over is the third where the mode has all its curvature. The weakest-segment answer is safe everywhere and wasteful by about as much. Stability

An average stiffness is not a safe stiffness

Euler's load belongs to a column of one EI. Give the same column two, and the temptation is to average them — which is wrong, and wrong in the unsafe direction by a quarter. Buckling weights stiffness by the square of the curvature of the mode, so the middle of a pinned column decides everything and the ends decide almost nothing.

 The same curve, computed twice and from opposite ends. A cantilever of 6 m, with its deflected shape drawn from a double integration of M/EI and the bending moment of its conjugate beam drawn on top of it. The conjugate is the same span carrying M/EI as a load, with its supports transformed — a real fixed end becomes a free end and a real free end becomes a fixed one, because a fixed end has no slope and no deflection and the conjugate therefore needs no shear and no moment there. The two curves agree to 4.7e-6 of the largest deflection, which is the trapezium rule and not the method. The largest deflection is 36.00 mm against the closed form's 36.00. The reaction of the conjugate beam is -9.000 milliradians, which is the real beam's rotation at that support — so the whole of a slope calculation is one reaction. Deflection

The beam whose moment is a deflection

A beam's bending moment is the second integral of its load. Its deflection is the second integral of M/EI. They are the same problem, so a deflection can be found by loading a fictitious beam with M/EI and asking a statics question — and the only thing to remember is the supports, which are not remembered but derived, one boundary condition at a time.

 A soap film over a hole, and its volume is the torsion constant. The stress function for Saint-Venant torsion of a square of 100 mm, relaxed on a 97 by 97 grid until it stopped moving — 385 sweeps. Its contours are the lines the shear stress runs along, its slope is the magnitude of that stress, and twice its volume is the torsion constant: 1.4053e+7 mm⁴ against a closed form of 1.4058e+7, an error of -0.035 per cent. The steepest slope is 67.515 at the boundary, at the point on it nearest the centre — which is why the peak shear in a solid section is at the middle of the longest side and never at a corner, where the film comes down to zero from two directions and its slope vanishes. Sections and stress

Two volumes, and both of them are torques

Torsion of a solid section that is not a circle has no elementary answer, and for thirty years after Saint-Venant posed it the only way to get one was to blow a soap film over a hole cut in a plate and measure it. The film is not an illustration of the solution. It is the solution, and so is a heap of sand poured on the same hole.

 Every pressure points at the pin, so the water lifts nothing. A radial gate of radius 8 m holding 6 m of water, with its pivot 6 m above the sill. The pressure on a curved surface cannot be obtained by multiplying anything by anything, so it is integrated round the arc: the horizontal component comes to 176.6 kN/m and the vertical to 110.5. Both are recoverable without any integral at all — the horizontal is the pressure force on the surface's own vertical projection, γH²/2 = 176.6, and the vertical is the weight of the water standing above it, 110.5. They agree to 0.000 per cent. And because every pressure is normal to a circle, every one of them passes through the centre: the moment of the whole 208 kN/m about the pivot is -3.4e-15 kNm, against 353 for a flat gate on the same hinge. Equilibrium

Every pressure points at the pin

Pressure acts normal to a surface, so on a curved one every element pushes in a different direction and no multiplication gives the resultant. Two free bodies recover it without an integral — and on a circular surface a third observation makes the whole force disappear from the equation a hoist has to satisfy.

 Two answers added, and the answer to the two together, drawn on top of each other. A 8 m beam under a 60 kN point load at mid-span (152.38 mm), under 12 kN/m of uniform load (152.38 mm), and under both at once (304.76 mm). The sum of the first two is 304.76 mm, and the residual between it and the third is zero — not small, zero, to the last bit of the arithmetic. That exactness is not a numerical accident: the governing equation is linear in the load, so the response is a linear operator applied to it, and a linear operator distributes over addition by definition. Every calculation that adds one load case to another is standing on that one line. Deflection

The addition everything else rests on

Influence lines add, the unit-load method adds, moment distribution adds, load combinations add, and a stiffness matrix is linear by construction. All of it stands on one sentence with three hypotheses in it — and when they fail, two of the failures point in opposite directions.

 Evenly spaced modes, so one of them is always where the feet are. The first 6 modes of a 120 m stay under 3.50 MN. A taut string's frequencies are an arithmetic progression — every one of them 1.006 Hz above the last — where a beam's go as the square of the mode number and spread out. That difference is the whole of why a cable is a lively member and a beam is not: a beam has a first mode and then a gap, and a cable has a mode every 1.01 Hz for ever. The shaded band is ordinary walking, 1.6 to 2.4 Hz, and mode 2 sits inside it. Nothing about the tension can move a mode out of the band without moving another one in. Dynamics

The force read off a frequency

Nothing can measure the tension in a stay cable directly — there is no gauge, no accessible end and no place to put a load cell. What there is, is a member whose frequencies are an arithmetic progression whose spacing is the square root of its own tension, so a phone taped to it for thirty seconds returns the force.

 Independence, and the floor it never goes below. The reduction factor a column is allowed, two ways. The Eurocode storey rule falls to 0.70 and stops. The independence argument — n bays each with a mean and a standard deviation, summed — gives (1 + zv/√n)/(1 + zv), which falls faster and stops at 0.503: a floor with no n in it at all, decided only by how variable the load is and how far out the fractile is drawn. The mean is never reduced away, because every bay really does carry its mean. At ten storeys the two differ by 10.0 points. Equilibrium

The load that is never all there at once

A column at the bottom of twenty storeys is designed for the imposed load of twenty floors added up, and twenty floors do not reach their own worst day together. The reduction that follows is not a discount on the safety margin. It is the central limit theorem, and it has a floor it never goes below.

 The moment a joint makes out of geometry. Bending stress in the chord divided by the axial stress already in it, against how far the diagonal's working line misses the node. A truss analysed as pin-jointed is drawn with its members as lines through their own centroids meeting at a point, because that is what makes the joint a pin with nothing but forces in it. A fabricated joint is not obliged to oblige, and the resultant then has a moment about the node — 16 kNm at the 60 mm drawn. It divides between the members meeting there in proportion to their rotational stiffnesses, 4EI/L or 3EI/L, which is the moment-distribution rule applied to an unbalanced moment that came from geometry rather than from load. The chord takes 48 per cent of it and carries 25 per cent of its axial stress again in bending. Ten per cent arrives at 28 mm, which on a chord of any depth is a detailing decision rather than a mistake. Connections

The joint that is not where it was drawn

A truss is drawn as lines through the centroids of its members meeting at points. A fabricated joint is under no obligation to oblige, and when the working lines miss each other the resultant has a moment about the node that has to go somewhere.

 Steel has a third direction and it is not as good. Through-thickness strain demand against weld size, on a 30 mm plate, with the ductility the plate can supply in each of its three directions drawn across it. Rolling stretches the plate's inclusions into flat stringers, so a bar cut along the rolling direction, one cut across it and one cut THROUGH it are three different specimens of one steel — 60, 45 and 15 per cent reduction of area, which converts exactly to a true fracture strain of ln(1/(1 − Z)): 0.916, 0.598 and 0.163. A factor of four in the reported percentage is 5.6 in the strain the material can take. The demand goes as the deposited area over the square of the thickness, so doubling the weld size quadruples it: the 12 mm throat drawn asks for 5.4 per cent, which an ordinary plate supplies and a plate with a bad inclusion cluster does not. Connections

The direction a plate was never tested in

A rolled plate is not one material. Rolling stretches its inclusions into flat stringers lying in the plane, so a bar cut along it, one cut across it and one cut through it are three different specimens of one steel — and every mill certificate reports the first.

 A weld is a force, and it is applied where the weld is. The bow a welded girder leaves the shop with, against how far its welds sit from the section's centroid. A weld cannot contract while the plate holds it, so it yields in tension and what is left when everything is cold is a locked-in force at about the yield stress: 312 kN for the 1.2 kJ/mm of heat drawn, over a shrinkage zone of 439 mm². Applied 210 mm off the centroid that is a moment, and a moment applied along a member is a curvature: the 12 m girder comes out bowed 12.5 mm, which is L/962 against a fabrication tolerance of L/1000. It also comes out 1.2 mm shorter. Welding symmetrically about the centroid puts the resultant on the neutral axis and the bow becomes 0.00 mm — the same heat, the same force, and no moment at all. Connections

The shape that came out of the shop

A weld cools by seven hundred degrees while the plate holds it, so it yields in tension and stays that way. What is left is a locked-in force of three hundred kilonewtons applied where the weld is, and if that is not on the centroid the member leaves the shop bent.

 A pile has no length until the ground gives it one. Deflection, bending moment and soil reaction down a 0.6 m pile carrying 150 kN at a free head, in ground whose modulus grows by 0.005 N/mm³ per millimetre of depth. The one length in the problem is T, the fifth root of EI over n_h, which is 1.89 m here; the head moves 20.5 mm, the worst moment of 219 kNm is at 2.50 m — 1.32 T — and below about four T nothing happens at all. The classical coefficients come out of the finite differences rather than a table: 2.430 against Matlock and Reese's 2.435, and 0.772 against their 0.772. Internal forces

A pile has no length until the ground gives it one

Almost every other structural member is handed a length by the drawing. A pile goes into the ground until it stops, and what decides how much of it is working is a fifth root of the ratio between its own stiffness and the soil's.

 A skew deck spans square, and the corner knows it. Plan of a 30-degree skew slab, 12 m along the road by 10 m wide, with the reaction per unit length of abutment drawn as a bar at each support point. The load takes the shortest route between the abutments, which is the square span of 10.4 m rather than the 12 m of carriageway — so the reaction runs to the two OBTUSE corners, where the abutments are closest, and drains away from the acute ones. Peak 2.15 times the average, least 0.16. Nothing about the loading is uneven; the geometry is. Internal forces

The deck that spans square

A slab bridge crossing a road at an angle is loaded uniformly and does not carry uniformly. Load takes the shortest route between the abutments, which is not the direction the carriageway runs, and the reaction piles up in two corners.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

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