Structural Engineering & Statics
where the load goes
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Series — page 10

A field says what an essay is about. A series follows one idea essay by essay — from the question that introduces it to the one that assumes all the others.
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.

Erection stability

  1. 1 The most dangerous day is before it is finished
1 essay · stability Eight slices is enough, and nobody would have guessed it. The error in a cracked section's moment capacity against the number of strips it was integrated with, for a 300 × 450 mm section with 1200 mm² of steel, measured against the same computation at 2048 strips. The point of the fibre method is that it contains no formula: slice the section, give every strip the strain the assumed curvature puts it at, move the neutral axis until the axial force balances, and sum. It handles a cracked section, a confined one, a prestressed one and a composite one with the same twenty lines. The discretisation costs 1.4% at 2 strips and 0.088% at 8 — and the convergence is not smooth, because what the error actually depends on is where the neutral axis falls relative to a strip boundary rather than on the strip count as such.

Fibre model

  1. 1 The section calculation with no formula in it
1 essay · sections 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.

Filled tube

  1. 1 Each one stops the other failing
1 essay · materials Five millimetres short, and a hundred kilonewtons in every member. An X-braced bay 6 m by 4 m in which one diagonal was fabricated 5 mm short, with the force that leaves in every member. Nothing is applied to this frame. The forces are the self-stress state the frame's one redundancy supports, scaled so that the diagonal is pulled back to the length it should have been: tension in both diagonals at 100 kN, compression in the four members round the outside, and the whole set in equilibrium with nothing. That is 25% of the force the diagonal was sized to carry, and it is there for the life of the structure. Take one diagonal out and the frame becomes determinate: the short member then simply puts the joint somewhere else, and the structure is in the wrong place instead of under stress. Redundancy is bought, and this is the price.

Fit-up

  1. 1 Built to the wrong length
1 essay · connections A transverse load with nothing applied. Web slenderness against web thickness, with the limit the flange's own curvature sets. A flange carrying 6213 kN and curved to a radius of 592 m needs 10.5 N per millimetre of radial force to stay on its curve, and the only thing available to supply it is the web. Nothing has been applied to the girder: the load comes from the deflected shape, which is why a straight beam has none of it and a beam at a plastic hinge has a great deal. Setting the radial force against the web's own plate-buckling resistance gives, in four lines, h_w/t_w ≤ k·(E/f_yf)·√(A_w/A_fc) — the form the codes use, arrived at without them. The constants differ: an elastic flange strain gives k = 1.34 and the rule uses 0.3, a factor of 4.5, and the gap is the curvature assumed. k goes as the inverse square root of the flange strain, so 0.3 is a flange strained to 3.4% — which is what a plastic hinge does to it. The rule is not conservative; it is written about a different beam.

Flange induced

  1. 1 The web that is crushed from inside
1 essay · stability The same sheet, twice, and a factor of ten thousand. A 3000 mm developed width of 3 mm sheet, covering 2400 mm in plan — so the legs sit at 36.9° and the fold is 300 mm deep. Flat, its second moment about its own mid-plane is 6750 mm⁴, which spans nothing. Folded, it is 67.50×10⁶ — 10000 times as much, which is exactly the depth in thicknesses squared. The material is identical, the plan cover has fallen by 20%, and the only thing that changed is where the material sits. What limits it is buckling of the leg: at this leg length the flat between the folds goes at 27 N/mm², well below the steel's 275.

Folded plate

  1. 1 Folded until it spans
1 essay · structures 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.

Follower force

  1. 1 The load it cannot buckle under
1 essay · stability Two different structures released, and one bending moment diagram. The bending moment in a continuous beam of 8, 10, 8 m under 12 kN/m, solved twice by the force method with different redundants. The first release puts a hinge over each interior support, so the released structure is a row of simple spans and the redundants are moments. The second removes each interior support, so the released structure is one simple span of the whole length and the redundants are reactions. The two released structures have nothing in common — different shapes, different deflections, different everything — and the diagrams they produce lie on top of each other to 9e-15 of the peak moment. Which restraints are released is a choice about the arithmetic and not about the structure, which is a fact worth trusting: it means a hand calculation can pick whichever release makes the sums easiest and be sure of the answer.

Force method

  1. 1 Choose what to take away
1 essay · deflection Three ways to move the same column, and they are not close. The same 2000 kN moved 3 m across 14 m, built three ways and drawn to one scale. The deep beam is 1.14 m of concrete, 24.1 tonnes, and settles 60.6 mm in the long term. The storey-deep truss takes the same moment as a couple at 3.6 m centres, so its chords carry M/h and it weighs 2.6 tonnes — a fifth of the beam — while settling 15.8 mm, and it does not creep. The wall is 72.8 tonnes and hardly moves at all, 1.71 mm, of which 36% is shear rather than bending — which is what a member as deep as it is long always does, and is why beam theory does not describe one. A wall as a deep beam is the stiffest of the three by a factor of 35.4.

Form selection

  1. 1 The same span, four ways
1 essay · structures The corner columns take what the middle ones did not. Axial stress in the columns across one flange face of a 30 by 40 m framed tube, at the base. Plane sections says the flat line: every column on the face at the same distance from the neutral axis, therefore at the same stress. The solved distribution is the curve — 69.4 N/mm² at the corner against 18.2 in the middle, a ratio of 3.81. The middle columns lag because the only route the axial force has into them is the in-plane shear of the spandrel frame, bay by bay from the corner. The face is carrying its resultant on an effective width of 51 per cent, and the tube deflects as though its second moment were 72 per cent of the gross.

Framed tube

  1. 1 The corner columns take more than their share
1 essay · structures The same beam, cut at x = 5. A beam separated at one station. On the exposed face a shear force and a bending moment appear, equal and opposite on the two pieces, with values obtained by summing the forces on whichever piece is easier.

Free body

  1. 1 The free body is a choice, and choosing it well is the whole skill
1 essay · equilibrium The check that everything adds up, and the error it cannot see. Four versions of the same 2-bay, 3-storey frame, with the global equilibrium residual each one produces — the sum of the reactions against the sum of the applied loads, as a fraction of the applied total. It is the first thing every analysis prints and it is worth having: a lost restraint and a load entered in the wrong unit both show up immediately, at 6% and 24%, because both change what the structure is carrying. The fourth bar is the point. A member whose stiffness is wrong by a factor of ten redistributes the internal forces completely — the second bar shows the change in the member forces, 18% — and the global residual is exactly zero, because the wrong answer is still in equilibrium with the same loads. Equilibrium is one equation per degree of freedom of the whole body, and a stiffness error lives entirely in the many equations underneath it. A model can satisfy every equilibrium check ever devised and be a model of a different structure.

Geometric stiffness

  1. 1 The stiffness the load takes away
1 essay · stability Four ways to make a cell resist being racked, and one that is not one. One cell of a grid shell under the membrane shear it has to carry, by the four mechanisms available for carrying it, with the racking each produces over a 30 m span under 1.2 kN/m² of asymmetric load. A serviceability limit of span/250 is 120 mm. Four pin-jointed bars in a quadrilateral have no in-plane shear stiffness whatever — the cell folds, and the answer is not a large deflection but a mechanism. Rigid nodes carry the shear by bending the members over a cell, which smears to 12EI/s³ and comes to 0.35% of what a continuous sheet of the same stretching stiffness gives: 1205 mm, ten times the limit. One diagonal per cell, or a third member direction, carries it axially instead and lands within a factor of two of the sheet. That is the whole difference between a grid shell and a row of arches.

Gridshell

  1. 1 A shell only if the grid takes shear
1 essay · structures The width nobody drew. A gusset plate with a brace bolted to it over 240 mm, and the width the profession has agreed to pretend is carrying the force. Everything else on this site arrives with a cross-section; a gusset does not, because it is a piece of steel with something attached somewhere in the middle of it and there is no geometry that says how much of it is working. The answer is the Whitmore section: assume the force spreads at 30° from the first fastener and take the width it has reached at the last, b_eff = w + 2L·tan30° = 367 mm. That is 4.08 times the width anything is actually attached to, and the rule comes from a 1952 master's thesis. It has since been checked against finite element work and holds to about ten per cent, which is fortunate, because moving the assumed angle by ten degrees moves the answer by 34%. On this plate the check that governs is not the stress the rule was written for: it is the Whitmore section buckles, at 721 kN against 1564.

Gusset

  1. 1 The width nobody drew
1 essay · connections Where the wind's energy is, and where the structure can reach it. The gust spectrum at a mean speed of 30 m/s, plotted as n·S(n) against frequency on a logarithmic axis so that equal areas are equal energy. Its total variance is 27 m²/s², which the closed form 6·K·U² gives as 27. Most of the energy is below a hundredth of a hertz — gusts lasting a minute or more. A structure at 0.2 Hz sits far out on the tail, and still takes 90% of its response from there, because the resonant part is amplified by π·f₀/(4·ζ) and the damping is 1.2%. A stiffer structure at 2 Hz takes 22%.

Gust response

  1. 1 The wind is a spectrum
1 essay · dynamics The windward guy tightens, the leeward one gives way. A 120 m mast on three guy levels, at a wind of 3 N/mm, with the deflection drawn 0.54 times its true size. The guys start at 160 kN each and end at 259 against 95, 299 against 83, 221 against 112 kN. The leeward guys still carry a real force — the lowest keeps 28 per cent of its partner's tension — and supply almost none of the restraint, because their tangent modulus has fallen to 54 per cent of the steel's. The mast top moves 100 mm, its worst bending moment is 554 kNm at 40 m, and it is carrying 801 kN of axial load that nothing but the guys put there.

Guyed mast

  1. 1 Held by something that goes soft
1 essay · structures 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.

Hierarchy

  1. 1 Every level is a longer span
1 essay · structures The free body that makes a hoop force a pressure times a radius. Half a ring cut along a diameter, with the pressure drawn normal to the wall wherever the wall is. Vertical equilibrium of the half ring is the whole derivation: the pressure acts over the projected width 2R whatever the shape of the arc, the two cut faces carry N each, so N = pR — 300 kN per metre here at 1 MPa on a 0.3 m radius. The result contains no wall thickness, no second moment, and no length along the pipe, which is why a hoop force is the one internal force in this collection that arrives with no lever arm attached to it. The stress does contain the thickness — 25 MPa at 12 mm — but the force does not, and a thicker wall carries exactly the same force at a lower stress.

Hoop tension

  1. 1 The force that is only a radius
1 essay · internal-forces A tie is worth all of itself and a strut is not, which is worth nine per cent. The same 20 floors carried two ways, with every member drawn at the width its own force requires. Hung, the loads accumulate upward, so the largest hanger is at the top: 18.0 MN at 355 N/mm² with no buckling reduction of any kind. On columns they accumulate downward and the largest column is at the bottom, at the same force — but every column above it is understressed by its own slenderness, worst at the top where a 900 kN column still has to be 4.8 × 10³ mm² to reach χ = 0.529. Over the height the hangers total 0.002 m³ of steel against 0.002: a saving of 9.0 per cent, which is the average χ and nothing else.

Hung structure

  1. 1 Hung from the top, and nine per cent lighter
1 essay · structures Three paths out of the same critical load. Load against sideways movement past the critical load, for three systems whose critical loads are identical. The stable one climbs, so a real structure with a small crookedness reaches nearly the full load and keeps going. The unstable one falls symmetrically, so the imperfect structure has a maximum below the critical load and it matters not at all which way it leans. The asymmetric one falls one way and climbs the other, so the direction of the imperfection decides everything. All three are drawn at an imperfection of 0.02 radians.

Imperfection sensitivity

  1. 1 A third of what the theory promised
1 essay · stability

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