Structural Engineering & Statics
where the load goes
Depth

Series — page 15

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.
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.

Stepped column

  1. 1 An average stiffness is not a safe stiffness
1 essay · stability A stiffener is a boundary condition, and it is bought at a threshold. The buckling stress of a 2400 × 12 mm plate with one longitudinal stiffener, against how rigid that stiffener is. Below γ the stiffener rides on the buckle and the plate takes the whole-width mode; at γ the stiffener stays straight and the plate buckles between stiffeners at 74 N/mm², 4.0 times the bare plate's 18.5. Above γ nothing further happens at all, because the sub-panel mode does not know the stiffener is there. The curve is a ramp and then a horizontal line, so a stiffener at twice γ is exactly as good as one at γ. Here γ = 31.5, which asks for an outstand of 144 mm; the 150 mm one drawn gives γ = 35.5, a margin of 1.13.

Stiffener rigidity

  1. 1 The rib that is a boundary condition
1 essay · stability Almost all of it is exactly zero. The stiffness matrix of a 2-bay, 3-storey plane frame: 36 freedoms, of which 16.2 per cent of the 36² entries are non-zero. The zeros are not small numbers; they are absences. A member reaches only the two nodes at its ends, so it can contribute nothing to any row belonging to a node it does not touch, and every such entry is zero exactly rather than nearly. The non-zeros therefore sit in a band of width 13 about the diagonal. Before the supports are applied the matrix is singular, and its null space has exactly three dimensions — the three rigid-body motions a plane frame has with respect to the ground, which is the same statement the null vector of the equilibrium matrix makes about a truss that is a mechanism, arrived at from the other end.

Stiffness method

  1. 1 The matrix that replaced the hand methods
1 essay · deflection Effective length is a property of the storey. The effective length factor of the one column that resists sway, against the total gravity load on the storey as a multiple of its own. At the left-hand end it carries the storey alone and its K is 1.99 — the 2.0 every chart gives a column fixed at the base and free to sway at the top, reproduced here by a route that never mentions a chart. Then columns are added that have pinned bases and therefore no lateral stiffness whatever. They contribute load and nothing else, so they cannot buckle on their own and they lower the load at which everything buckles together. K rises as the square root of the load ratio, exactly, and at the storey drawn — three leaning columns carrying 69% of the gravity load — it is 3.57. That is off the end of every published alignment chart, and the leaning columns themselves, which a designer would take at K = 1.0 for pinned ends, are at 2.54.

Storey buckling

  1. 1 The column that leans on its neighbours
1 essay · stability How fast the strain arrived, which a quoted strength does not record. The dynamic increase factor on strength against strain rate, over eight decades. Steel follows Cowper and Symonds' fit, whose constant D = 40.4 s⁻¹ is not an arbitrary parameter — it is the rate at which the material is exactly twice as strong. Concrete in tension follows the model code's two-branch curve and is steeper. The four marked regimes are the argument: a testing machine works at about 10⁻⁴ per second, an earthquake at 5 × 10⁻³, a vehicle impact at a half, a blast at a hundred, and the enhancement across them runs 1.00, 1.08, 1.32, 2.04. So this is a correction that is either negligible or decisive with very little in between, which is why no seismic code carries it and every blast code does. What does not rise is the modulus, which is a lattice property, and the ultimate strength rises only a third as much — so the ultimate-to-yield ratio closes from 1.56 to 1.23 and the material has less warning left in it than it started with.

Strain rate

  1. 1 The steel that is stronger in a millisecond
1 essay · materials Wrong in shape, right in two integrals. The compression zone of a C30 section with its neutral axis 150 mm down, drawn twice. The curved outline is the real parabolic-rectangular stress distribution — the material's own law read off the linear strain profile plane sections supplies. The rectangle over it is what every design office uses instead: intensity η f_cd = 16.5 MPa over a depth λx = 125 mm. The two shapes are visibly different and give the same answer, because a bending calculation asks a stress distribution only two questions — how much compression there is, and where its resultant acts. Both are 619 kN at 62.4 mm from the face. The factors are α = 0.8095 and β = 0.4160, and λ = 2β follows from wanting the same centroid. A triangle and a full rectangle match neither integral and are nowhere near.

Stress block

  1. 1 Deliberately the wrong shape
1 essay · sections 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.

Superposition

  1. 1 The addition everything else rests on
1 essay · deflection 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.

Support flexibility

  1. 1 The deflection that belongs to the support
1 essay · deflection Where to put the supports. Peak sagging and hogging moment for a uniformly loaded beam, against how far the supports are moved in from the ends. The best arrangement is where the two curves cross, and it is nowhere near the ends.

Support layout

  1. 1 Where to put the supports, which is not at the ends
1 essay · internal-forces The bearing is one length and the web is loaded over another. A load applied over a stiff bearing of 200 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 659 mm — 3.3 times the bearing, and 70% 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.

Support width

  1. 1 The support that is not a point
1 essay · internal-forces The strength that does not keep pace. Concrete's mean tensile strength against its characteristic compressive strength, with the ratio of the two on the same axis, scaled. The tensile strength goes as f_ck^⅔, so it rises from 1.57 to 5.04 N/mm² over a sixfold rise in the compressive strength, and the ratio between them falls from 11.1% at C20 to 6.0% at C80 — a factor of 1.83. Nothing in a bending or a column calculation ever uses the lower curve, and everything that decides a transition does: when the section cracks, how much minimum steel it needs, how far a bar has to be lapped, what a member without links carries. So a stronger concrete needs more minimum reinforcement, longer laps and a bigger crack-control check, in a member whose ultimate capacity has barely moved. Its real scale is a length: E·G_F/f_ct² is 299 mm here, which is the size at which a member stops behaving plastically and starts behaving like a fracture problem.

Tensile strength

  1. 1 The strength that is never used
1 essay · materials The beam is stiffer than its cracked section and softer than its gross one. Moment against mid-span deflection for a 300 × 550 mm beam spanning 8.0 m, with the two bounds it lies between. The uncracked line is what the gross transformed section gives; the cracked line is what the section at a crack gives; and the curve between them is the member, because between the cracks the concrete is still carrying tension and the average curvature is not either section's. At the service load the deflection is 23.2 mm — span over 345 — against 8.3 uncracked and 25.2 fully cracked, a factor of 3.03 between the bounds. The interpolation ζ = 1 − β(M_cr/M)² sits it 88 per cent of the way across, and β falls from one to a half under sustained or repeated load because the bond that does the dragging deteriorates.

Tension stiffening

  1. 1 Stiffer than its cracked section says
1 essay · deflection Steel and concrete happen to match, and nothing else on the list does. The mismatch strain a 40 degree change produces in seven pairs of materials that engineering bonds together, which is the difference of their coefficients of expansion times the temperature. Steel against concrete is 80 microstrain — 17 per cent of the larger coefficient, and by far the smallest on the list. It puts 0.223 N/mm² of tension into the concrete, 7.7 per cent of its tensile strength and 1.9 per cent of the 12 N/mm² a fully restrained member would have carried. Reinforced concrete works because of a coincidence in the third significant figure of two numbers nobody chose, and the same bar in aluminium would put in two and a third times as much.

Thermal compatibility

  1. 1 The coincidence reinforced concrete stands on
1 essay · materials 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.

Thermal gradient

  1. 1 The stress nobody restrained
1 essay · sections It moves, or it pushes. Never both, and never neither. A 30 m steel member 30 °C warmer than it was built, in three conditions. Free, it grows 10.8 mm and carries nothing. Held, it moves nothing and carries 75.6 MPa in compression — which is E·α·ΔT and contains neither the length nor the area of the member, so the identical stress arises in a two-metre strut. Held by a spring it does some of each: 3.2 mm of movement and 53.2 MPa, and the split is decided by the spring rather than by the member.

Thermal movement

  1. 1 The movement nobody applied
1 essay · deflection Three legs, three equations, one answer. A rigid top on three legs carrying 120 kN at (0.3, 0.2) m. The three equilibrium equations available — one vertical and two moments — leave three unknowns, so the system is exactly determinate and the reactions are 56.8, 13.6, 49.6 kN. Move the load anywhere and the answer moves with it; nothing about the legs' stiffness enters.

Three-dimensional equilibrium

  1. 1 Six equations, and the drawing shows three
1 essay · equilibrium The tie is a redundancy, so its stiffness decides the thrust. Thrust and rib bending for a 60 m tied arch of 0.15 rise ratio, against the stiffness of its tie. Cut the tie and the structure is a curved simply supported beam, so the tie force is the one redundant and the force method gives it: with a rigid tie the answer is 2227 kN, within 1.0 per cent of the funicular wL²/8f, and the shortfall is the arch's own axial shortening. A real tie stretches 68 mm and returns 2166 kN — 2.7 per cent of the flexibility is the tie — and whatever thrust the arch does not get, it carries as bending: 760 kNm at 30 m. A tenth of the tie stiffness is not a tenth of the problem; it is a different structure.

Tied arch

  1. 1 The thrust that never reaches the ground
1 essay · structures A buckling load with no compression anywhere in the member. A tube of radius 81.6 mm and wall 5 mm under pure torque — no axial load at all — buckles into a helix at T = 2πEI/L, which for the section drawn is 1877 kNm. Two things are absent from that expression and both are surprising: the length appears to the first power rather than the second, and the shear modulus does not appear at all, so the torsional stiffness of the member has nothing to do with the torque at which it buckles in torsion. The comparison is the torque that yields the same tube, 42.9 kNm — a factor of 43.8 below it. The mode arrives first only past a length of 263 m, which is πE/τ_y = 3219 radii and contains no thickness whatever.

Torque buckling

  1. 1 The buckling load with no compression in it
1 essay · stability A torque diagram is a shear diagram about a different axis. A torque of 40 kNm applied 2 m along a member of 6 m held against twist at both ends. The two ends take 26.7 and 13.3 kNm, in inverse proportion to their distances, because the two halves are springs in parallel and torsional stiffness is GJ over length. The diagram steps at the applied torque and closes at the far end, exactly as a shear diagram does — the only difference is which axis the arrows turn about.

Torsion

  1. 1 The internal force with no diagram
1 essay · internal-forces The depth is decided by how far it moves, not by what it can carry. A column carrying 6000 kN landing 3 m into a 12 m transfer member. The free body is the member itself, cut under the column: M = P·a(L − a)/L = 13500 kNm, with 4500 kN of shear on one side of the cut and 1500 on the other. At an allowable stress that moment asks for 1.94 m of depth — the dashed outline — and keeping the settlement it causes inside the floors' own bending asks for 2.55 m, which is the member drawn solid. 31% more depth is bought by nothing the strength calculation can see. The depth grows as √(P·a), so four times the load is exactly twice the depth, and depth in a transfer member is a storey nobody occupies.

Transfer structure

  1. 1 The column that stops
1 essay · structures

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