Structural Engineering & Statics
where the load goes
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Which failure arrives first — page 5

Essays 97 to 120 of 360 on this thread, in the same order.
One drift, two motions, opposite curvatures. The sideways movement of a 120 m building under a uniform wind, drawn as the sum of the two mechanisms that produce it. The bending curve is a cantilever's: flat at the base, steepening upward, concave one way. The racking curve is a stack of parallelograms: steepest at the base and flattening, concave the other. They add to 366 mm at the roof, of which 61% is bending. The one group that decides the split is αH = H√(GA/EI) = 2.48: below one the racking dominates and the building behaves as a frame, above about six the bending does and it behaves as a cantilever, and everything interesting is in between. Deflection

Two motions with one name

A tall building's sway is two movements added. A frame racks like a stack of parallelograms, worst at the bottom; a cantilever bends about its base, worst at the top. The total at roof level says nothing about which storey is worst, and on this building it is neither.

 The same concrete, held sideways. Two stress-strain curves for one concrete. The lower is a cylinder test: it peaks at 30 N/mm² near a strain of 0.002 and has nothing left by 0.0035, because it fails by splitting apart sideways. The upper is the same material inside a 12 mm hoop at 100 mm centres, which cannot stop it expanding but can make the expansion stretch steel: the lateral pressure of 2.48 N/mm² — 8% of the strength it is multiplying — takes the peak to 44.4 and the ultimate strain to 0.028. The strength gain is 1.48 times and the strain gain 8.0; the area under the curve, which is the toughness, goes up by 11. It is the third number the confinement is provided for. Materials

Squeezed sideways into a different material

Concrete in a cylinder test fails by splitting apart sideways under a load pushing it down. Put a hoop round it and the splitting has to stretch steel — and a lateral pressure of a twelfth of the strength raises the strength by half and the ultimate strain by eight.

 A strength that is a property of the specimen. Nominal strength against size for geometrically similar specimens of one material. On the left the specimen is too small for a crack to run and the strength is a plateau — a plastic limit, and the regime laboratory specimens sit in. On the right a crack releases more energy than it consumes as soon as it starts and the strength falls as the inverse square root of size, which is the regime real structures sit in. The turn happens at D₀ = 120 mm. A 100 mm specimen reads 3.10 N/mm² and a 1500 mm member of the same material carries 1.14: the test overestimates the structure by a factor of 2.71. Materials

The bigger one is the weaker one

Two geometrically similar beams of the same concrete should fail at the same nominal stress, because a strength is supposed to be a material property. They do not. The large one fails at less, and the reason is that a crack releases energy in proportion to a volume and consumes it in proportion to an area.

 An eccentric load is three load cases, and only two of them are checked. A line load of 40 N/mm at 1.5 m from the axis of a 3.0 by 2.0 m box, replaced by the three cases it is equivalent to. Bending is the load on the axis. The torque 60 kNm per metre then splits into a set of edge forces that drives Bredt's shear flow and distorts nothing, and a set with the flange forces reversed — 10.0 kN/m up one web and down the other, 15.0 kN/m across the flanges — which carries no torque at all and squashes the rectangle into the rhombus drawn behind it. Its generalised load is exactly half the torque, so a box girder spends half of an eccentric load's torsion on changing its own shape, and no torsion calculation contains that half. Sections and stress

The section that will not keep its shape

A box girder is closed, so torsion costs it almost nothing. What an eccentric load actually does to it is something a torsion calculation contains no term for — the rectangle becomes a parallelogram, in its own plane, along the whole length of the span.

 Pull it along the girder and it just unfolds. One period of a 30° trapezoidal corrugation, 300 mm of flat and 260 mm of incline, and the same period pulled along the girder's axis. The fold opens by bending the inclined panels out of the web's own plane, so the axial flexibility contains the plate's t³ where a flat web's would contain t — and the effective modulus that comes back from solving the cell as a frame is 222 N/mm², which is 10.6 parts in ten thousand of the steel's 210 GPa. A web with a thousandth of the stiffness carries a thousandth of the stress, which is why the flanges of a corrugated girder carry the whole moment and why the section has 9 per cent less second moment than the flat-webbed girder it replaces. The fold buys freedom from stiffeners and pays for it here. Sections and stress

The web that carries no bending

A corrugated web needs no stiffeners, because the folds give it in one direction a depth it does not have in its thickness. In the other direction the same folds make it an accordion — and a web that cannot be stretched cannot carry a bending stress at all.

 The neutral axis obeys neither the load nor the moment. A 305 × 102 mm I-section carrying a moment 5° out of the plane of its web. The moment vector is the short arrow; the neutral axis is the long line, at 69.7° to the strong axis. They do not line up, and the reason is that the neutral axis follows the moment ratio scaled by the stiffness ratio: tan α = (M_z/M_y)(I_y/I_z), and I_y ÷ I_z is 30.8 here. So a 5° tilt of the load puts the neutral axis 70° over, the corner that ends up furthest from it carries 489 N/mm² against the 258 the straight-down case would give, and the section has lost 47 per cent of its capacity to a misalignment nobody would draw on a detail. Sections and stress

Two moments and a neutral axis that obeys neither

Tilt the load on a rolled beam by five degrees and the neutral axis swings by seventy. The section is doubly symmetric, its product of inertia is exactly zero, and none of that helps — because what decides the axis is the moment ratio multiplied by a stiffness ratio of thirty.

 The same beam, the same load, and one of them has to lift it. Two lower-bound models of one beam, differing only in which chord the 600 kN is applied to. The chord forces are identical in both — the moment diagram does not know where the load arrived — and so are the struts. What is not identical is the vertical at the load: nothing in the top-loaded model, and the whole 600 kN in the hung one, because a bottom-chord panel point touches no strut and the load has nowhere to go but up. That tie is 1200 mm² of steel against the 96 mm² the shear calculation asks for over the same length — a factor of 12.5, in the same place, and additional to it. Internal forces

The load that has to be lifted

Every shear calculation on this site assumes the load arrives on the top of the beam and walks down a diagonal to the support. Hang the same load from the soffit and the diagonal cannot start — the load has to be carried up to the compression zone first, by a tie no sectional calculation contains.

 Three minima, and only two of them get a check. Elastic buckling stress against half-wavelength for a 200 × 65 × 15 × 1.5 mm lipped channel in uniform compression. The local minimum is at 200 mm and 41 N/mm²; the distortional at 689 mm and 287; the global curve falls away to the right and reaches 489 at the 1.5 m member. The distortional branch is a strut on an elastic foundation — the flange and lip rotating about the web junction, restrained by the web's own bending at 627 N·mm per radian per millimetre — so its minimum is at π(EC_w/k_φ)^¼ and its value is (2√(EC_wk_φ) + GJ)/I₀, the same closed form a continuously braced strut has. The elastic stresses are in the order local, distortional, global, and the mode that governs the strength is not the lowest of them, because they have very different amounts of post-buckling reserve. Stability

The mode between the two that get checked

A thin-walled strut has three ways of buckling and two of them have design rules. The third has a half-wavelength several times the section depth, a shape in which the fold lines themselves move, and an elastic stress that no effective-width calculation can produce.

 The deflection goes on growing, and sometimes it does not stop. Second-order deflection of a sustained-loaded concrete column against age, on a log time axis. Creep takes the effective modulus down, which takes the buckling load down with it — from 11580 kN on the day to 3309 in the long term, 29 per cent of it — so the amplifier 1/(1 − N/N_cr) grows even though nothing was added to the load. At 2200 kN the column settles: 25 mm of eccentricity on the day and 60 mm at the end, a factor of 2.4 for a load that never changed. At 5294 kN — still only 46 per cent of the day-one critical load — it does not settle, and the divergence arrives at 55 days for no new reason at all. Stability

The column that fails years later

A concrete column under sustained load goes on straining at constant stress, so its deflection grows — and because the second-order moment is the load times that deflection, the demand grows with it. There is a load below which the two settle and one above which they never do.

 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. Deflection

Stiffer than its cracked section says

At a crack the concrete below the neutral axis has gone and the steel carries the tension alone. Between the cracks it has not gone — bond drags it back into tension, the steel strain drops, and the curvature averaged over a length of beam is neither section's.

 One criterion inside the other, touching at six points. The two yield criteria in principal stress space with the third principal stress zero, both normalised by the yield stress. Von Mises is the ellipse — σ₁² − σ₁σ₂ + σ₂² = f_y², which is a circle seen at an angle — and Tresca is the hexagon inscribed in it, touching at the six points where one principal stress is zero or the two are equal. Everywhere else Tresca is the smaller, by up to 15.5 per cent, and the widest gap is at pure shear, where σ₁ = −σ₂ and the two answers are 205 and 178 N/mm². The ratio there is exactly 2/√3, computed rather than quoted, and it is the whole reason a web is checked against f_y over root three. Materials

The shear strength nobody measured

Every web on this site is checked against the yield stress divided by the square root of three, and no test produced that number. It is a consequence of a decision about what makes a metal yield, and the alternative decision gives a different answer by fifteen per cent.

 Strength at an angle, and the straight line that is not it. Compressive strength against the angle between the load and the grain. Hankinson's formula — f₀f₉₀ ÷ (f₀sin²α + f₉₀cos²α) — is an interpolation rather than a failure theory, and what makes it worth having is how far it sits from the straight line anyone would otherwise draw between 21 and 2.5 N/mm². At forty-five degrees it gives 4.5 N/mm² against the line's 11.8: 38 per cent of it, and 21 per cent of the strength along the grain. The curve drops away in the first twenty degrees because the weak direction starts governing as soon as it has any component at all, which is the same arithmetic as a section's weak axis and the reason a skewed bearing detail is a real loss rather than a small one. Materials

The material that has a direction

Every material in this collection so far has had one modulus and one strength. Timber has three of each, differing by more than an order of magnitude, and the consequence is not a correction to steel design — it is a different set of checks with a different one governing.

 The bond stress is crowded against the loaded end. A 20 mm bar embedded 806 mm, with the force in it and the bond stress on it plotted along the embedment. Uniform bond — the assumption behind every development length ever tabulated — is a flat stress and a straight line of force. An elastic bond of the same peak strength is neither: the slip is largest where the bar is pulled and dies away over 1/α = 471 mm, so the far end of the bar is doing almost nothing. At the design rule's length of 40 diameters the elastic bond is 55 per cent used. The uniform answer is what the bond looks like after it has yielded along the whole length, which is a statement about ductility rather than about strength. Internal forces

The force that arrives along a length

A bolt takes its force at a hole and a weld along a line. A reinforcing bar has no such place — it is a smooth cylinder in a hole of its own shape, and the only thing stopping it sliding out is a stress smeared over its surface. So the force in it is not a number, it is a function of position.

 What is left after the first fibre yields, which is a property of shape. The shape factor — plastic modulus over elastic — for six sections, computed by finding each one's equal-area axis and summing ±f_y over it. The numbers contain no dimension, no stress and no material: a rectangle is exactly 3/2 whatever its size, a diamond exactly 2, a circle 16/3π. The spread is the argument. An I-section keeps only 13 per cent in reserve past first yield, because nearly all its material is already at the extreme fibre and there is nothing further in to recruit; a diamond keeps 100 per cent, because most of its material is near the middle and doing very little elastically. So the section shapes that are best at elastic bending are the ones with the least left afterwards, which is exactly backwards from the way the reserve is usually described. Sections and stress

What is left after the first fibre yields

The elastic section modulus stops at the moment the outermost fibre reaches yield. Nothing else in the section has, so it goes on taking load — and how much more it takes turns out to be a property of the shape alone, with no dimension, no stress and no material anywhere in the answer.

 One slit, and the torsional stiffness falls by a factor of hundreds. A 200 by 200 box of 8 mm wall, drawn closed and then slit along its length. Closed, the torque runs round the wall as a shear flow and the torsion constant is 5.66×10⁷ mm⁴; slit, the loop is broken and only each wall's own thickness resists, giving 1.31×10⁵ mm⁴. The ratio is 432 to one, so the same torque twists the slit section 432 times as far and raises a peak shear stress 36 times as high. Nothing about the material changed. Sections and stress

The slit that costs a factor of six hundred

Bending stiffness cares where the material is, and changes by a factor of two or three between sensible sections of the same area. Torsional stiffness cares whether the material forms a closed loop, and the penalty for not doing so is an order of magnitude squared.

 The point the rafter turns about, which is off the frame. A pitched portal of 8 m span and 4.0 m to the eaves, with a 1.5 m rise, collapsing. Each rigid part of the mechanism rotates about some point: the left column about its base hinge, the right about its own. The rafter between them does neither, and its centre is found by one rule — two bodies joined at a hinge share that hinge, so the second body's centre lies on the line through the first body's centre and the hinge, extended. Two hinges give two lines and they cross at (8.0, 11.0) metres, which is 5.5 m above the ridge and outside any drawing of the frame itself. From there the whole collapse is two ratios of lengths and no trigonometry: the load factor is 1.339. Flatten the roof and the centre descends; make the two lines parallel and it goes to infinity, which is the statement that the rafter translates instead of turning. Equilibrium

The point the mechanism turns about

A collapsing frame is a chain of rigid pieces, and every piece is rotating about some point. Find those points and the whole collapse load reads off two ratios of lengths, with no trigonometry anywhere — and for a pitched roof the point in question is well above the top of the drawing.

 The end bolts do the work and the middle ones very nearly nothing. A lap of 8 bolts at 70 mm pitch transferring 800 kN between two plates, with the force each bolt actually carries drawn above it and the flat line a division by the bolt count would have given drawn behind. The end bolts carry 1.09 of their nominal share and the middle ones 0.94. The reason is not in the bolts: at the leading end the first plate is carrying everything and the second nothing, so the two strain at different rates and the slip between them is largest there. In the middle they strain alike, there is almost no slip, and a bolt with no slip across it transfers almost no force. The mean over the worst is 0.918, and the end bolt has to slip 1.36 mm before the rest catch up. Connections

The bolts that do not share

Every bolted connection in this collection has divided a force by a number of bolts. That is right for a short joint and wrong for a long one, and the reason has nothing to do with the bolts — it is that the plates they join are elastic, and stretch by different amounts at different points along the lap.

 Six ways for one dowel to fail, and the capacity is the smallest. Johansen's single-shear mechanisms for a 12 mm dowel through 40 and 40 mm members, each drawn as the shape it is: the dowel straight and the timber crushing, the dowel rotating rigidly, one plastic hinge, then two. The capacity under each is that mechanism's own, and the joint's strength is the smallest — 5.02 kN by mode c, which is the dowel rotates rigidly and both members crush. That is the kinematic theorem of plasticity: every mechanism gives an upper bound and the true collapse is the lowest of them. The crushed timber is shaded, and the circles are plastic hinges in the steel. Connections

The smallest of six failures

Everything else in this collection that fails does so in one way at a time. A dowel through timber does not — the wood can crush while the steel stays straight, or one plastic hinge can form in it, or two — and the capacity is the smallest of the six, which is the kinematic theorem of plasticity applied to a joint rather than to a frame.

 The worst speed is not the fastest one. Peak deck acceleration against train speed, for a 20 m span at 6.25 Hz under 10 axles 18 m apart. The spikes are not a numerical artefact and they are not about how heavy the axles are: a regularly spaced train is a forcing function with a frequency v/d, and where a multiple of it lands on the bridge's own frequency each coach arrives in step with the motion the last one left. The arithmetic is v = d·f₁/k, which puts peaks at 405, 203, 135, 101 km/h — all of them operating speeds. What fails first is the acceleration rather than any stress: ballast loses its interlock at about 3.5 m/s², and strength does not appear in the equation at all. Here the limit is first passed at 376 km/h. Dynamics

The train that arrives in time with itself

A single load crossing a span is a mild problem. A train is not one load — its axles are regularly spaced, so the forcing has a frequency of its own, and where a multiple of it lands on the bridge's frequency each coach arrives exactly in step with the motion the last one left behind.

 The demand falls and the movement rises, by the same factor. One elastic spectrum read twice: as an acceleration on the left and as the displacement that goes with it on the right. A fixed-base building at 0.5 s sits on the plateau and is asked for 1.05 g. Put it on bearings soft enough to make its period 2.54 s and the demand falls to 0.103 g — a base shear 10.2 times smaller, bought with no strength whatever. The same shift on the right-hand plot goes the other way: displacement is S_a T²/4π², so the demand rises from 65 mm to 165. That number is the design. It is a gap all the way round the building, a moat every service has to cross, and a detail that a later contractor will fill in unless somebody says what it is for. Dynamics

Made weaker on purpose

Everything else in this collection resists a load by being stiff or strong enough for it. A base-isolated building resists an earthquake by refusing to hear it — a layer of bearings under the whole structure with a lateral stiffness a twentieth of the frame's, bought with almost no strength at all.

 The cheapest way out of being round. A ring under uniform external pressure, drawn in its first four buckling modes with the pressure each one needs underneath it, in N/mm². The pressure has no direction: it stays normal to the wall wherever the wall goes, so it does work on any change of shape that reduces the enclosed area, and the ring buckles into whichever shape is cheapest. Bare, that is the oval — n = 2 at 3EI/R³ — and the modes rise as n² − 1, so three lobes cost 2.67 times as much. Nothing in the drawing prefers any orientation, which is the point — a column has an axis to buckle about and a ring has none. Stability

The pressure that needs no direction

Every buckling problem in this collection has had a load with a direction — a column pushed along its axis, a plate along its edge, an arch by what is on it. A buried pipe has none. The pressure is the same everywhere, it stays normal to the wall as the wall moves, and it does work on any change of shape that reduces the area inside.

 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. Materials

The steel that is stronger in a millisecond

Every strength quoted anywhere in this collection was measured at about a ten-thousandth of a strain per second, because that is what a testing machine does, and nothing on a drawing says so. Load the same steel a million times faster and its yield stress rises by a third.

 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. Structural form

A shell only if the grid takes shear

A curved surface carries load in its own plane at a fraction of the material a flat one needs, and every gridshell ever built is an attempt to buy that with members instead of with a surface. The attempt succeeds or fails on one property nobody draws — whether four bars meeting at a corner can resist being racked — and a pinned quadrilateral grid cannot resist it at all.

 A fourth power, and then a cliff. The factor of safety against rolling, against beam length, for one section hung from a roll axis 0.9 m above its centre of gravity. Nothing about the section changes along this axis. z̄ goes as the fourth power of the length — 0.236 m at 30 m becomes 0.747 m at 40 — and the factor of safety is proportional to (y_r − z̄), so it does not decline gently: it falls away and then stops existing. The working factor of 1.5 is lost at about 41 m, and past 42 m there is no hook height at all at which this beam hangs stably. Which is why long girders are lifted with the picks moved inboard, or with the beam braced, or not in one piece. Stability

Hung from above and still unstable

A rigid body hanging from a point above its centre of gravity is a pendulum and cannot fall over. A beam is not rigid, and tilting it puts a component of its own weight sideways — which bows it, which moves its centre of gravity further out. Past a length there is no hook height at which it hangs stably at all, and the length arrives as a fourth power.

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