Structural Engineering & Statics
where the load goes
The collection

Every essay — page 16

Essays 361 to 384 of 728, in the same order.
Two per cent of the traffic and most of the damage. A 120-year traffic spectrum on one detail of category 71, with each band's share of the cycles and its share of the damage. The two bars have almost nothing to do with one another, and the reason is the slope of three: life goes as the inverse cube of the stress range, so a cycle twice as large does eight times the damage and a cycle a third as large does a twenty-seventh of it. A full train is 2% of the crossings and 58% of the damage; and three of the five bands — an ordinary lorry, a van, a car, 90% of the crossings — sit under the cut-off and do none of it at all. The equivalent constant range that would do the same damage in the same number of cycles is 25.5 N/mm², which is the one number a designer is usually handed — and it is a cube-weighted average, so it is nearer the heaviest vehicle than to the average one. Materials

The cycles that do not count

A fatigue spectrum has to be reduced to one number, and the reduction is a cube-weighted average rather than an ordinary one. Two per cent of the traffic does most of the damage, ninety per cent of it does none at all, and the equivalent range that comes out is nearer the heaviest vehicle than the average one.

7 figures · Fatigue Throat stress round a fillet weld group. A l shape weld group carrying 150 kN at 200 mm from its centroid. The peak throat stress is 2.17 kN per mm of throat, at (139.53, 100); the worst point at maximum radius from the centroid carries 1.99. Checking by radius is wrong here by 9.17%, and points at identical radius differ by a factor of 1. Connections

The radius rule, and where it fails

A weld group under an eccentric load is checked at the point furthest from its centroid, on the reasoning that the stress from the twist grows with the radius. That reasoning ignores the direction the two stresses point in, and for one common shape it misses the peak by nine per cent.

8 figures · Weld group Prying against flange thickness. The ratio of bolt force to applied force, for a tee stub carrying 140 kN per bolt, as the flange thickness varies. Prying disappears above 31.91 mm and the flange has become a mechanism below 22.75 mm, where the shaded region begins and the bolt has stopped being the thing that decides. Connections

The thickness that decides who fails

A bolt in a tee stub carries more than the load applied to it, because the flange bends and levers against its own edge. How much more, and whether the bolt or the flange is the thing that gives way, are both decided by one dimension — and the two regimes it separates fail in completely different ways.

8 figures · Prying A preloaded joint, before and after it slips. Four preloaded bolts at 172 kN each, on one friction face at μ = 0.5. The joint carries 344 kN by friction with the bolts in tension and not in shear at all; past that it slips into bearing and carries 362 kN with the bolts now in shear. Two different mechanisms, one joint. Connections

The hole made bigger so the steel would fit

A preloaded joint carries load by friction, and the friction is reduced by the shape of the hole the bolt passes through — not by how much steel the hole removes, but by a coefficient in a table. An oversize hole costs fifteen per cent of the resistance; a long slot costs thirty-seven. Both are provided because the steel would not otherwise line up.

7 figures · Slip-critical Block shear: the metal between the holes. Three bolts in a 9 mm plate end connection. The shaded block tears out along a shear plane 180 mm long and a tension plane 55 mm long. Shear ruptures first, and the capacity is the sum of two different strengths on two different planes: 524.79 kN, of which the shear plane carries 63.03%. Connections

The end that is only a plate

Cut one flange off a beam's end and what is left is a tee. Cut both and what is left is a plate with holes in it — no flanges, no section modulus worth the name, and none of the checks the beam was selected by. Three plate checks replace them, and the one that governs depends on dimensions that appear in no section table.

7 figures · Coped beam Where the drift went. Storey drift at the target displacement, for the same frame with and without a soft ground storey at 50 per cent of the others' stiffness and strength. The regular frame spreads 219 mm over every storey; the soft one reaches 266 mm and puts 5.58 per cent of it into the ground storey against 0.53 next to it — a concentration of 5.9 against 1.7. The roof goes 22 per cent further, and where that extra displacement lands is the whole of the difference between the two buildings. Dynamics

Weaker in one place, and better on every average

Take an eight-storey frame and make its ground storey half as stiff and half as strong. Its ductility demand falls, its first mode carries more of the mass, and its period lengthens into a gentler part of the spectrum. Three global numbers all improve, and the building is the one that collapses.

9 figures · Pushover A check made on a perimeter, not on a section. One bay of a flat slab, 7.2 m square, on a 400 × 400 mm column. The heavy closed line is the control perimeter, 2d from the column face with its corners rounded at that radius — 4427 mm long against 1600 mm round the column itself. The shaded area inside it delivers no shear across it and is subtracted from the load; everything outside arrives through the perimeter. At 12 kN/m² that is 604 kN across 4427 × 225 mm, a shear stress of 0.697 N/mm² against a resistance of 0.658. Internal forces

Turn the column, and the slab passes

A flat slab that is comfortable under gravity fails its punching check the moment a moment arrives at the column, and nothing about the load has changed. The fix is not more concrete. It is the column's plan shape and, at equal area, which way round it is turned — worth more than adding half again as much column.

8 figures · Punching shear A built-up column has a second way to bend. A 24 m column of two chords 800 mm apart, joined by single lacing. On the left it buckles the way a solid column does, by bending; on the right the chords stay straight and the lattice racks, which a solid column cannot do at all. Neither happens alone, and the two flexibilities add rather than the two stiffnesses — so the critical load is 3832 kN against an Euler load of 4629 kN, which is 83% of it, and the column behaves as though its slenderness were 66 rather than 60. Stability

The lacing decides the force it has to carry

A perfectly straight column carries no shear, so the diagonals of a built-up column are resisting a force that exists only because the column is crooked. Computing it turns out to be circular — the shear is amplified by the very flexibility the lacing supplies — and the rule of thumb that replaces the calculation is wrong by a factor that grows as the lacing gets worse.

7 figures · Built-up column The worst force in a pile is not at the top of it. A 0.75 m pile 30 m long through ground that is settling, carrying 1200 kN at its head. Above the neutral plane the soil moves down past the shaft and the friction acts downward, so the axial force grows with depth; below it the friction acts upward in the ordinary way and the force falls again to the 600 kN the base takes. The maximum is 2331.39 kN at 18.86 m — 1.94 times the load applied, and it is at a depth where nothing is applied, nothing is connected and nothing can be inspected. A pile section chosen for the head load is under-sized by that factor over the middle third of its length. Internal forces

The coating that takes the resistance with it

The cure for downdrag is to make the pile slippery, and it works — a bitumen slip layer takes the drag on this pile from 1,131 kN to 34. It also removes the shaft friction that was holding the pile up, in the same proportion and over the same length, and past a certain smoothness there is no neutral plane to find because there is no equilibrium.

7 figures · Downdrag Cut the throat, and the face carries a moment and a tension at once. A crane hook of trapezoid section, 50 to 120 mm radius and 40 to 15 mm wide, carrying 100 kN, drawn beside the section at the cut and the stress across it. The load hangs on a line through the centre of curvature, so cutting the throat and taking everything below the cut as the free body leaves a face carrying a direct tension of 100 kN and a moment of N·R = 7.970 kN·m about the section's own centroid, which sits a full R = 79.70 mm from the load line. The stress is a hyperbola, zero at r = 75.04 mm rather than at the centroid 4.65 mm outside it, reaching 497.5 N/mm² of tension at the inner fibre and 281.3 of compression at the outer. The straight-beam formula, drawn dashed, reports 323.4 N/mm² for the bending part against the true 445.5, and leaves the 51.9 N/mm² of direct tension out altogether — between them, 54% under the real peak, at the fibre where a hook actually breaks. This is why a hook is trapezoidal: both effects are worst inside, so the material goes there. Sections and stress

The wide side goes inside

A crane hook's section is a trapezoid with its broad face towards the centre of curvature, and that is not a casting convenience. Turn the same section round — same area, same depth, same moment — and the stress at the fibre that breaks rises by thirty-nine per cent. The shape is doing two things at once, and only one of them is in a straight beam's arithmetic.

9 figures · Curved beam The shear goes round the corner instead of across it. A 6-panel Vierendeel girder, 21 m by 6000 mm, under 300 kN at mid-span. There is no diagonal in it, so each panel's 150 kN of shear is carried as bending in the chords: the curves drawn along them are the chord moments, and every one passes through zero at the middle of its own panel. The local moment is the panel shear times the panel length over four, 131.3 kNm, and it adds to an axial force of 263 kN from the global moment at the same point. The girder deflects 109.09 mm against 6.53 mm for the same members triangulated — 16.70 times — and 98% of that movement is chord bending that a diagonal would have removed entirely. Structural form

The frame is a girder stood on end

Every unbraced building frame is a Vierendeel girder turned through ninety degrees, and the identification is not an analogy — it is the same equations with the axes swapped. Which means the frame inherits results that read as absurd for a building — more bays is stiffer, a wider building is not, and doubling one section property halves the sway.

7 figures · Vierendeel A buckled panel is a truss that nobody drew. A 1500 × 2000 panel of 8 mm web, at d/t = 188. It buckles in shear at 41.0 N/mm², which is 492 kN — and it then carries 1187 kN, 2.41 times as much, because the tension diagonal takes over from the compression one that has gone. The band runs at 18.5° with a membrane stress of 348 N/mm² over a width of 788 mm, and it pulls on the flange at 280.2 N per millimetre of its length. A web that never buckled at all would have reached 2460 kN, so the panel ends at 48% of a stocky web's capacity on a fraction of its steel. Stability

The tension has to pull on something

A buckled web carries its shear on a diagonal band of membrane tension, and the band pulls sideways on the flanges and stiffeners that bound it. That pull is the design output nobody plots — it runs from 72 to 603 newtons per millimetre across ordinary panel proportions, it is largest exactly where the panel is most efficient, and at the end of the girder there is nothing beyond to take it.

7 figures · Tension field One of these two curves is a stiffness and the other is a statement of statics. The torque a spandrel beam carries, against how much of its torsional stiffness is left. The rising curve is compatibility torsion — a floor beam framing into the side of the spandrel, which shares its fixed-end moment of 197 kNm between the spandrel's torsional stiffness and its own flexural one. Uncracked, the spandrel takes 51% of it, or 100 kNm; at a quarter of that stiffness it takes 21%, or 41 kNm, and the floor beam picks up what was shed. The flat line is equilibrium torsion — a canopy cantilevering 2.2 m off the same spandrel, whose 116 kNm is fixed by statics and contains no stiffness at all. The first can be designed away by accepting a rotation. The second cannot be designed away by anything. Internal forces

The torque that should not be shed

A compatibility torque can be let go, because the load has somewhere else to go. What the rule does not say is what it costs the somewhere else — a twenty per cent rise in a floor beam's midspan moment, a crack width nobody limits, and a rotation the spandrel has to actually deliver. There is a size of torque past which shedding is the wrong answer, and no code states it.

7 figures · Compatibility torsion The distortion runs the length of the span, and a diaphragm stops it. Longitudinal stress at a corner of the box from distortional warping, along a 60 m span carrying 45 N/mm at 2.0 m off the axis. With no interior diaphragm it peaks at 73 N/mm², which is 83 per cent of the bending stress the girder was designed for. Two interior diaphragms take it to 18. The governing length is Winkler's: the distortion decays over 23.6 m, so a diaphragm helps its neighbours only if it is closer than that, and past it the spacing stops mattering. Sections and stress

One diaphragm is nearly none

A box girder's distortion decays over a length the section decides, and on a sixty-metre span that length is twenty-four metres. So a single diaphragm at midspan sits further from each end than the distortion can reach and removes a third of the problem; three diaphragms remove nine tenths. The spacing rule is not span over five — it is a property of the plates.

7 figures · Box distortion Removing each member in turn. Every member of a 8-panel pratt truss removed one at a time, with the worst demand on the survivors plotted against the member removed. Four of the 35 leave a mechanism — the bars drawn to the top of the frame — and for those there is no redistribution to compute, because there is no structure left. The rest redistribute, and the worst of them asks a survivor for 2.04 times what it carried before. A single number for robustness does not exist: it depends on which member goes. Structural form

A determinate truss has no robustness at all

Remove any one of a Warren truss's thirty-one members and what is left is a mechanism. Not weakened — gone, with no set of forces that holds the load in any position. Robustness is not a property a structure has by degree; it is bought by adding members that carry nothing until something else stops carrying, and a truss without them has none of it to measure.

7 figures · Robustness One pulse, two blocks of the same shape. Rotation as a fraction of the toppling angle, under a single 0.8 s sine pulse of 1.00 g, for two blocks of identical proportion whose sizes differ by a factor of 3. Both lift off at the same instant, because uplift depends on the shape alone. The small one reaches the toppling angle and goes over; the large one reaches 46%. The kinks are impacts: there is no dashpot anywhere in this model, and the only energy the block loses is lost when it lands on its other corner, at a velocity ratio of 0.926 per landing — 14% of the energy each time, decided by the block's shape and by nothing else. Dynamics

The only damping is the landing

A rocking block has no dashpot in it. The only energy it ever loses is lost at the instant it lands on its other corner, and how much is a property of the block's proportions — 14 per cent for a slender one and 38 for a stocky one. That single number decides whether it settles or goes over, and a real base does not deliver the value the theory computes.

7 figures · Rocking What a torsional brace buys, and the stiffness it takes. The critical moment of an 8 m beam against the stiffness of a single rotational restraint at midspan — a cross-frame or a stiffened connection to a secondary beam, resisting the twist rather than the sideways movement. It climbs from an unbraced 143 kNm to the same two-half-wave plateau of 447 a lateral brace reaches, and gets to 99% of it at 253 kNm/rad. Nothing in the calculation refers to a height, which is the difference that matters: a torsional brace cannot be put on the wrong flange because it is not attached to a flange in the sense the lateral one is. Stability

Twice the moment, four times the brace

A lateral brace has to be on the right flange and its demand is very nearly linear in the load. A torsional brace has no flange to be wrong about, and its demand is exactly quadratic — so the restraint that is indifferent to where it is attached is the one that gets expensive fastest.

7 figures · Beam bracing The stub column a bearing stiffener makes, in plan. A plan through the girder at the bearing. The 8 mm web runs across; the pair of 100 × 12 stiffeners stands off it; and the shaded strip of web either side — 15ε t_w, or 98 mm each way — is the width that buckles with the stiffeners rather than independently of them. Together they are an area of 3962 mm² with a second moment of 9.01·10⁶ mm⁴ about the web's centreline, a radius of gyration of 48 mm over a buckling length of 900 mm — 0.75 of the depth, because the flanges hold the ends. That is a slenderness of 0.25, at which the column curve returns 0.98: the stub column reaches 98 per cent of its squash load, and the section's own strength is very nearly the whole answer. Stability

A column nine hundred millimetres long

The patch-load check asks how much of a web a flange can spread a wheel over, and answers in a plate-buckling reduction that throws seven tenths of it away. A pair of stiffeners does not improve that answer. It replaces the question with a different one, from a different family, with a different failure in it.

7 figures · Patch loading The same chord, buckling under a constant force and under a varying one. Two buckled shapes of the same 40 m compression chord on the same continuous restraint of 0.35 N/mm per mm, at the same scale. Under a constant force the buckle fills the member — 8 half-waves over 98 per cent of the length — and the critical force is 1881 kN, which is the length-free answer the closed form gives. Under the parabolic force a uniformly loaded deck delivers to it, the buckle LOCALISES: 5 half-waves over 56 per cent of it, gathered where the force is largest, and the peak force at buckling is 2113 kN. The shaded curve is the force distribution the second shape is buckling under. Stability

The buckle that will not spread out

A compression chord on a continuous restraint chooses its own number of half-waves and forgets how long it is. Give it the force it actually carries — a parabola, largest at midspan — and the number barely moves while the shape changes completely, which is the half that decides where the restraint has to be.

6 figures · Continuous restraint The curve the machine drew and the curve the material was on. The same tensile test twice. Force over the ORIGINAL area against extension over the original length is the engineering curve, which peaks at 431 MPa and 24.6 per cent and then falls. Force over the ACTUAL area against the natural logarithm of the length ratio is the true curve, which passes 538 MPa at the same instant and keeps rising. Nothing softens anywhere on this figure: the descending branch is the original area still being divided by, and the material is hardening the whole way. The two separate at the very first plastic strain and the gap between them is exactly e^ε, which is 1.25 at the ultimate load. Past that instant the deformation stops being uniform, so the engineering curve is drawn dashed: the real one falls faster than this, because the extension is now happening in one short length of the bar and the strain axis is still dividing it by the whole gauge. Materials

The curve was rising the whole time

Every tensile curve ever printed turns over and falls, and nothing about the material does. The fall is the original area still being divided by after the specimen has stopped having it — and correcting the two denominators turns a strength, a ductility and a failure into one exponent.

6 figures · Gauge length What a crack makes of a notch, against how sharp the notch is. The fatigue notch factor against the root radius, at a fixed elastic factor of 3 in a 430 MPa steel. K_t is a property of the shape and does not move along this axis at all — the dashed line — while the factor fatigue actually feels climbs toward it from below. At a 1 mm root the answer is 2.40, which is 30 per cent of the notch relieved; at 0.1 mm it is 1.38, and a notch that concentrates by 3 elastically is barely felt. Nothing has changed about the stress field: what has changed is that the peak is confined to a smaller volume than the material's own process size, so the crack starts against an average rather than against a maximum. Materials

The notch a crack does not feel in full

The elastic concentration factor is a property of shape and knows nothing about size, which is what makes it so useful and so misleading. A fatigue crack starts against an average over a volume the material owns, so two notches with the same factor and different radii have different fatigue strengths — and the stronger the steel, the less of that relief it gets.

6 figures · Stress concentration The fire is one curve, and the steel in it is several. Gas temperature and steel temperature against time in a standard fire. The gas curve is the same for every member in the compartment; the three bare steel curves are section factors of 76, 160 and 323 per metre, and they reach 558 °C at 17, 11, 8 minutes — a spread of a factor of two from geometry alone. The fourth curve is the middle section with 15 mm of board on it, which reaches the same temperature at 46 minutes. The kink near 735 °C on the bare curves is not a numerical artefact: steel's specific heat spikes there as its crystal structure changes, and the member spends several minutes absorbing heat at almost constant temperature. Materials

The temperature is a shape

Two members of the same steel in the same compartment, under the same fire and the same load ratio, fail eight minutes apart. Nothing about the material differs and nothing about the fire does. What differs is a perimeter divided by an area, and it is the only number in the whole calculation that a designer chooses.

6 figures · Fire The four corners of an opening, on the tee's own interaction diagram. The plastic interaction of the tee left above and below a 400 × 300 mm opening — every point on it computed by sweeping the plastic neutral axis through the section rather than from an interaction formula. Its squash load is 1519 kN and its plastic moment 32.2 kNm. At the working load the global moment puts 301 kN into each tee — compression above the hole, tension below — and the shear puts the same Vierendeel moment of 6.3 kNm into all four corners. The line is the path the demand takes as the load rises, and it reaches the surface at a load factor of 3.89 against 1.49 for first yield at one corner: the elastic check is finding one corner and the mechanism needs all four. Sections and stress

Four corners and a mechanism

The check on a hole in a web adds an axial stress to a local bending stress and compares the sum with the yield stress at one corner. What ends the opening is four hinges arriving together — and the gap between the two is a factor that varies along the span from two and a half to exactly one.

6 figures · Web opening Two webs of one box, carrying different amounts of the same load. The equivalent stress on each web of a 3.0 × 2.0 m box along the half span, under 40 N/mm at an eccentricity of 1500 mm, with the bending stress alone drawn for comparison. The two webs carry the same bending and opposite torsion, so one gets the sum and the other the difference: at the quarter point the loaded web is at 116 N/mm² and the far web at 11, a ratio of 10.9. A check made on the section rather than on the wall reports the average of two numbers that differ by that much. Sections and stress

Three actions on one web

A load applied off the centreline of a box girder is three actions at once, and every textbook decomposes it into them. What the decomposition does not say is that all three land on the same piece of plate — added on one web and subtracted on the other, so the two walls of one section differ by a factor of ten.

6 figures · Multicell torsion