Structural Engineering & Statics
where the load goes
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Which failure arrives first — page 12

Essays 265 to 288 of 495 on this thread, in the same order.
A plate welded to a loaded beam carries only the load that comes after it. The bending stress across the depth of a 457 mm beam coped 200 mm long and 50 mm deep, carrying a reaction of 300 kN, at the end of the cope, repaired with a doubler 8 mm thick on the web. With 60 per cent of the reaction already on the beam when the plate is welded, the original steel carries that share on its coped section and the rest on the repaired one. At the web's free edge it reaches 137 N/mm², against 164 as coped and 97 if the plate had been there from the start. The plate carries only the later load, 39 N/mm² at its outer edge. The repair has removed 40 per cent of the stress it would have removed on an unloaded beam. Connections

The plate that arrives after the load

A repair plate welded to a beam already in use carries only the load that arrives after its weld has cooled. The elastic checks remember that order — a doubler welded under 60 per cent of the reaction removes 40 per cent of the stress it would remove from an unloaded beam — while tear-out, a plastic mechanism, gets the whole of the repair. Propping the beam is what hands the plate the load already there.

 The humidity through a slab's depth, and the strain it asks for. A slab 200 mm thick drying through its top face only into air at 50 per cent, with its underside sealed by a steel deck. Left, the pore humidity against depth at 28 days — 50 per cent at the surface and 100 at the base; 90 days — 50 per cent at the surface and 100 at the base; 1 year — 50 per cent at the surface and 99 at the base; 5 years — 50 per cent at the surface and 74 at the base; 20 years — 50 per cent at the surface and 51 at the base. Right, the free strain each of those profiles asks for, taken as shrinkage at the local humidity. At 28 days the top wants to be 461 microstrain shorter than it was and the bottom 0; at 20 years the two are 461 and 456, and the gradient that produced the curl has gone. Not one of the profiles is straight, and a section that stays plane cannot deliver any of them. Materials

The slab that dries from one face

A creep coefficient is one number for a member, and a member drying through one face does not have one. Give every depth its own humidity and the section a strain profile it cannot deliver, and two things follow that no single coefficient contains — a six-metre slab lifts eleven millimetres at its edges with nothing on it, and its top surface is past cracking before it has been loaded.

 Where a tendon's force goes, over fifty years. The loss of stress in a tendon stressed to 1300 N/mm² and released at 7 days into a member of notional size 300 mm at 70 per cent humidity, with the three causes stacked. At 28 days the total is 61.9 N/mm²; at a year 131.5; at fifty years 190.8, which is 14.7 per cent of what the tendon started with. Creep supplies 107.7 of that, drying and autogenous shrinkage 50.6, and the steel's own relaxation 32.5. Half the loss has happened by 119 d and nine-tenths by 8 y. Materials

The prestress the member takes back

A tendon is stretched, locked off against the concrete, and then has to hold that extension while the concrete underneath it shortens by itself. Fifteen per cent of the force goes, most of it to creep, and how much goes is decided by the shape of the member and the air it stands in rather than by anything about the steel.

 One bracket, three neutral axes, three sets of bolt forces. A bracket 300.0 mm deep with three rows of two bolts at 75.0 mm pitch, carrying 15.0 kN·m about an axis in the plane of the bolts. Taking the axis at the group's centroid puts the outer rows at 50.0 kN of tension and 50.0 of compression, with the middle row idle. Taking it at the plate's compression edge puts every row in tension — 7.1, 14.3, 21.4 kN from the bottom up — with the top row at 21.4. Solving for it instead, with the plate bearing over 200.0 mm of width and the bolts as areas, puts it 40.2 mm above the edge and the top row at 24.6 kN. All three make the applied moment exactly. The top row differs between them by a factor of 2.33. Connections

The bolt group has no neutral axis

A bolt group carrying a moment about an axis in its own plane has some bolts in tension and something in compression somewhere else. Where that somewhere else is decides the answer, nothing in the group decides it, and the two defensible choices give the worst bolt 50 kilonewtons and 21.

 The coefficient is an envelope, and its scallops are whole half-waves. The plate buckling coefficient against aspect ratio α = a/b. Each faint branch is one half-wave count m: k = (m/α + α/m)², a curve whose own minimum is exactly 4 at α = m. The plate takes whichever branch is lowest, so the answer is the bold envelope — four touching 4 at α = 1, 2, 3 and 4, with cusps between them at α = √(m(m+1)) = 1.41, 2.45, 3.46, 4.47, where the plate is indifferent between m and m+1 half-waves. The first cusp reaches k = 4.50 and every later one is lower — 4.17, 4.08, 4.05. Past α = 1 the envelope never exceeds 4.49, which is why the length of a plate drops out of a formula that is otherwise entirely geometry. Stability

Four was never a fact about plates

The coefficient every plate calculation starts from is quoted as 4, derived nowhere and remembered by everyone. It is the minimum of a quantity that has nothing to do with plates in it — and what the plate supplies is not the four but the restriction that produces the scallops around it.

 The coefficient a web gets depends on where its neutral axis is. The plate buckling coefficient for an internal element against ψ = σ₂/σ₁, the ratio of the stresses at the two edges of the panel. Uniform compression is ψ = 1 and k = 4; a gradient running from compression to zero is ψ = 0 and k = 7.81; pure bending is ψ = −1 and k = 23.92, six times the value a column's flange gets. The 1800 × 12 mm web drawn here starts at ψ = −1.000 and k = 23.92 and ends at ψ = −0.910 and k = 21.63, because losing width from the compressed half drops the neutral axis and deepens the compression zone. The curve is steepest exactly where a bending web sits, so a small movement of the neutral axis costs 2.29 of coefficient. Stability

Classified by a gradient it does not have

A web in bending is the one plate whose buckling coefficient cannot be looked up. It depends on the stress gradient, the gradient depends on where the neutral axis is, and the neutral axis depends on how much of the web the coefficient has just taken away — so the answer is a fixed point, and the calculation everyone does is its first term.

 Two collapses for the same plate, and only one of them fits. The same three bolt rows at 90 mm pitch, folding two ways. On the left each row folds on its own pattern — 249 mm of hinge line round each bolt — and the patterns overlap, because 249 mm of fold cannot fit in a 90 mm pitch. On the right the plate does what it can actually do: the hinges run straight from one row to the next, and the whole group folds on 429 mm rather than 746. The arithmetic follows the drawing — 377 kN for the group against 657 for the rows added up. Connections

The rows have to share one fold

Each bolt row of an end plate is checked on its own, and the answers are added up. Two rows ninety millimetres apart cannot each fold the plate on a pattern two hundred and fifty millimetres long, because there is only one plate — so the group folds on one shorter pattern, and the sum was never available.

 The bolt force is the larger of two lines. What an M20 bolt in a 25 mm tee flange actually carries, against the tension applied to the flange. Preloaded to 171 kN it starts there and climbs at Φ = 0.185 — the bolt's own stiffness over the bolt's plus the clamped plates', 857 against 3781 kN/mm — so 18 per cent of every kilonewton applied reaches it and the rest is unloading the contact. At 138 kN the contact runs out and the line joins the one an ordinary bolt has followed from the start, climbing at 2.12. The two lines meet, so the strength is the same either way; what differs is the slope, by a factor of 11.5. The flange's own mechanism is at 172 kN, comfortably past the crossing. Connections

The bolt that was already stretched

Prying is a lever that needs the flange to lift before it can act, and a preloaded bolt does not let it. The bolt carries a fifth of every kilonewton applied until the plates part, and everything after that is the ordinary calculation — which is why preload changes the fatigue answer by a factor of a thousand and the strength answer by nothing at all.

 The cross term has the sign of the two contributions. Four responses of the floor, each combined three ways and divided by the complete quadratic combination at ρ = 0.50. Base shear: modal 927 and 839 kN, so the root-sum-square is 0.82 of it and the absolute sum 1.15. Base torque: modal -8640 and 8640 kNm, so the root-sum-square is 1.41 of it and the absolute sum 2.00. Flexible edge: modal 60 and -6 mm, so the root-sum-square is 1.05 of it and the absolute sum 1.15. Stiff edge: modal -5 and 47 mm, so the root-sum-square is 1.05 of it and the absolute sum 1.15. Where the two modes push the same way the root-sum-square is short; where they push opposite ways it is long. Dynamics

The twist the combination rule invents

Two modes close together respond together, and the square root of the sum of squares assumes they do not. The error has the sign of the two modal contributions: where they agree, as they do in base shear, the rule comes up short, and where they oppose, as they always do in torque, it comes up long — by a factor of five for a floor whose stiffness sits twenty centimetres off its mass, a torque the building does not have.

 The damping that stops helping. The storey drift's white-noise root-mean-square as the bearings' damping is raised from 2 to 50 per cent, each divided by the exact value at 2 per cent. The classical analysis promises that every increment helps, and at 50 per cent predicts 0.20 of the lightly damped drift. The exact analysis flattens: 0.34 at 20 per cent and 0.29 at 50, because the damping force at the bearings is transmitted into the superstructure's own mode, which the classical analysis has decoupled from it. Dynamics

The damping that belongs to no mode

Give every mode its own damping ratio and throw the rest of the damping matrix away, and an isolated building's periods and damping come out right to within a fifth of a per cent. Its storey drift at the superstructure's frequency comes out six times too small. The bearings' dashpot pushes on both modes at once, and the more of it there is the less extra damping buys: the classical analysis promises the drift keeps falling, and it stops.

 The push that tips it, in every direction. The tipping push in each plan direction, drawn as a distance from the centre, for a 6000 kN body with its weight 15 m up and the push at 18 m. On rigid ground the curve is the hull's: weakest along the axes at 1333 kN. On four equal pads it shrinks, and more toward the corners. With the pads as built — stiffnesses 20000, 20000, 8000, 20000 kN/m — it is no longer symmetric: the weakest direction is 15°, on the soft pad's side, at 1039 kN. Equilibrium

It tips inside its own hull

On rigid ground a body tips when its resultant reaches the edge of its base, and how stiff its supports are has nothing to do with it. On pads that settle, the body leans as it is pushed, the lean moves its weight, and the push that tips it falls by one number a site engineer already has: settlement times the height of the weight, over the square of the half-width. Toward a corner the loss doubles, and one soft pad makes the weakest direction one nobody checks.

 Three routes to one deflection, and the one that is wrong. Two 2500 mm bars of 250 MPa proof stress meeting at a loaded apex, the drop of the apex against the load. The line is geometry: each bar's extension from its own stress, divided by the sine of its slope. The dots are the derivative of the total complementary energy with respect to the load, and lie on it. The dashed line is the derivative of the strain energy — Castigliano's theorem applied to a material that is not linear — which leaves the truth by ten per cent at 99 kN and at 170 kN gives 308 mm for a deflection of 46.0. Deflection

The other area under the curve

Castigliano's theorem says a deflection is the derivative of the strain energy with respect to the load, and it is true only while the material is linear. Past that, the right energy is the area on the other side of the stress–strain curve. On two aluminium bars at their proof stress the strain energy gives a deflection four times too large, and on a redundant truss minimising it picks a set of forces in perfect equilibrium that no deformed shape can produce.

 Settling down or walking away, cycle by cycle. The total plastic hinge rotation of the beam after each cycle of loading — span 1, both, span 2, neither — with the midspan load at 0.98, 1.02, 1.05, 1.10 times the shakedown load of 126.3 kN. At 0.98 it stops at 0.59 mrad. At 1.02 it grows 4.57 mrad a cycle. At 1.05 it grows 11.43 mrad a cycle. At 1.10 it grows 22.86 mrad a cycle. Nothing collapses in any single cycle; above the shakedown load the beam walks. Materials

The load it can carry once

A two-span beam whose loads come and go span by span collapses at 150 kN under any one arrangement, and walks at 127. Between the two it can carry every arrangement once and none of them forever: each cycle leaves a few more milliradians of rotation at the support and a midspan fifteen millimetres lower. Melan's theorem finds the limit as the last residual moment line that fits, Koiter's as a mechanism no single load state can drive, and a cycle-by-cycle calculation walks exactly where both say it will.

 Capacity against load direction, layout by layout. The same capacities as a function of direction on straight axes, for a load through (150, 150) mm. Three rows of two is weakest at 138°, 135.8 kN. Two rows of three is weakest at 312°, 135.8 kN. A ring of six, r = 70.4 mm is weakest at 132°, 150.3 kN. The curves cross: no layout is strongest in every direction, and the one to choose is the one whose lowest point is highest. Connections

The bracket pushed from the wrong side

A six-bolt bracket checked for a load straight down carries 198.5 kN. Push the same load through the same point at 138 degrees and it carries 135.8 — the direction changes the torque as well as the shear, and the worst direction is one no drawing shows. The capacity of a group whose load can turn is a closed curve, the bolt that governs it changes as the load turns, and what the curve rewards is not a larger polar moment but a smaller distance to the furthest bolt.

 The two methods, direction by direction. Capacity against the direction of the load for three rows of two bolts of 100 kN each, loaded through (150, 150) mm. The elastic method bottoms out at 135.8 kN at 138°; the instantaneous centre at 168.3 kN at 136°, 23.9 per cent higher. Both spike to 600.0 kN where the load aims at the centroid and every bolt carries an equal share. The gap between the curves is not constant: it is widest where the elastic distribution is most uneven, which is the same place the group is weakest. Connections

The better method flatters the worse layout

The instantaneous centre method finds capacity the elastic method cannot see, and how much it finds depends on the layout it is applied to. On a six-bolt rectangle under a load of unknown direction it is worth 24 per cent; on a ring of six with a smaller polar moment it is worth 14. The two methods agree on which layout to choose and disagree by a factor of five about the margin — and the reason is that a layout the elastic method already likes is one with nothing left to redistribute.

 What each missing bolt costs. The weakest-direction capacity of six bolts at 75 by 75 mm, loaded through a point (150, 150) mm from the centroid, with the whole group and with each bolt in turn left out. The full group carries 135.8 kN. Leaving out bolt 3 leaves 90.2 kN, a loss of 34 per cent; leaving out bolt 2 leaves 131.1 kN, a loss of 4. One sixth of the bolts is not one sixth of the capacity, and which sixth it is matters by a factor of 10. The dashed line is the capacity a group that lost a proportional share would have, 113.2 kN. Connections

The bolt that was never fitted

A six-bolt bracket found with five bolts in it has lost a sixth of its fasteners and between four and thirty-four per cent of its capacity, depending which one is missing. The share is the smallest of the three things that changed: the centroid moves away from the gap, which lengthens the load's own lever arm, and the polar moment falls by more than the count does. The bolt whose absence costs most is not the bolt that governed the check.

 An unsymmetric frame, and a section with a product of inertia. A portal of 8.0 m span whose columns are 5.0 m and 3.0 m, carrying 10.0 kN/m on its beam, and beside it the analogous column — the frame's own centreline at a width of 1/EI. The section is no longer symmetric about a vertical line, so it has a product of inertia of −18.4 against Ix = 31.6 and Iy = 168.0, and its principal axes are tilted 7.54° from the horizontal — the dashed pair through the elastic centre. A section with a product of inertia does not bend about the axis it is loaded about, so the thrust and the redundant shear come out of a two-by-two rather than out of two divisions. Deflection

The axes that have to be turned first

Make one column of a portal shorter than the other and the analogous column stops being symmetric about a vertical line. It acquires a product of inertia, its principal axes tilt seven and a half degrees, and the thrust and the redundant shear stop being two separate divisions. Using the elastic centre and nothing else — which is what the symmetric construction looks like from outside — reports 17.0 kN·m at the left foot where the frame carries 27.8.

 A frame bent by nothing at all. A portal of 8.0 m span and 5.0 m columns at EI = 20000.0 kN·m², with its right foot settled 10.0 mm and no load on it anywhere. The moment diagram is drawn on the members: −3.95 kN·m at the left foot, −3.95 at the left knee, −0.00 at the crown, 3.95 and 3.95 on the right. The settlement is drawn hugely magnified; at true scale it is 10.0 mm on an 8.0 m frame. The diagram is antisymmetric, the vertical force the settlement develops is 0.99 kN, and every one of those numbers is proportional to EI. Deflection

The load that is not a load

Settle one foot of a portal frame by ten millimetres and the frame develops moments with nothing applied to it anywhere. In the analogous column the case is simpler than a load case, not harder — the section carries no direct stress at all, and the whole answer is one bending stress. And it scales the wrong way: the moments are proportional to EI, so the stiffer the frame, the more a settlement costs it.

 Half the life is spent growing the first half-millimetre. The crack length against time for a detail starting with a 0.50 mm flaw under 800 cycles a day of the same five-band traffic the S-N calculation used, integrated by Paris's law. It reaches 1 mm after 91.2 years, 2 mm after 131.3, 10 mm after 172.9 and its critical length of 125.7 mm after 195.8. The curve is nearly flat and then nearly vertical, because the rate goes as the cube of ΔK and ΔK goes as the square root of the crack: the crack spends most of its life being too small to find and the rest being too large to ignore. The dashed lines are the crack lengths at which each band of traffic starts doing anything at all. Materials

The loops a crack grows on

A Miner sum reduces a hundred years of traffic to a number and throws away everything below the cut-off — on this bridge, a third of the vehicles doing exactly none of the damage. Integrate the same spectrum as a crack instead and that third grows thirty-seven per cent of the crack, because a cut-off is a statement about a constant-amplitude test and a crack's threshold is a length rather than a stress. The two calculations disagree about the life by a third and about which vehicles matter entirely.

 The same cycles, and two different lives. Two sequences made of exactly the same cycles — blocks of 6.00M at 40 N/mm² alternating with blocks of 0.22M at 90, on a detail starting with a 0.50 mm flaw. Taking the small cycles first, the crack reaches its critical 87.7 mm after 12.36M cycles; taking the large ones first, after 6.36M. Miner's sum at the moment of failure is 0.99 for the first and 0.68 for the second, so a rule that predicts failure at a sum of one is right to within a per cent about the first and 48 per cent unconservative about the second. The mechanism is on the axes: ΔK rises with the crack, so a large block met late finds a longer crack and does more with it. Materials

The record played backwards

Miner's rule adds damage, and a sum has no order. A crack does, twice over: a large block met late finds a longer crack and does more with it, and an overload leaves a plastic zone that slows everything after it. The same cycles rearranged fail at 12.4 million or at 6.4, and a Miner sum that is right to one per cent about the first is out by half about the second. One cycle in four million can add fifty-four per cent to a life.

 The magnification a crack sees is a ratio, not a depth. The stress magnification at a weld toe against the crack's depth as a fraction of the plate's thickness, on BS 7910's two-branch fit. It is a function of a/t alone, because a weld's own size scales with the plate it is on, so the elevated field is geometrically similar. The dots are the same absolute starting flaw of 0.20 mm in plates of 12, 16, 25, 40, 60, 80, 100 mm: the flaw does not move and its magnification runs from 1.81 to 3.50. A fixed flaw in a thicker plate is a smaller fraction of it, which puts it deeper inside the raised field rather than nearer the edge of it. Materials

The rule that points sideways

Every fatigue code puts the same detail in a thicker plate into a lower category, by a factor of (25/t) to the power 0.2, and explains nothing. It is a strange rule: a detail's strength made to depend on a dimension at right angles to the crack. Integrate a crack through a weld toe's own stress field and the rule falls out — same form, same sign, and an exponent of 0.13 against the design code's 0.2. Remove the toe's magnification and the effect reverses.

 The two knees the environment takes away. The design S-N curve for a category 71 detail and the two shapes a corrosive environment leaves. In air the slope is three to the constant-amplitude limit at 52.3 N/mm², five to the cut-off at 28.7, and nothing below it. With the cut-off removed the second branch continues. Under free corrosion there is one slope of three all the way down and no knee at all. The faint lines are the five bands of the traffic: three of them sit below the cut-off — an ordinary lorry, a van and a car, 90 per cent of the crossings — which is why they do nothing to a detail in air and something to every other curve on the figure. Materials

The cut-off belongs to the water

An S-N curve's cut-off is the most consequential thing on it: on a category 71 detail under ordinary bridge traffic it deletes ninety per cent of the crossings and leaves two bands doing all the damage. It is a property of steel in air. A crack tip that seawater or de-icing salt can reach has no threshold, no endurance limit and no knee, and the same bridge's life runs from 300 years to 45 depending on which of six defensible calculations is asked.

 Bearing damping that shakes the roof harder. The roof's white-noise root-mean-square acceleration as the bearings' damping rises from 3 per cent to 58 per cent, exact and classical, both divided by the exact value at 3 per cent, for the six-storey building. The classical analysis falls all the way, to 0.33. The exact one falls to a minimum of 0.44 at about 28 per cent and then rises, to 0.50 at 58 per cent: past the minimum, every extra per cent of damping at the bearings shakes the top floor harder, while the classical analysis says it is still helping. Dynamics

The top floor the bearings shake

On a building of several storeys, the damping at the isolation bearings is tied less tightly to each higher structural mode than to the one below. It still does more harm in each, because the ground barely reaches those modes on its own. The error the usual analysis throws away gathers in the upper storeys, and past about a third of critical damping at the bearings the roof shakes harder while the analysis says it is still helping. How much damping is best depends on how tall the building is.

 The same mode's curvature, which can. The first mode's drift in each storey — the difference between the floors above and below it, which for a shear building plays the part a beam's curvature plays — intact, solid, and with storey 3 10 per cent less stiff, dashed, each scaled to the roof. The damaged storey's drift rises by 9.6 per cent; the storeys above it change by at most 1.8 and below it by at most 1.5. The largest change is in storey 3: the drift finds the damage and says where it is, which the frequency's fall of 0.91 per cent cannot. Dynamics

What a mode shape notices that a frequency does not

Take a tenth of the stiffness out of one storey of a ten-storey building and its first frequency falls by a per cent at most, and by almost nothing if the storey is near the top. The mode shape looks unchanged, its match to the old one 0.99997. But the mode's storey drift — its curvature — rises by about a tenth in the damaged storey and hardly anywhere else, wherever that storey is. The instrument that sees the damage is the one almost nobody installs.

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