Structural Engineering & Statics
where the load goes
The collection

Every essay — page 17

Essays 385 to 408 of 728, in the same order.
The worst section of a haunched rafter is inside the haunch. Utilisation along a 15.3 m portal rafter carrying 8 kN/m, with an eaves moment of 500 kNm and an apex moment of 150, haunched over 3 m from 906 mm deep down to the rafter's own 453. The moment is largest at the eaves and the depth is largest there too, so the eaves is at 0.42; the apex is at 0.31. The peak is 0.46 at 2.98 m — the haunch tip, where the section has just become the bare rafter and the moment is still 226 kNm. The dashed curve is the same rafter with no haunch, which reaches 1.02 and does not pass. Sections and stress

The section that governs is inside the haunch

A tapered cantilever has its worst section somewhere along it because the moment grows linearly and the modulus quadratically. A haunched rafter has the same competition with a step in it, and the step is where the check lands — at neither end of the member, at a station no formula names.

6 figures · Tapered member What a damper at the anchorage can do, and the ceiling it cannot pass. Modal damping against damper size for a 200 m stay at 4500 kN, with the damper 4 m from the anchorage — 2.0 per cent of the length. Each curve is a mode, found as a complex root of the taut string with a viscous damper in it rather than from a formula. Every one of them peaks at 1.00 per cent of critical, which is x/2L exactly, and the peaks are at different damper sizes — a higher mode wants a smaller damper, because it moves faster at the same amplitude. The curves are flat near their peaks: half the optimum coefficient gives 80 per cent of the ceiling, and so does twice it. Dynamics

The damper that is too near the end

A stay cable has almost no damping of its own, so it is given a damper — and the damper cannot go where the motion is, because the middle of a two-hundred-metre stay is a hundred metres above the road. What it can supply is then decided by one length, and no amount of damper changes it.

5 figures · Cable dynamics Damping that is computed rather than measured. Radiation damping of a 2.5 m block on soil with a shear wave speed of 178 m/s, against how heavy the block is made — the horizontal axis is a multiple of the 150 tonne block drawn. At that mass the three modes are at 47, 29 and 13 per cent of critical. None of this is dissipation: the energy leaves as waves travelling away into the half-space, so the quantity is a geometrical coupling and it can be computed from the size of the footing, the density of the soil and its shear modulus. Every curve falls as the block gets heavier, because a heavier block moves less for the same wave field — which is the one counter-intuitive thing here: mass buys frequency and costs damping. Dynamics

The damping that is radiated

Every response in this field is quoted with a damping assumption attached, because damping is measured rather than designed and the measurement has a factor of two in it. A machine block on the ground is the exception: its damping is not dissipation at all, and it can be computed from three numbers none of which is a material property of anything that dissipates.

5 figures · Vibration isolation Where two buildings touch, and what is there when they do. Closure between an 8-storey building 24 m tall and a 4-storey one 15 m tall, drawn against height, with the 50 mm gap between them. The two swaying out of phase close on each other more the higher up they are, so they first touch at 7.6 m and are in contact above it. The dots are the taller building's floors: its storeys are 3.00 m and the other's are 3.75, so two of them arrive part way up a column rather than at a slab. A blow at mid-height of a column asks it for a shear of half the impact and an end moment of Fh/8, neither of which is a demand any part of the design contains. Dynamics

The floor that arrives at a column

The gap between two buildings is computed from their roof displacements, which is where each of them moves most. It is not where they touch, and it is not what is there when they do — a slab edge meeting a column part way up its height is a different event from two slabs meeting, and it is the one that appears in the photographs.

5 figures · Pounding The same bridge by the two theories it might have been designed by. Deck moment along a 500 m suspended span with half of it loaded, computed twice. Elastic theory treats the deck as a beam and applies the cable's extra tension as an upward load: 80.1 MN·m. Deflection theory keeps the cable's total tension acting on the deck's own deflected shape — a geometric stiffness — and returns 56.3, which is 30 per cent less. The extra cable tension is very nearly the same in both (6.20 against 6.11 MN), so nothing about the cable is in the difference: it is entirely the H·v″ term the older theory drops. Structural form

The tension that was left out

A suspension bridge's deck sits on a cable pulling hard along it, and a member with a large tension in it is stiffened by that tension. Leaving the term out of the deck's own equilibrium is what elastic theory does, and on a long span it asks for fourteen times the girder.

5 figures · Stiffening girder What the second, third and fourth arms are worth. Top drift removed against the number of outriggers, each arrangement at its own optimum levels, on a 40-storey core 200 m tall. One arm at 59 per cent of the height removes 82.3 per cent of the drift. A second, with both moved to 35 and 71, takes it to 91.6 — a gain of 9.3 points, which is half of what was left. The third is worth 3.0 and the fourth 1.5, and each one costs a storey of the building's most valuable height. Structural form

What the second arm is worth

One outrigger at its best height removes five sixths of a tall core's drift, which sounds like the end of the argument. A second removes half of what is left, a third half of that, and each of them costs a storey of the most valuable floor area in the building — so the question is not where to put an outrigger but how many the arithmetic still justifies.

6 figures · Outrigger Two restraints, and only one of them cares how much was imposed. Crack width against the restrained strain, for the same 20 m wall restrained two ways. Edge restraint — a wall cast on a base — gives a width proportional to the strain, because the concrete has to accommodate the movement and the cracks are where it does. End restraint — a bay cast between two that have hardened — gives 0.32 mm at every strain on the axis, because the crack opens only until the steel can push the cracking force back into the concrete, and that force is a property of the section. At the 304 microstrain this wall is asked for, the two are 0.17 and 0.32 mm, and the movement is divided into 35 cracks and 19. Internal forces

The bay that is cast last

A wall cast on a base cracks in a way the steel controls: more movement gives wider cracks, and reinforcement decides how many. A bay cast between two walls that have already hardened cracks in a way that has nothing to do with how much movement there was — and the threshold that separates a controlled crack from a single wide one is a quantity of steel rather than a limit on anything.

5 figures · Restraint cracking Two load sets with the same resultant and different work. The two ways of putting a uniform load of 10 kN/m onto a beam element 6.00 m long. Both put 30.0 kN at each node, so both have the same resultant and the same moment about any point — they are equivalent for a rigid body. The consistent set adds a couple of 30.0 kN·m at each end, in opposite senses, which is what makes it do the same virtual work over the element's shape functions as the real load does. The couples cancel in the resultant, which is exactly why the resultant cannot see them, and they are the whole difference between an exact answer and one that is a third out. Equilibrium

Equivalent in work, not in resultant

Two force systems with the same resultant and the same moment about every point are interchangeable — for a rigid body. A finite element is not a rigid body, and substituting one for the other on a beam element leaves the tip of a cantilever a third too low with no warning of any kind.

5 figures · Force couple The part of a base plate that is delivering anything. A plan of the plate: the 260 × 260 column in the middle, and the shaded region it can reach — its own outline grown by 43 mm, which is t√(f_y/3f_jd) for a 20 mm plate on grout at 20 N/mm². That is 78176 mm² of a 200000 mm² plate, or 39 per cent of it. The corners are outside it and are carrying nothing: a plate cantilevers from the column's perimeter and runs out of bending capacity at c, so making it larger in plan beyond that changes nothing at all, and making it thicker changes everything. Connections

The plate that is only as big as it is thick

A base plate's design is presented as a bearing calculation: an area, a bearing strength, and a check that the pressure fits. The area in that calculation is the plate, and the plate cannot deliver it — a 20 mm plate on a 500 square reaches 43 mm past the column and the corners carry nothing at all.

5 figures · Base plate Force and capacity round a weld group, which do not vary together. Utilisation round the c shape group under 100 kN at 150 mm, walked from one end of the weld to the other, with the force scaled onto the same axis for comparison. The capacity is 1460 N/mm where the force runs along the weld and 1789 where it runs across it — √1.5 more — so the utilisation is not simply the force in different units. The worst point is at 47 degrees to the weld, where the capacity is 1612 N/mm, and the group carries 182 kN against the 165 the along-the-weld value would allow. Connections

The weld that is stronger where it is pulled

A fillet weld pulled across its axis is √1.5 stronger than the same weld pulled along it. On a group under an eccentric load the direction of the pull varies from point to point, so the capacity does too — and the gain that follows is worth twenty per cent at one eccentricity and nothing at all at another.

5 figures · Weld strength Where the span moment and the support moment cross. Span moment and support moment against the joint's stiffness, for a 6 m beam under 30 kN/m. They move in opposite directions because they add to a constant — the simple-span 135 kN·m is fixed by statics and the joint only decides how it is split. They cross at 68 kN·m, where the joint is delivering 75 per cent of the fixed-end moment, and neither the crossing nor the fraction depends on the beam, the span or the load: it is the point where f·wL²/12 equals wL²/8 − f·wL²/12, which is f = 0.75 for every beam there has ever been. Connections

The joint that was chosen

A joint's stiffness decides how a beam's moment divides between its span and its supports, and the two add to a constant. So there is a stiffness at which they are equal, the beam is sized by the smaller of two numbers rather than the larger of one, and the design moment is half what a simple connection leaves behind.

5 figures · Joint classification The deflection a building can take, and which way it is bending. The limiting deflection ratio — the sag or hog of a building divided by its length — at a critical tensile strain of 0.075 per cent, against how long the building is compared with its height. Two curves, and the gap between them is the whole finding: hogging is worse than sagging by exactly 2.0 times, because the neutral axis of a hogging building sits near the bottom and the tension face is the full height of the wall. The minimum is at L/H = 1.6, where the two mechanisms cross: a squat building cracks diagonally in shear and a long low one cracks in bending at the extreme fibre. Deflection

The crack is a strain, not a slope

Angular distortion is what every table of settlement limits is written in, and it is a proxy. What cracks a building is a tensile strain in it, and reading the building as a deep beam says which of two mechanisms supplies it, why hogging is exactly twice as damaging as sagging, and why a horizontal stretch of a few millimetres takes half the capacity away before the settlement starts.

5 figures · Differential settlement A portal on a stepped base — the two passes added. Bending moments on a single-bay portal with columns of 5 m and 3.5 m under one horizontal beam of 9 m, carrying 10 kN/m down and 60 kN across, with fixed bases. This frame is the two passes added. The corner moments are 5.8 and 99.1 kNm, and the short column's top carries 17.01 times what the tall one does. The diagram is drawn on the tension side of each member. Deflection

Every joint balanced, and the frame still leaning

Moment distribution enforces one equation per joint, and a frame free to translate has one more equation than it has joints. So a table that balances perfectly can describe a structure held up by a prop nobody built — and finding the prop, then removing it, is a second pass whose unknown is a distance rather than a rotation.

10 figures · Moment distribution Four things a far end can be doing, and what each is worth. The rotational stiffness of a member at one end, for four conditions at the other, each drawn as the shape the member takes when the near end is rotated through one unit. They are the same expression evaluated four times — M = (2EI/L)(2θ_a + θ_b) — and the only thing that changes is what the far end is known to be doing. A held far end gives 4EI/L and carries over a half; a free one gives 3EI/L and carries over nothing; a far end rotating equally and oppositely gives 2EI/L and carries over minus one, which is what a symmetric structure does to a member crossing its axis; and a far end rotating equally and in the same sense gives 6EI/L and carries over one. None of the four is an approximation. Each removes a freedom that was going to be discovered by iteration. Deflection

Told what the far end is doing

Moment distribution discovers, cycle by cycle, that the pinned end of a beam carries no moment — a fact known before any arithmetic started. Telling it instead changes one stiffness from 4EI/L to 3EI/L and the work from thirty numbers to eight, for the identical answer. Cutting the beam on its own axis of symmetry gets it in two.

8 figures · Moment distribution Every number in the table is a rotation, and none of them is a moment. The rotation contributions of a three-span beam of 8, 10, 8 m under 24 kN/m, sweep by sweep. There are six of them, one per member end, and not one is a bending moment: the moment is assembled at the end from M = FEM + 2m′ + m′ of the far end, and until that is done the table holds quantities that mean nothing on their own. That is the trade. A moment distribution stopped after two cycles hands over moments that are wrong by a known amount and can be used; this table stopped after two sweeps hands over nothing that can be read at all — and it gets there in four sweeps against Cross's own count on the same beam, writing six numbers a sweep rather than one per distributed member end plus a carry-over. Deflection

The table that cannot be read halfway

Kani's method converges at exactly the rate moment distribution does, sweep for sweep and digit for digit, because it is the same iteration. What it changes is what is written in the boxes — rotations rather than moments — and that buys a shorter table that repairs its own mistakes and cannot be stopped early.

9 figures · Moment distribution The whole line works, and the axis is not a choice. A weld group 200 mm deep carrying 15 kNm about an axis in its own plane, together with 100 kN of vertical shear. The bending force per unit length runs linearly from 1125 N/mm at one extreme to the same the other way at the other, through zero at the group's own centroid — and the centroid is where the neutral axis is because a weld carries compression across its throat as readily as tension. The 250 N/mm of shear runs along the weld and is uniform over the whole 400 mm, so the two components are perpendicular to one another and are combined on the throat rather than added as vectors in the plane. Connections

The eccentricity at right angles to the drawing

A bracket's load stands off the plane of its welds as well as being offset within it, and the second eccentricity produces a completely different object — bending about an axis through the group rather than torsion about a point in it. The weld line has no compression zone to argue about, so its neutral axis is its own centroid, and the whole length works.

7 figures · Weld group Three fasteners, three completely different clocks. What each of three shear fasteners carries against how far the joint has moved. A 400 kN fillet weld is linear to 0.4 mm and then gone — it is stiff and it is not ductile. A 380 kN bolt in a hole 2 mm larger than itself carries nothing until the hole closes and then rises over several millimetres of hole elongation. A 250 kN preloaded bolt is at its slip resistance in under a tenth of a millimetre and holds it until it slips. Two of these in one joint are at the same displacement, so the one that gets there first carries the load — and the weld gets there 23 times sooner than the bearing bolt does. Connections

Two fasteners that never arrive together

Every steel code forbids adding a weld's capacity to a bolt's in one shear joint, and states it as a rule rather than deriving it. It is derivable. Two fasteners in parallel are at the same displacement rather than the same force, and a weld has ruptured at four tenths of a millimetre while a bolt in a standard hole has not yet touched the side of it.

7 figures · Weld group Two ways for the same plate to fold. An end plate 200 mm wide with a bolt 45 mm from the web face and 55 mm from the edge, and the two families of yield line it can collapse along. The circle closes round the bolt and is 283 mm of hinge — a circle round the bolt. The straight pattern runs out to the plate's free edges and is 249 mm — hinges to the plate edges. The plate folds along whichever is cheaper, which here is the fan, and the 200 mm that comes out is the length of the equivalent tee stub — a dimension that is nowhere on the plate and is 100 per cent of its width. Connections

How much of the plate is bending

A tee stub is an object nobody builds, and the whole component method rests on replacing a real end plate with one. The length of the substitute is not a dimension of the plate — it is the length of the cheapest fold the plate can collapse along, and two families of fold compete for it on a criterion with no strength in it at all.

7 figures · Prying Three quantities converging at three different rates. How much of the exact answer a truncated modal analysis of a ten-storey building reaches, against how many modes it keeps. The mass count is the rule — ninety per cent, reached at two modes. The base shear is ahead of it: one mode carries 85 per cent of the mass and 98 per cent of the base shear. The force in the top storey is behind it, at 80 per cent on one mode and 94 on two. The rule is written in the quantity that converges fastest, and it is checked against a quantity nobody plots. Dynamics

The modes that were left out

Nobody runs every mode a model has, and the rule for how many is a mass count — enough to account for ninety per cent of the structure. The rule is written in the one quantity that converges fastest. On a twenty-storey frame two modes give the base shear to within one per cent and the force in the top storey to within twenty.

7 figures · Mode shapes Two capacity curves that agree, from two answers that do not. Base shear against roof drift for the same eight-storey frame with a soft ground storey, pushed with a fixed triangular pattern and with one recomputed from the tangent stiffness at every step. At a roof displacement of 426 mm they differ by 8.7 per cent — which is the number a capacity-spectrum procedure reads, and it is the number that is nearly the same. The storey drifts underneath these two curves differ by a factor of 3.8 at the fourth floor, and the drift is what the analysis was run for. Dynamics

The pattern that stopped describing the building

A pushover analysis pushes with a load pattern chosen to resemble the first mode, and by the time the structure has done anything worth analysing it no longer has that mode. Recomputing the pattern as the frame softens changes the base shear by nine per cent and the drift at the fourth floor by a factor of four.

6 figures · Pushover Both checks pass, and the contact lets go. A contact pressed together by 1000, with μ = 0.4. Every tangential force the contact can supply lies inside a disc of radius μN = 400.0, because the friction law bounds the length of the force and not its components. The contact is asked for 300.0 one way and 300.0 the other way — 75% and 75% of the radius taken one at a time — and 424.3 together, 106% of it. Each one-direction check passes and the force does not fit: the square those checks describe reaches √2 times further at its corners than the contact can. Equilibrium

Seventy-five per cent each way

A contact asked for friction in two directions at once can supply a force of a certain length pointing any way it likes, so its limit is a disc and not a square. Two checks made one direction at a time, each passing at seventy-five per cent, describe a contact that has already let go.

7 figures · Friction The pier grips, gives, and grips again. A sliding bearing carrying 3000 kN on a pier head of 20 kN/mm, dragged by a deck expanding at 1.7 mm an hour, with a static coefficient of 0.05 and a kinetic one of 0.03. The force in the pier climbs while the bearing grips, reaches 150 kN, and falls in a fraction of a second to 30 kN: the pier springs back under only the kinetic friction, overshoots the 90 kN that friction would hold it at, and grips again. The swing is 120 kN — 2.00 times the 60 kN between the two coefficients — and the pier head jumps 6.00 mm each time, three times in 12 hours. Equilibrium

The pier that moves in jumps

A sliding bearing whose static friction is larger than its kinetic friction does not release a slow thermal movement as a drift. It grips, gives and grips again, and each time the force in the pier swings by twice the difference between the two coefficients — whatever the pier is made of.

7 figures · Friction Same deck, same load, and two pier forces. A deck bearing on a pier of 20 kN/mm with μ = 0.03, taken to the same final state two ways: the deck moves 4 mm over the pier, and the bearing's load rises from 2000 to 4000 kN. Moved first, while the bearing carries 2000 kN, the pier force reaches the limit of 60 kN and the bearing slides for the rest of the movement; the load arriving afterwards raises the limit and changes nothing, and the pier is left carrying 60 kN. Loaded first, the limit is 120 kN before the deck moves, the bearing grips throughout, and the pier carries 80 kN. Both states are at the same displacement under the same load, and both satisfy equilibrium and the friction bound; the order is the only difference, and it appears in neither. Equilibrium

The order the loads arrived in

Statics allows a contact with friction a whole range of forces and has no way to choose between them. A real structure does choose, and what it chooses by is the order in which things happened to it — so the force in a pier under a sliding bearing is a record of its history, not a function of its loads.

7 figures · Friction The resultant has left the base, and it tips. A body on three supports weighing 60 kN, pushed sideways by 16 kN at a height of 3 m in the plan direction 270°. The push moves the resultant of weight and push 0.80 m from under the weight, and the base — the convex hull of the supports, shaded — lets it go 0.75 m that way before the edge drawn heavy becomes a tipping line: a factor of 0.94, found both along the ray and by moments about that edge. The dashed rosette is the same reach in every direction, from 0.75 m toward the middle of the nearest edge to 1.50 m toward the furthest support. To hold it the support opposite the tipping edge would have to pull 1.3 kN, which a support standing on the ground cannot do, so it lifts and the body turns about that edge. Equilibrium

Half as far between the legs

A body standing on feet, legs or pads has for its base the polygon its supports enclose, and how far its weight can be pushed before it tips depends on which way it is pushed. A three-legged stand pushed toward the gap between two legs has exactly half the reach it has pushed toward one of them.

7 figures · Overturning