Structural Engineering & Statics
where the load goes
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One support too many — page 4

Essays 73 to 96 of 186 on this thread, in the same order.
Five millimetres short, and a hundred kilonewtons in every member. An X-braced bay 6 m by 4 m in which one diagonal was fabricated 5 mm short, with the force that leaves in every member. Nothing is applied to this frame. The forces are the self-stress state the frame's one redundancy supports, scaled so that the diagonal is pulled back to the length it should have been: tension in both diagonals at 100 kN, compression in the four members round the outside, and the whole set in equilibrium with nothing. That is 25% of the force the diagonal was sized to carry, and it is there for the life of the structure. Take one diagonal out and the frame becomes determinate: the short member then simply puts the joint somewhere else, and the structure is in the wrong place instead of under stress. Redundancy is bought, and this is the price. Connections

Built to the wrong length

A redundant structure's members do not have independent lengths. Choose all but one and geometry decides the last, so a member made a different length has to be pulled or pushed into place — and the force required stays in the structure for as long as the structure does. Nothing has been applied to it, there is no load case and no factor, and the members are carrying real force.

 Restraint is a fraction, and the length decides how far up it reaches. A 20 m wall 3.0 m high cast against a base that has already hardened — a length-to-height ratio of 6.67. The base holds the bottom of the wall at R = 0.50 and the top of it at 0.304, decaying as 0.609 to the power of the height in wall heights. The free contraction is 380 microstrain, of which 84 per cent is the wall cooling from its own hydration peak and the rest is drying; the concrete's own strain capacity is 50. Everything to the right of the dashed line cracks, which here is the bottom 3.00 m of it. Nothing has been loaded. Internal forces

The steel decides how many, not how much

A wall cast on a base that has already set cools, tries to contract, and is not allowed to. What follows is not a stress problem with a strength on the other side of it. The movement is going to happen; the only question the reinforcement gets to answer is how many pieces it is divided into.

 The web down the middle carries no torsion at all. A 2-cell box 3.0 m wide and 1.5 m deep under 4000 kNm of torque. Statics gives one equation, T = 2 Σ q_i A_i, and there are 2 unknown flows — so the section is torsionally redundant and the missing statements are that every cell twists by the same amount. Solving that system gives 444.4 and 444.4 N/mm in the cells, so the internal web carries 0.00 N/mm — 0.00 per cent of the outer wall's. J is 2.3679e+12 mm⁴ against 2.3679e+12 for the same outline with no internal web at all, a ratio of 1.000000. The same walls slit open would give 7.413e+10, so closure is worth 32 times and the internal web is worth what the ratio says. Sections and stress

Two cells, one equation, and a web with nothing in it

Bredt's formula answers a single closed cell because a single closed cell has one unknown and one equation. Put a web down the middle and there are two unknowns and still one equation — and the answer, when the missing statement is supplied, is that the new web carries exactly nothing.

 Two answers added, and the answer to the two together, drawn on top of each other. A 8 m beam under a 60 kN point load at mid-span (152.38 mm), under 12 kN/m of uniform load (152.38 mm), and under both at once (304.76 mm). The sum of the first two is 304.76 mm, and the residual between it and the third is zero — not small, zero, to the last bit of the arithmetic. That exactness is not a numerical accident: the governing equation is linear in the load, so the response is a linear operator applied to it, and a linear operator distributes over addition by definition. Every calculation that adds one load case to another is standing on that one line. Deflection

The addition everything else rests on

Influence lines add, the unit-load method adds, moment distribution adds, load combinations add, and a stiffness matrix is linear by construction. All of it stands on one sentence with three hypotheses in it — and when they fail, two of the failures point in opposite directions.

 The load path has a kink in it, and the kink is a plate thick. Two 10 mm plates lapped over 60 mm and pulled with 60 kN. The two load paths are offset by the thickness of a plate, so the joint carries a moment nobody applied: P × 10.0 mm / 2. Taken at face value that gives a peak stress 4.00 times the mean. The joint rotates under load and the moment falls to 86 per cent of it, leaving 3.57 times — a saving of 11 per cent and not, on a plate this thick, a rescue. The bolt is bent as well as sheared: 382 N/mm² of bending against 191 of shear. Connections

The joint that is crooked by construction

Lap two plates and fasten them and the two load paths are offset by the thickness of a plate. The joint carries a moment nobody applied, the peak stress is four times the mean, and the rotation that is supposed to straighten it out saves eleven per cent — because the rescue works for thin sheet with a long lap and a bolted structural joint is neither.

 The moment a joint makes out of geometry. Bending stress in the chord divided by the axial stress already in it, against how far the diagonal's working line misses the node. A truss analysed as pin-jointed is drawn with its members as lines through their own centroids meeting at a point, because that is what makes the joint a pin with nothing but forces in it. A fabricated joint is not obliged to oblige, and the resultant then has a moment about the node — 16 kNm at the 60 mm drawn. It divides between the members meeting there in proportion to their rotational stiffnesses, 4EI/L or 3EI/L, which is the moment-distribution rule applied to an unbalanced moment that came from geometry rather than from load. The chord takes 48 per cent of it and carries 25 per cent of its axial stress again in bending. Ten per cent arrives at 28 mm, which on a chord of any depth is a detailing decision rather than a mistake. Connections

The joint that is not where it was drawn

A truss is drawn as lines through the centroids of its members meeting at points. A fabricated joint is under no obligation to oblige, and when the working lines miss each other the resultant has a moment about the node that has to go somewhere.

 A tenth of a per cent of the thrust is all of the moment. The thrust a two-hinged arch loses to its own axial shortening, against rise-to-span. The flexibility equation's denominator has two terms — ∫y²ds/EI for bending and ∫cos²θ ds/EA for shortening — and their ratio is about (15/8)(i/f)², the square of the radius of gyration over the RISE. The loss is a little more than that ratio, because the released rib also shortens under its own shear, and at the 10 per cent rise drawn it is 0.11 per cent of the thrust, which sounds like a rounding error and is not: a parabolic arch under a uniform load is funicular, so the rigid solution has NO crown moment at all, and the 0.11 per cent that the rib shortening removes from the thrust leaves 29 kNm behind. The correction that is a tenth of a per cent of the thrust is a hundred per cent of the bending. At a two per cent rise the loss is 2.6 per cent, because a shallow arch's thrust is enormous and its lever arm is not. Deflection

The arch that gets shorter

A parabolic arch under a uniform load is funicular, so the perfect solution gives it no bending at all. Then the rib shortens under its own thrust by a tenth of a per cent, and every kilonewton-metre of moment the arch will ever carry comes from that.

 3 continuous spans against 3 simple ones. The bending moment in a continuous beam, solved by the stiffness method, drawn over the moment in the same spans made simply supported. The peak sagging moment falls from 28.1 to 18.0, and a hogging moment of 22.5 appears over the supports where there was none. Structural form

The chord is a continuous beam

A truss analysis reports one number per member, and for a chord running through eight panels that number is an axial force with no bending in it. The chord is a single piece of steel spanning between its own nodes, and anything landing between them bends it.

 Influence line for the shear force at x = 10.5. The shear force at one fixed station, plotted against the position of a unit load walking across the span. The beam was re-solved at 301 load positions. The worst position is x = 10.56, giving 0.560. Structural form

Two diagonals, one of which is absent

A truss diagonal is sized for the shear in its panel, and near mid-span that shear changes sign depending on where the load stands. A member that can only pull cannot carry the reversed case, so the panel gets a second diagonal — and at any instant one of the pair is not there.

 Flattening the truss saves stirrups and crushes the web. Two capacities against the angle of the cracks, for a web 350 mm wide with a lever arm of 630 mm. The rising line is the stirrups: a cut along the crack severs z·cot θ/s of them, so flattening the crack from 45° to cot θ = 2.5 takes the 315 kN they carry to 787 — 2.5 times as much from the same steel. The falling line is the concrete strut, whose stress is V(cot θ + tan θ)/b_w z and therefore least at 45°. They do not cross in this range, so the stirrups govern throughout and the angle is a free choice. Internal forces

The angle is a choice, not a property

The truss inside a cracked concrete web has a strut angle, and nothing measures it. The designer picks it, the stirrup requirement falls as it flattens, the web stress rises, and every choice in between is a different structure that carries the same load.

 The same load, two diagrams, both in equilibrium. One span of a pair of 9 m spans under 30 kN/m, drawn twice. The elastic solution puts 304 kNm over the support and 171 in the span. Reducing the support moment by 30% and taking what statics then gives leaves 213 and 207: the section the beam needs falls from 304 kNm to 213, a saving of 30%. Both curves are in equilibrium with the same load — the mid-span ordinate plus half the support moment is the free moment 304 kNm for either — and the second is legitimate for that reason alone. What it costs is 13.0 milliradians of rotation at the support, which the section has to be able to deliver. Internal forces

The moment that was shed has to land

Redistribution takes a moment off a beam's support and pays for it with rotation. On a beam that is the whole story. In a frame the support is a column, the shed moment does not vanish, and it arrives at a member whose section was chosen from the diagram it has just left.

 A reaction with no load, and the moment it bends the beam with. The prestress moments in a 2-span beam. The primary moment is −P·e, the tendon acting on its own section, and it reaches 1440 kNm over the middle support. The secondary moment is what is left when the primary is taken off the total, and it is 720 kNm — 50% of the primary, with the same sign, so it does not cancel anything. It comes from the middle support refusing to let the beam lift: 102.9 kN pressing down there and 51.4 kN lifting at each end, a reaction set that sums to 0e+0 because nothing external was applied. Its diagram is straight between supports to 3.6e-13% of its own peak, which it has to be: reactions are point forces and a point force puts no curvature in a span. Internal forces

The tendon that can be moved

Lift a continuous beam's tendon at its interior support without changing its drape and nothing about the beam's total moment changes. The primary falls, the secondary rises by exactly as much, and the pressure line stays where it was — which turns a parasitic effect into a quantity a designer can place.

 The load being amplified is not the load doing the amplifying. The sway amplifier 1/(1 − ΣP/P_cr) against the storey's total gravity load, as columns are added that carry load and provide no lateral stiffness. The critical load of the storey is fixed at 8203 kN by the bracing that exists, and every leaning column moves the structure along the axis without changing it. At the storey drawn the amplifier is 1.57, so the second-order sway moment is 57% on top of the first-order one — and none of that 70% of the load which is causing it appears in any stability calculation done column by column. The storey stays stable across the whole of this axis. That is the practical reason a gravity-only column is drawn on the frame model rather than designed on its own: it is not being checked, it is being counted. Stability

Counted, not checked

A column with pinned ends and no bracing cannot buckle on its own, so nothing about it fails a stability check. It still carries load, and load with no stiffness attached lowers the buckling load of everything around it — which is why a gravity-only column is put on the frame model and never designed by itself.

 A base plate, and when the bolts start working. A 550 × 450 mm plate carrying 900 kN and 140 kN·m, so the resultant sits 155.56 mm from the centre against a kern of 91.67 mm. The plate is in partial contact: bearing over 358.33 mm at a peak of 11.16 N/mm², with the holding-down bolts carrying 0 kN. The plate lifts at 82.5 kN·m and crushes at 192.95 kN·m, and the bolts are not needed until 247.5 kN·m. Stability

The pinned base that is not pinned

A column base drawn as a pin is a plate bearing on grout, and a plate in contact over its whole length resists rotation whether anybody wanted it to or not. The stiffness it delivers depends on the axial load, so the assumption is one a frame can leave and re-enter as its loads change.

 Maxwell's reciprocal theorem. A load at one point and the deflection it causes at another, against the same load moved to the second point and the deflection read at the first. Both integrals return 307.5006, and neither calculation was told about the other. The two deflected shapes are entirely different; the two readings are identical. Deflection

An influence line is a deflected shape

Finding where a load has to stand to be worst means solving the structure once for every position it could stand in. Reciprocity says the answer is a single deflected shape — release the quantity being asked about, move it by a unit, and the shape the structure takes is the influence line.

 The answer arrives in instalments. The hogging moment at support 1 of a three-span beam, cycle by cycle. It starts at the fixed-end moment of 98.0 kNm — the value with every joint clamped — and settles at 156.9 kNm against an exact 156.9. The error falls by about a factor of four per cycle: 21.03, 5.92, 1.54, 0.60 kNm after one, two, three and four. Two cycles is an engineering answer and nobody had to invert anything. Deflection

Why it converges, and how fast

Moment distribution is an iteration, and iterations do not always converge. This one always does, at a rate the beam's own proportions fix — about a factor of four per cycle on a regular beam and considerably worse on an irregular one, which is where the method's reputation for two cycles being enough comes from and where it stops being true.

 Which of them stops moving. Three load cases on the same rectangle, each a constant moment plus a temperature profile cycled from nothing to a peak and back, over sixteen cycles. At 20% of the plastic moment with a 40°C profile it never yields at all; At 50% of the plastic moment with a 150°C profile it shakes down; At 90% of the plastic moment with a 260°C profile it ratchets, at 51.0% of the first-yield curvature per cycle. The ratcheting case never collapses and never returns: it simply arrives somewhere further round every cycle, which is a serviceability failure that no collapse calculation contains. Materials

The map with three regions

A structure carrying a constant load and a cycling temperature has three possible fates and only one of them is a collapse. It can stay elastic, it can yield once and then stop, or it can gain a little more deformation every cycle for ever — and the third has no failure load at all.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

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