Theme

One support too many — page 5

Essays 97 to 120 of 186 on this thread, in the same order.

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.

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.

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.

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

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.

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.

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.

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.

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.

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.

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.

Deflection

The shear that moves the moments

In a statically determinate beam, shear deformation adds movement and changes no force. In a redundant one the forces come from how the beam deforms, so a stubby beam soft in shear carries less moment at its wall and more in its span — and the carry-over factor, the one half every hand method passes along, falls to nothing at φ = 2 and then changes sign.

A fixed-ended member's end moments are the edge stresses of a column. A prismatic member, fixed at both ends under a point load 0.333 of the span from its left end, drawn three ways. At the top, the member. In the middle, the column the analogy puts in its place: a strip as long as the member and as wide at each point as 1/EI there, with its centroid, the elastic centre, 0.500 of the span from the left. At the bottom, the simply supported moment diagram, dashed, which is the load on that column; the straight line, which is the stress that load produces, P/A + M(x − x̄)/I; and the member's own moment diagram, which is the difference. The end moments are the column's edge stresses: 0.148 PL at the left and 0.074 at the right, which are Pab²/L² and Pa²b/L², and the largest sagging moment is 0.099 PL.

Deflection

The fixed-end moment is a column stress

A fixed end forbids a change of slope and a deviation, so on a member fixed at both ends Mohr's two theorems both have the answer nothing. Written with 1/EI as a width, those two conditions are the P/A + My/I of a short column under an eccentric load — the end moments are its edge stresses, and a haunched member's are read from where its elastic centre has moved.

Connections

The joint that has to keep turning

A beam on partial-strength joints collapses at 8(Mp + Mj)/L² however stiff the joints are. Stiffness decides only which yields first, and a joint that yields first has to go on rotating at full moment until the span catches up. A weak joint on a stiff connection has the most turning to do — more than a rigid full-strength one.

$A pin's force does not pass through its centre. Left, a pin of radius 0.15 m in its hole, with a coefficient of friction of 0.15. The reaction at the contact is inclined by φ = 8.5° to the radius through it, because the friction it can develop is that fraction of the force pressing the surfaces together, and the perpendicular distance from the pin's centre to that inclined line is R sin φ = 0.022 m. Every position the contact can take gives a line tangent to the same circle, shaded. Right, a link 0.48 m long pinned at both ends, at the same scale — 3.2 pin radii, which is a stubby linkage rather than a structural tie, drawn that way because at the thirty radii an ordinary tie has, the two circles are smaller than the pencil: its force is a common tangent to the two circles — the two solid lines for the two senses of rotation, the two dashed ones for the senses in which its ends turn oppositely — and the dashed centre line every construction in this collection draws is none of them. A pin of this size carrying 5.0 MN delivers a couple of 111.3 kN·m to whatever it is pinned to, which is the same offset read as a moment rather than as a distance.$ Equilibrium

The pin that is not a point

Every line of action drawn so far passes exactly through a pin's centre, which is true of a frictionless pin and of nothing else. A real one carries its force tangent to a small circle instead, so a link's line is a band, a construction's answer is a range, and a support drawn as a hinge hands a couple of a hundred kilonewton-metres to whatever it is pinned to.

Materials

The support that had no moment when it was cast

Two beams are set on their bearings, carry their own weight for a month, and are then stitched together over the middle support. Nothing about the loading changes afterwards. Twenty years later the stitch is carrying 155 kilonewton-metres, because the concrete went on creeping and the joint would not let it — and how much arrives is decided by a crane schedule.

The column under the first interior support is given a quarter more. Two equal spans of 8.0 m carrying 5.0 kN/m, continuous over three supports. The tributary rule gives each interior support one span's worth of load, 40.0 kN, and each end support half of that. The continuous beam gives 15.0 kN, 50.0 kN, 15.0 kN — ratios of 0.75, 1.25, 0.75 to what the areas say. The reactions still add to the whole load, because they must; what has moved is which support gets it. The end supports are relieved because the span next to them hogs over the first interior support, lifting their end of it.

Equilibrium

The column given more than its rectangle

The tributary rule draws a rectangle round each column and hands it whatever stands inside. A floor is continuous over its columns, and a continuous beam does not give each support the load above it — the first interior one takes a quarter more and the end ones a quarter less. On a grid the two directions multiply, and two columns on the same floor differ by a factor of nearly three.

A portal frame, and the section it is a drawing of. A single-bay portal of 6.0 m span and 5.0 m height with fixed feet, carrying 10.0 kN/m on its beam, with the beam's second moment equal to the columns'. On the right, the analogous column: the frame's own centreline drawn as a section as wide as 1/EI at every point, so it is narrow where the frame is stiff. Its area is 16.000 and its elastic centre sits 1.56 m below the beam — inside the frame, on no member at all. The released moments loaded onto that section give a direct stress of 33.8 kN·m and a bending stress whose gradient is the horizontal thrust, 6.4 kN. Together they give 10.6 kN·m at the feet, −21.2 at the knees and 23.8 at the crown, against a free moment of 45.0. The same frame solved by stiffness gives 10.6, −21.2 and 23.8.

Deflection

The elastic centre is not on the frame

Cross's analogy turns a member fixed at both ends into a short column and its end moments into edge stresses. A closed frame's analogous column is the frame's own outline, its elastic centre is a point hanging in mid-air inside it, and the horizontal thrust a gravity load produces is that section's bending stress about the axis through that point.

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.

A unit load carried by the prop taken away. A beam fixed at its left end and propped at its right, under 4 kN/m over 8 m, asked how far it moves at 4 m. The real moment is the propped cantilever's own, with -32.0 kNm at the wall. The unit load is carried by the prop taken away, whose moment diagram peaks at 4.00. Their product has 98.92 of area on one side and -13.33 on the other, and the net, divided by EI, is 85.33 — the propped cantilever's closed-form deflection, 85.33, which no part of this calculation was given.

Deflection

Any structure will carry the unit load

Virtual work has two readings and each is free exactly where the other is bound. A unit load needs only something to stand on in equilibrium, so the deflection of a beam statics cannot solve comes out of a cantilever statics can. A virtual displacement needs only to fit together, so a reaction comes out of pushing a mechanism — and on a redundant beam the unknown cancels out of the equation and nothing is found at all.

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.

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.

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.

A fixed arch, and the section it is a drawing of. A parabolic arch of 30.0 m span and 6.0 m rise, fixed at both springings, carrying 20.0 kN/m over the span. On the right the analogous column: the arch's own centreline drawn as a section of width ds/EI, so it is narrow where the arch is stiff. Its area is 32.99 and its elastic centre sits 3.86 m above the springings — 2.14 m below the crown and on no part of the arch at all, which is the point of it. The released moments loaded onto that section give a direct stress of 3054.0 kN·m and a bending stress whose gradient is the horizontal thrust, 375.0 kN.

Deflection

The centre that hangs in the air

Cross's analogy turns a closed frame into a short column and reads its redundant moments off as that column's stresses. Nothing in it requires the frame to be straight. Run round a fixed arch, the analogous column's centroid is a point two thirds of the way up the rise, on no part of the arch at all, and the bending stress about it is the arch's horizontal thrust. For a parabola under a uniform load the answer it returns is a diagram of nothing, which is exactly right.

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.

All themes