Structural Engineering & Statics
where the load goes
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One support too many — page 3

Essays 49 to 72 of 186 on this thread, in the same order.
At thirty degrees each leg carries the whole load. A 100 kN lift on two legs at 60 degrees to the horizontal. Each leg is a two-force member, so its force is the vertical share it carries divided by the sine of its angle: T = W/(2 sin β) = 57.7 kN, which is 0.58 times the whole load in each. The curve on the right is that division, and it is the reason the rule about sling angles exists rather than a convention: at sixty degrees a leg carries 0.58 W, at forty-five 0.71, at thirty exactly 1.00, and at fifteen 1.93. And the horizontal components do not disappear — they run through the thing being lifted, which here carries 57.7 kN of compression between the two pick points. That is what a spreader beam is for: it takes the compression as a designed strut so the legs above it can stand up. Equilibrium

The angle that doubles the force

A crane hook takes the weight and nothing else. What holds the load is a set of legs running down to it at an angle, and each leg is a two-force member — so at thirty degrees from the horizontal each of two legs carries the entire load, and the difference goes into the thing being lifted as compression.

 The end bolts do the work and the middle ones very nearly nothing. A lap of 8 bolts at 70 mm pitch transferring 800 kN between two plates, with the force each bolt actually carries drawn above it and the flat line a division by the bolt count would have given drawn behind. The end bolts carry 1.09 of their nominal share and the middle ones 0.94. The reason is not in the bolts: at the leading end the first plate is carrying everything and the second nothing, so the two strain at different rates and the slip between them is largest there. In the middle they strain alike, there is almost no slip, and a bolt with no slip across it transfers almost no force. The mean over the worst is 0.918, and the end bolt has to slip 1.36 mm before the rest catch up. Connections

The bolts that do not share

Every bolted connection in this collection has divided a force by a number of bolts. That is right for a short joint and wrong for a long one, and the reason has nothing to do with the bolts — it is that the plates they join are elastic, and stretch by different amounts at different points along the lap.

 The demand falls and the movement rises, by the same factor. One elastic spectrum read twice: as an acceleration on the left and as the displacement that goes with it on the right. A fixed-base building at 0.5 s sits on the plateau and is asked for 1.05 g. Put it on bearings soft enough to make its period 2.54 s and the demand falls to 0.103 g — a base shear 10.2 times smaller, bought with no strength whatever. The same shift on the right-hand plot goes the other way: displacement is S_a T²/4π², so the demand rises from 65 mm to 165. That number is the design. It is a gap all the way round the building, a moat every service has to cross, and a detail that a later contractor will fill in unless somebody says what it is for. Dynamics

Made weaker on purpose

Everything else in this collection resists a load by being stiff or strong enough for it. A base-isolated building resists an earthquake by refusing to hear it — a layer of bearings under the whole structure with a lateral stiffness a twentieth of the frame's, bought with almost no strength at all.

 Almost all of it is exactly zero. The stiffness matrix of a 2-bay, 3-storey plane frame: 36 freedoms, of which 16.2 per cent of the 36² entries are non-zero. The zeros are not small numbers; they are absences. A member reaches only the two nodes at its ends, so it can contribute nothing to any row belonging to a node it does not touch, and every such entry is zero exactly rather than nearly. The non-zeros therefore sit in a band of width 13 about the diagonal. Before the supports are applied the matrix is singular, and its null space has exactly three dimensions — the three rigid-body motions a plane frame has with respect to the ground, which is the same statement the null vector of the equilibrium matrix makes about a truss that is a mechanism, arrived at from the other end. Deflection

The matrix that replaced the hand methods

Moment distribution passes moments round a frame until they stop moving. Virtual work computes one deflection at a time. Both are exact and both stop scaling in the low tens of members. What replaced them adds no physics at all — the whole of the invention is the bookkeeping.

 Two cantilevers, or one wall, and the beams decide which. The deflected shape of a coupled pair of 6 m walls, drawn against the two limits it lies between. Release the coupling beams entirely and the pair is two independent cantilevers, deflecting 111 mm. Make them rigid and it is one composite wall of the full width, deflecting 16 mm — 6.8 times stiffer, because the lever arm between the wall centroids is 8.40 m and everything inside either wall is smaller than that. Real beams of 600 × 350 mm over a 2.4 m opening land at 23 mm and carry 63% of the base overturning as an axial couple rather than as wall bending. The degree of coupling never reaches one, because a beam of finite depth cannot suppress the walls' curvature entirely. Internal forces

Two walls that agreed to be one

A pair of shear walls with a row of doors between them is the commonest lateral system there is, and it has two readings that differ by a factor of seven. What decides which one applies is a beam 600 mm deep over a 2.4 m opening — and most of the overturning ends up as an axial couple that no bending diagram contains.

 Eight slices is enough, and nobody would have guessed it. The error in a cracked section's moment capacity against the number of strips it was integrated with, for a 300 × 450 mm section with 1200 mm² of steel, measured against the same computation at 2048 strips. The point of the fibre method is that it contains no formula: slice the section, give every strip the strain the assumed curvature puts it at, move the neutral axis until the axial force balances, and sum. It handles a cracked section, a confined one, a prestressed one and a composite one with the same twenty lines. The discretisation costs 1.4% at 2 strips and 0.088% at 8 — and the convergence is not smooth, because what the error actually depends on is where the neutral axis falls relative to a strip boundary rather than on the strip count as such. Sections and stress

The section calculation with no formula in it

Every ordinary section result is a closed form, and each was derived once for one arrangement of material. Slice the section instead, give each strip the strain a curvature puts it at, and move the neutral axis until the axial force balances — and the same twenty lines answer for a cracked section, a confined one, a prestressed one and a composite one, having been told nothing about any of them.

 Two different structures released, and one bending moment diagram. The bending moment in a continuous beam of 8, 10, 8 m under 12 kN/m, solved twice by the force method with different redundants. The first release puts a hinge over each interior support, so the released structure is a row of simple spans and the redundants are moments. The second removes each interior support, so the released structure is one simple span of the whole length and the redundants are reactions. The two released structures have nothing in common — different shapes, different deflections, different everything — and the diagrams they produce lie on top of each other to 9e-15 of the peak moment. Which restraints are released is a choice about the arithmetic and not about the structure, which is a fact worth trusting: it means a hand calculation can pick whichever release makes the sums easiest and be sure of the answer. Deflection

Choose what to take away

The other machine for a redundant structure works by removing restraints until what is left can be solved by statics, then putting back exactly enough force to close the gaps that opened. Which restraints are removed does not change the answer at all, and changes the arithmetic completely — one choice gives a tridiagonal matrix a person can solve on paper, and another gives a full one.

 The force falls, the drift rises, and the damping goes the wrong way. What a compliant foundation does to a 0.6 s building on a 8 × 8 m footing, against the stiffness of the ground under it. Three curves, all normalised to the fixed-base answer. The period lengthens — 1.31 times at 200 m/s — because the swaying and rocking of the foundation are flexibilities in series with the structure's own, and the rocking term carries an h², so it is the tall building that feels it. The base shear falls with the period, which is why a fixed base is usually called conservative. The displacement rises, by 51% here, and that is what breaks the cladding, the services and the gap to the building next door. And the effective damping falls rather than rises: the structure's own is divided by the cube of the lengthening — 2.2% of an original 5% — while a slender building's foundation radiates only 0.28% back, because rocking radiates almost nothing at these frequencies. Dynamics

The ground is a spring

Every dynamic result in this collection has assumed a structure rising from something that does not move. Nothing does. A foundation can slide and it can rock, both are flexibilities in series with the structure's own, and the rocking one carries a square of the height — so the period lengthens, the force falls, the drift rises, and the damping goes the wrong way.

 The check that everything adds up, and the error it cannot see. Four versions of the same 2-bay, 3-storey frame, with the global equilibrium residual each one produces — the sum of the reactions against the sum of the applied loads, as a fraction of the applied total. It is the first thing every analysis prints and it is worth having: a lost restraint and a load entered in the wrong unit both show up immediately, at 6% and 24%, because both change what the structure is carrying. The fourth bar is the point. A member whose stiffness is wrong by a factor of ten redistributes the internal forces completely — the second bar shows the change in the member forces, 18% — and the global residual is exactly zero, because the wrong answer is still in equilibrium with the same loads. Equilibrium is one equation per degree of freedom of the whole body, and a stiffness error lives entirely in the many equations underneath it. A model can satisfy every equilibrium check ever devised and be a model of a different structure. Stability

The stiffness the load takes away

Buckling is usually taught as an event — a critical load, a bifurcation, a mode. Written as a matrix it stops being an event at all. A compressive load subtracts a stiffness from the structure, the subtraction grows with the load, and the critical load is simply where what is left reaches zero.

 A force may be moved anywhere, at the price of a couple. A 80 kN force applied 250 mm off the centreline of a body, and the same force applied ON the centreline together with a couple of 20 kNm. The two systems are equivalent: they have the same resultant force and the same moment about every point in space, so no equilibrium equation written about the body can tell them apart. What they are not is the same loading — the stresses inside the body differ, and they differ over a distance of about the body's own depth. The offset is drawn to a scale that keeps the arrow on the body; the number beside it is the real one. Internal forces

The moment the beam left behind

A beam reaction is drawn arriving on a column's centreline. It arrives on a cleat a hundred millimetres out from the face, and the difference is a couple that goes into the column and has to be shared between the lengths above and below it. Nothing about it appears in a frame model whose members meet at nodes.

 One member's stiffness, scattered into the freedoms it touches. A member's own six-by-six stiffness matrix relates the forces at its two ends to the displacements there, and it is written in the member's own axes. Assembly is two operations and no physics: rotate it into the structure's axes, then add each of its thirty-six entries into the row and column of the global freedom that entry belongs to. Every member does the same, and the sum is the structure. The shaded rows and columns are the six freedoms this one member reaches; every other entry it contributes is exactly zero, and that is the whole reason a global stiffness matrix is sparse. Nothing here is an approximation — the result is the same equilibrium and the same compatibility a hand method writes, in an order a machine can follow. Deflection

The answer that depends on how it was divided

Every computed answer in this collection came out of a structure chopped into pieces — elements, strips, stations, trial positions. The chopping is invisible in the result and it is not neutral: some divisions give the exact answer, some give one that is always too stiff, and one of them changes nothing but the cost of getting there.

 The end bolts do the work and the middle ones very nearly nothing. A lap of 10 bolts at 75 mm pitch transferring 800 kN between two plates, with the force each bolt actually carries drawn above it and the flat line a division by the bolt count would have given drawn behind. The end bolts carry 1.16 of their nominal share and the middle ones 0.89. The reason is not in the bolts: at the leading end the first plate is carrying everything and the second nothing, so the two strain at different rates and the slip between them is largest there. In the middle they strain alike, there is almost no slip, and a bolt with no slip across it transfers almost no force. The mean over the worst is 0.861, and the end bolt has to slip 1.16 mm before the rest catch up. Connections

The joint that has to be as good as the member

A splice exists because members come in lengths and structures do not. It has to deliver the same force, at the same stiffness, in the same distribution across the section, through a discontinuity — and each of those three requirements is met by a different feature of the detail, with the third one usually left to look after itself.

 Two curvatures of opposite sign, which is what makes it a structure. A cable net over a 36 m square, drawn as the two families of cables that are also the two rulings of the surface. One family sags and carries downward load by hanging; the other rises and carries upward load — wind uplift, and a load reversal anywhere — by the same mechanism upside down. Neither can do anything alone. A single family of cables is a mechanism: it changes shape freely under any load pattern it was not tensioned for, and the shape it moves to is decided by the load rather than by the designer. Put the two together and each is the other's restraint, but only if they are pulled against one another first — the pretension of 520 kN in the sagging family and 715 in the hogging one is a self-equilibrating state that exists with no load on the roof at all, and it is what turns two mechanisms into one structure. The curvatures are drawn four times their true value: a real net of this span sags 2.2 m over 36, which is flatter than it looks anywhere. Structural form

Two curvatures of opposite sign

A single family of cables is not a structure. It is a mechanism that takes whatever shape the load asks for, and it will do that under any load pattern it was not tensioned for. Cross it with a second family curved the other way, pull the two against each other, and the pair becomes stiff — with no bending anywhere and no material property involved in the stiffness at all.

 Two drawings of one deck, and they are not the same structure. A 112 m viaduct on five supports, articulated two ways. Above, the fixed point is at the left abutment: the far end has to be given 45 mm of movement, and the friction of every sliding bearing runs one way, so the fixed support takes 660 kN before any wind or braking is applied. Below, the fixed point is at the middle pier: the largest joint halves to 22 mm and the friction now cancels across the fixed point, leaving 0 kN. The movement arrows are drawn at 900 times the scale of the deck, because a 45 mm movement on a 112 m span is thinner than the line the deck is drawn with. Nothing about the deck, the loads or the ground has changed between the two. Structural form

Where the structure is allowed to move

One drawing decides how big every movement joint on a bridge is and where every horizontal force goes, it takes an afternoon, and it appears on no calculation sheet. Move the fixed point from an abutment to the middle pier and the largest joint halves and the horizontal force on that support drops from the whole of the friction to none of it.

 Effective length is a property of the storey. The effective length factor of the one column that resists sway, against the total gravity load on the storey as a multiple of its own. At the left-hand end it carries the storey alone and its K is 1.99 — the 2.0 every chart gives a column fixed at the base and free to sway at the top, reproduced here by a route that never mentions a chart. Then columns are added that have pinned bases and therefore no lateral stiffness whatever. They contribute load and nothing else, so they cannot buckle on their own and they lower the load at which everything buckles together. K rises as the square root of the load ratio, exactly, and at the storey drawn — three leaning columns carrying 69% of the gravity load — it is 3.57. That is off the end of every published alignment chart, and the leaning columns themselves, which a designer would take at K = 1.0 for pinned ends, are at 2.54. Stability

The column that leans on its neighbours

A column with a pinned base and a pinned top has no lateral stiffness at all and cannot stand up alone, and yet thousands of them do. What holds them is the rest of the storey, and what it costs is paid by whichever columns do have stiffness — whose effective length rises as the square root of the load being leaned on them, straight off the end of every chart.

 The check that everything adds up, and the error it cannot see. Four versions of the same 3-bay, 4-storey frame, with the global equilibrium residual each one produces — the sum of the reactions against the sum of the applied loads, as a fraction of the applied total. It is the first thing every analysis prints and it is worth having: a lost restraint and a load entered in the wrong unit both show up immediately, at 8% and 32%, because both change what the structure is carrying. The fourth bar is the point. A member whose stiffness is wrong by a factor of ten redistributes the internal forces completely — the second bar shows the change in the member forces, 24% — and the global residual is exactly zero, because the wrong answer is still in equilibrium with the same loads. Equilibrium is one equation per degree of freedom of the whole body, and a stiffness error lives entirely in the many equations underneath it. A model can satisfy every equilibrium check ever devised and be a model of a different structure. Equilibrium

The check that cannot see the error

Every analysis prints a global equilibrium residual, and it is the first thing anybody looks at. It catches a lost restraint and a load entered in the wrong unit immediately. It is structurally incapable of catching a member whose stiffness is wrong by a factor of ten, because the wrong answer is still in equilibrium with the same loads.

 Every reason a building is stiffer than its model, added up. The computed natural frequency of a floor, and the same frequency after each source of stiffness that was deliberately left out is put back. Not one of them is a modelling error. Cladding and partitions are stiffness nobody is allowed to rely on for strength; a nominally pinned connection is never pinned; a slab acts with its beam whether or not shear connectors were provided; and concrete between the cracks is stiffer than a cracked section assumes. Together they multiply the stiffness by 1.83 and the frequency by 1.35, because a frequency is the square root of a stiffness, and every factor is halved on the way through. The asymmetry is the finding: leaving stiffness out makes a deflection conservative and a vibration check unconservative in the direction that matters, since a stiffer floor has a higher frequency and sits further from the footfall range. The model here reads 4.40 Hz against a 5.2 Hz criterion and fails it; the floor reads 5.95 Hz and passes. The correction that would have got it right is exactly the stiffness nobody is willing to count on. Deflection

Stiffer than the model said

Measured natural frequencies of finished buildings come out between ten and sixty per cent above the values computed for them, consistently and in one direction only. Nothing on the list of reasons is a modelling error: every one is a real source of stiffness deliberately left out — and leaving stiffness out is conservative for deflection and unconservative for vibration.

 The bearing that is drawn as a roller. The horizontal force a sliding bearing delivers, against the vertical load it is carrying, with its coefficient of friction on the same picture. The coefficient is not a constant: PTFE's falls as the contact pressure rises, and the standard fit is μ = 1.2/(10 + σ), so the bearing drawn is at 30.0 N/mm² and μ = 0.030 while the same bearing at a fifth of the load is at 0.075 — 2.5 times as much. The force curve is therefore strongly non-linear: a fifth of the load gives 50% of the force. Two readings follow and only one of them is usually taken. The largest force is at full load, 108 kN, and that is what the pier is designed for. The largest nuisance is at light load, where 54 kN of friction is 39% of the 140 kN of wind the bearing was put there to release the structure from. Cold makes it worse again: below about −5 °C the same bearing delivers 216 kN. A roller symbol on a drawing means this, and it is a pair of load cases rather than one, because friction opposes whichever way the deck happens to be going. Connections

The roller that is not a roller

A sliding bearing is drawn as a roller and detailed as a sheet of PTFE, and it delivers a horizontal force of a few per cent of whatever it is carrying. The coefficient everybody quotes is the one at full design pressure, and PTFE's coefficient rises as the pressure falls — so the bearing is at its freest exactly where nobody checks it.

 The hinge is at mid-height in exactly no storey. The height of the point of contraflexure in each column of a 4-storey, 3-bay frame under lateral load, as a fraction of the storey height, against the portal method's assumption that it is at the middle. The exact solution is a plane-frame stiffness analysis of the same frame. In the bottom storey the zero sits at 0.609 of the height, because a fixed base is stiffer than the joint above it and takes more of the column's moment; in the top storey it sits at 0.359, because there is no column above to share that joint. The average over the whole frame is 0.475, which is why the assumption survives — it is right on average and wrong everywhere. The worst error in the storey shear share is 25%, and the column shears still add to the storey shear to 0e+0 of it, because the method is exact statics applied to an assumed structure. Internal forces

The analysis that assumes the answer

A rigid frame is indeterminate, so statics cannot finish it. The hand methods finish it anyway, by assuming where the bending moment is zero and treating those points as hinges. That is not a shortcut around the analysis — it is a different kind of answer, exact in equilibrium and wrong in compatibility, and knowing which half is which is what makes the error a bound rather than a mystery.

 Two of these move and the third cannot. The first span of a 3-span beam under 5 kN/m, with the stiffness of the middle span swept over a factor of 25. The support moment and the mid-span moment both move — that is what redundancy does, and it is the whole reason a continuous beam has to be analysed rather than read off. Their combination does not: the mid-span ordinate plus the average of the two end moments is 30.6 kNm at every point on this axis, which is wL²/8 for this span and this load and contains nothing else. The largest departure anywhere on the sweep is 2.3e-16 of the value, which is the arithmetic of the stiffness solution rather than a property of the beam. Continuity buys a distribution and not a capacity, and this is the line that says so. Internal forces

Two of these move and the third cannot

Cut one span of a continuous beam free and add up the forces on it. What comes out is that the mid-span moment plus the average of the two end moments equals the free bending moment of that span, with nothing else in it — no stiffness, no support settlement, no analysis at all. Continuity moves moment about. It does not reduce the total, and it never has.

 The prop load is decided by the digging, not by the hole. Prop forces in a 12 m excavation propped at 3 levels, with the force each prop reaches at any stage of the sequence drawn thick and the force the finished arrangement gives it drawn thin. The pale dots are the individual stages. The middle prop reaches 214 kN while the dig is at 9.5 m and finishes at 88 — a factor of 2.44 between the two, and the larger one is not in the final analysis anywhere. Terzaghi and Peck's apparent pressure diagram, a rectangle of 49.4 kN/m², reproduces the total of the staged maxima to 2% — which is what it is: an envelope of measured prop loads, back-figured into a pressure, and a shape nothing on a wall is ever loaded with. Structural form

Every prop has its own worst day

A braced excavation has no finished state worth analysing. It is dug in stages, a level of props goes in at each stage, and a prop's force is largely fixed the moment it is installed — so the force to design it for is the largest it sees during a sequence that appears on no calculation sheet, and which for the middle prop here is nearly two and a half times what the finished arrangement gives.

 The group is not weaker; it is very much softer. A 3 × 3 pile cap on the left, with each pile's share of 9.0 MN and 4.5 MNm in meganewtons — N/n plus M·y/Σy², the same three terms in the same order as a bolt group under an eccentric load and a section under biaxial bending. The corner piles take 1.25 times the average and a pile added at the centroid would change that by nothing at all, because it adds to neither second moment. On the right is the effect a bolt group cannot have: the piles share ground, so the stress bulbs overlap and the group settles 3.9 times as much as a single pile at the same load per pile, rising to 14.2 for 144 of them. The capacity check everyone makes — block failure against the sum of the piles — comes out at 4.54 here and does not govern at all. The check nobody tabulates is the one that does. Structural form

Nine piles, and four times the settlement

A pile cap divides its load between its piles by the same three terms a bolt group uses and a section under biaxial bending uses. What a bolt group does not have is neighbours it shares ground with — and the group effect that matters is not the strength check everybody makes, but a stiffness effect nobody tabulates.

 The lining that carries less for being weaker. Bending moment and hoop thrust in a circular lining, against the lining's own bending stiffness, both as fractions of the free-ring values. The ground arrives already stressed — 500 kPa vertically and 300 horizontally at K₀ = 0.6 — and the difference between them tries to squash the hole into an ellipse. A lining stiff enough to refuse absolutely collects the whole distortion pressure, p₂R²/3 = 300 kNm/m; one flexible enough to go with the ground collects nothing, because there is no curvature change left to resist. The thrust is the flat line: it comes from the mean stress rather than the difference, so it does not move at all. Putting 8 joints in this ring drops the moment to 34% of the solid one and leaves the thrust exactly where it was, which is why a segmental lining is jointed and why the intuition carried over from a beam is inverted here. Structural form

The lining that is stronger for being weaker

A tunnel lining is not loaded. The ground arrives already stressed and the hole wants to squash into an ellipse; the lining's only job is to refuse, and how much moment it collects depends entirely on how hard it refuses. Make it stiffer and it takes more. Make it flexible — put joints in it, make it thin — and it takes almost none, while the hoop thrust it carries does not move at all.

 The same curve, computed twice and from opposite ends. A cantilever of 6 m, with its deflected shape drawn from a double integration of M/EI and the bending moment of its conjugate beam drawn on top of it. The conjugate is the same span carrying M/EI as a load, with its supports transformed — a real fixed end becomes a free end and a real free end becomes a fixed one, because a fixed end has no slope and no deflection and the conjugate therefore needs no shear and no moment there. The two curves agree to 4.7e-6 of the largest deflection, which is the trapezium rule and not the method. The largest deflection is 36.00 mm against the closed form's 36.00. The reaction of the conjugate beam is -9.000 milliradians, which is the real beam's rotation at that support — so the whole of a slope calculation is one reaction. Deflection

The beam whose moment is a deflection

A beam's bending moment is the second integral of its load. Its deflection is the second integral of M/EI. They are the same problem, so a deflection can be found by loading a fictitious beam with M/EI and asking a statics question — and the only thing to remember is the supports, which are not remembered but derived, one boundary condition at a time.

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