Structural Engineering & Statics
where the load goes
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Drawing as calculation — page 3

Essays 49 to 72 of 162 on this thread, in the same order.
The point the rafter turns about, which is off the frame. A pitched portal of 8 m span and 4.0 m to the eaves, with a 1.5 m rise, collapsing. Each rigid part of the mechanism rotates about some point: the left column about its base hinge, the right about its own. The rafter between them does neither, and its centre is found by one rule — two bodies joined at a hinge share that hinge, so the second body's centre lies on the line through the first body's centre and the hinge, extended. Two hinges give two lines and they cross at (8.0, 11.0) metres, which is 5.5 m above the ridge and outside any drawing of the frame itself. From there the whole collapse is two ratios of lengths and no trigonometry: the load factor is 1.339. Flatten the roof and the centre descends; make the two lines parallel and it goes to infinity, which is the statement that the rafter translates instead of turning. Equilibrium

The point the mechanism turns about

A collapsing frame is a chain of rigid pieces, and every piece is rotating about some point. Find those points and the whole collapse load reads off two ratios of lengths, with no trigonometry anywhere — and for a pitched roof the point in question is well above the top of the drawing.

 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.

 Wrong in shape, right in two integrals. The compression zone of a C30 section with its neutral axis 150 mm down, drawn twice. The curved outline is the real parabolic-rectangular stress distribution — the material's own law read off the linear strain profile plane sections supplies. The rectangle over it is what every design office uses instead: intensity η f_cd = 16.5 MPa over a depth λx = 125 mm. The two shapes are visibly different and give the same answer, because a bending calculation asks a stress distribution only two questions — how much compression there is, and where its resultant acts. Both are 619 kN at 62.4 mm from the face. The factors are α = 0.8095 and β = 0.4160, and λ = 2β follows from wanting the same centroid. A triangle and a full rectangle match neither integral and are nowhere near. Sections and stress

Deliberately the wrong shape

Concrete in compression follows a curve, and no design office has ever integrated it. Every code in the world replaces it with a rectangle of reduced depth and reduced intensity, and the answer is right to a fraction of a per cent — not because the shapes are similar, which they visibly are not, but because a bending calculation only ever asks a stress distribution two questions.

 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.

 The line, and the stone it has to stay inside. A masonry pier 9 m high, 1.6 m thick at the top and battered 12% on its outer face, taking a thrust of 40 kN per metre of run at 25° to the horizontal. The line drawn through it is the locus of the resultant on each horizontal cut: everything above the cut is the free body, and the resultant's position is the moment divided by the vertical force. The dashed pair is the middle third, inside which no tension is implied anywhere on the joint. The line stays inside the stone throughout and reaches the base at 0.503 m from the centre, against a half-width of 1.34 m — but outside the middle third, so part of the base joint is open and the toe is carrying a triangle. Nothing about the strength of the masonry appears anywhere in this figure, and that is the point. Structural form

The weight that makes it safer

Every load in this collection makes a structure worse. A pinnacle does not. A masonry pier fails when the line of compression leaves the stonework, and adding weight at the top rotates that line back towards the vertical without adding anything the pier cannot carry — so the stone is not being strengthened, it is being aimed.

 The wind pushes on one face and pulls on three. A 30 × 20 m building in plan, with the measured pressure coefficient on each face and the arrows drawn in the direction the pressure acts. Only the windward face is pushed; the other three are sucked, and the side faces are sucked hardest of all at c_p = -0.7. The horizontal resultant is 842 kN at a velocity pressure of 0.9 kPa, and the arithmetic of it is the whole point: the leeward suction pulls the building downwind, so it ADDS, supplying 38% of the answer, while the two side faces cancel each other exactly and supply none of it. The coefficients are wind tunnel data; what is computed is the free body they are applied to. Equilibrium

Most of it is suction

A wind load is drawn as arrows pressing on the windward face, which is where about three fifths of it comes from. The rest is a pull on the back. The two side faces carry the largest suctions on the building and contribute nothing at all to the answer — and the inside of the building, which nobody draws, decides whether the roof stays on.

 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 corner is where the plate twists, and it has to be held down. The twisting moment M_xy over a 6 × 6 m simply-supported panel under 10 kN/m², from Navier's double series. It is zero along both centrelines and largest at the corners, which is the opposite of the bending moments and is why no strip reading contains it: 31% of the load crosses this panel in twist, and a strip can only bend. At a free corner the twisting moment is statically equivalent to a downward point force of 2M_xy — 26.7 kN here, 7.4% of the whole load on the panel, applied at a point — so the corner has to be held down. A panel whose corners are free lifts there, deflects more, and cracks diagonally across them. The deflection at the centre is 2.53 mm, against 4.05 mm for the strip that ignores all this — 60% more, which is the size of what the twist is carrying. Deflection

A third of the load crosses sideways

A slab spanning both ways is usually explained as two beams sharing a load by a fourth power, and the explanation is not merely approximate — it is missing a mechanism. A real plate carries load three ways, and the third one has no beam strip in it: it is twisting, it accounts for a third of the load on a square panel, and it is why the corners lift.

 Three lines through one point, and three different winners. Modulus against density on logarithmic axes, with a guide line for each of three indices drawn through mild steel. A performance index E^(1/n)/ρ is a straight line of slope n on these axes, so ranking materials by it means sliding the line up and to the left and seeing what it leaves behind. The three lines have three different orders, which is why a table of properties cannot answer the question on its own: for a tie the answer is carbon fibre, for a beam it is timber at 4.28 times steel, and for a plate timber wins by more still. The construction is Ashby's; the arithmetic on it is this site's. Materials

The ranking belongs to the load case

Every table of material properties ever printed ranks by strength, and every one of them answers a question nobody asked. What a structure wants is the least mass for a stated performance, and the combination of properties that gives it changes with the shape of the member — so timber beats steel four to one as a beam and loses to it as a tie, without either material changing.

 The width nobody drew. A gusset plate with a brace bolted to it over 240 mm, and the width the profession has agreed to pretend is carrying the force. Everything else on this site arrives with a cross-section; a gusset does not, because it is a piece of steel with something attached somewhere in the middle of it and there is no geometry that says how much of it is working. The answer is the Whitmore section: assume the force spreads at 30° from the first fastener and take the width it has reached at the last, b_eff = w + 2L·tan30° = 367 mm. That is 4.08 times the width anything is actually attached to, and the rule comes from a 1952 master's thesis. It has since been checked against finite element work and holds to about ten per cent, which is fortunate, because moving the assumed angle by ten degrees moves the answer by 34%. On this plate the check that governs is not the stress the rule was written for: it is the Whitmore section buckles, at 721 kN against 1564. Connections

The width nobody drew

Every other member in this collection arrives with a cross-section. A gusset plate does not — it is a piece of steel with a brace bolted to it somewhere in the middle, and no geometry says how much of it is working. The profession's answer is a thirty-degree spread from a 1952 master's thesis, it invents three quarters of the area being checked, and the check it was written for is not the one that governs.

 The snow that left the roof is standing against the wall. A roof in section with a 1.0 m obstruction at its downwind end, drawn with the vertical scale exaggerated 4 times because a metre of snow on twenty metres of roof is thinner than the line it would be drawn with. The balanced layer is 0.48 kN/m² everywhere; the wedge against the wall holds the snow that 30% of the 20 m upwind gave up, so its area is fixed by conservation of mass rather than chosen. That makes it 0.85 m deep and 3.4 m long, with a peak load of 2.60 kN/m² — 5.4 times the balanced value, and still only 27% of the snow on the roof. An intensity several times the design load, made of a quantity nobody would notice had moved. Equilibrium

The load that arrives where the wind stops

Snow is the one load a structure is given rather than subjected to. What falls is spread evenly over a whole region; what a member carries is whatever the wind left above it, and the wind piles it against whatever gets in the way. The heaviest patch on a roof is usually a quarter of the snow on it.

 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.

 The bearing is one length and the web is loaded over another. A load applied over a stiff bearing of 200 mm on the flange of a girder with a 1200 × 8 mm web. The flange bends under it and the yield lines that form spread the load along the web over 659 mm — 3.3 times the bearing, and 70% of the yield resistance is that spread rather than the bearing. The effective length is not a decision anybody made: it is what the flange's own bending stiffness against the web's own strength works out to. Internal forces

The support that is not a point

A reaction is drawn as a single arrow because the equilibrium equations only need its total. Underneath the arrow is a bearing of some width, delivering a pressure over that width, and almost everything a designer would like to know about the region near a support is a consequence of the width the arrow does not have.

 The neutral axis is wherever the first moment vanishes. A 300 by 600 section with 1800 mm² of steel at a depth of 540, carrying 250 kNm after the tension side has cracked. The neutral axis is no longer at mid-depth: it has risen to 234.5 mm from the top, which is where the first moment of the compression zone plus the transformed steel vanishes. The compression is 15.4 N/mm² at the top fibre and the steel carries 301 N/mm²; the resulting couple is 541 kN on a lever arm of 462 mm, which multiplies back to the 250 kNm applied. The uncracked section would have had 6673×10⁶ mm⁴ against the cracked 3809×10⁶ — a loss of 43% of the stiffness. Sections and stress

Where the steel is, not how much of it

A section in bending resists a moment with a couple, and a couple is a force times a distance. The force is bought — it is an area of steel at a stress. The distance is free, decided by where the bars were put, and it is the variable almost nobody optimises because it does not appear on an order.

 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 characteristic strength, which nothing was measured at. A lognormal population of strengths with a mean of 30 N/mm² and a coefficient of variation of 0.15. The characteristic value is the 5% fractile — 23.2 N/mm², which is 77% of the mean, and which need not be the strength of any specimen that was tested. Dividing it by 1.50 gives 15.5, and the shaded sliver below that is the fraction of the population that would fail to reach it: 6.4e-6, or one in 155,818. A factor applied to a fractile is not covering the scatter, because the scatter has already been spent getting to the fractile. Materials

The strength no specimen had

A material property is written into a calculation as a number, and a material does not have one. It has a population of strengths with a mean and a spread, and the number used is a low fractile of that population — a value that need not have been measured, that most of the material exceeds, and whose distance below the mean is decided entirely by the scatter.

 The middle third, computed. The kern of a 300 × 600 mm rectangle, computed by asking, for every direction, how far the resultant can move before the far face would be pulled. It reaches ±100.0 mm vertically and ±50.0 mm horizontally, which are h/6 and b/6 exactly, and the region between is a rhombus rather than the ellipse the two numbers suggest — a resultant on a diagonal has less room than either axis allows. Materials

The strength thrown away on purpose

Masonry, concrete and soil are all analysed as though they had no tensile strength whatever. Each of them has some. The decision to set it to zero is the single most consequential modelling assumption in the subject, it is safe for one kind of check and unsafe for another, and almost nothing that uses it says which.

 The tendon is a load, pointing the other way. A 14 m beam with a parabolic tendon dropping 260 mm to midspan, stressed to 1440 kN after losses. Its curvature pushes the beam up along its whole length with an intensity of 8Pe/L² = 15.28 kN/m, against an applied 17.63 kN/m — so 2.34 kN/m is left to bend anything, and the beam carries 57.4 kNm where an unstressed one carries 432 kNm. What the section then feels is 6.40 MPa of uniform compression and very little else. Internal forces

The load that comes from changing direction

A force that travels in a straight line asks nothing of anything. Bend its path and it asks for a transverse load of F over R along every millimetre of the curve, and that load is real, is nowhere on the load schedule, and is the same statement behind a prestressing tendon, a hoop force, an arch thrust and a web that buckles with nothing applied to it.

 The answer is continuous and the catalogue is not. Capacity bought against capacity required, over a real rolled series. The straight line is what a continuous section would give — exactly the moment asked for, and nothing can be bought on it. The staircase is what a catalogue gives: each tread is one section, each riser is the step to the next, and the vertical gap between the two is steel that is paid for and does nothing. The steps in this series run from 23% to 59% in plastic modulus, so the average waste is 15.0% and the worst is 46% — just above a riser, where the section below has been missed by a kilonewton-metre. The 1200 kNm marked buys a 686×254×125 at 1418 kNm, which is 85% utilised. Two things follow that a continuous treatment cannot see: the sensitivity of a design to an assumption is zero over most of a tread and enormous at a riser, and an optimisation that returns three significant figures is answering a question with twelve answers in it. Sections and stress

The answer is continuous and the catalogue is not

Every optimisation in this subject returns a number with three significant figures in it, and nothing with three significant figures can be bought. What can be bought is a rolled series whose steps are a quarter to a half apart, so the member that goes on the drawing is on average a tenth stronger than the one that was calculated and can be a third stronger for no reason at all.

 A section has two areas and the tables give one of them. Peak shear stress divided by the mean, for four sections of exactly the same gross area and depth. The mean is V/A and is the number a first calculation uses; the peak is what the material actually sees, and the ratio between them is a property of shape alone. A rectangle's is 1.5 — the parabola's peak over its average — and it is one of the few numbers in this subject that is exactly derivable and universally ignored. An I-section's is near 1.98, and the reason is on the second bar: 97% of the shear is inside a web that is 56% of the area. So the flanges carry the moment and almost none of the shear, and the web carries the shear and almost none of the moment — which is why a shear check on an I-section uses the web area and a moment check uses the whole section, and why the two checks are about two different pieces of steel. Sections and stress

The section has two areas

A shear force divided by the area of the section is not the shear stress anywhere in it. A rectangle's peak is exactly one and a half times that number and an I-section's web carries nearly all of the shear over a fifth of the area, which is why a moment check and a shear check on the same member are checks on two different pieces of steel.

 A bearing capacity is a mechanism, and here it is. Prandtl's collapse mechanism under a 3.0 m footing in a soil of 32° friction. A rigid wedge is driven down with the footing at 61° to the horizontal; a fan of radial shear turns the stress through exactly ninety degrees on a logarithmic spiral whose growth rate is tanφ; and a passive wedge at 29° has to be pushed up and out of the way. Nothing here is empirical — every angle is a function of φ alone — and the mechanism reaches 15.9 m from the centre, which is 10.6 times the footing's half width. That is why two footings closer together than about four widths do not have separate bearing capacities. Equilibrium

The ground is a mechanism

Bearing capacity is met as a formula with three terms and a table of coefficients, and that presentation hides what it is. Underneath is a plastic collapse mechanism — a rigid wedge, a fan of radial shear on a logarithmic spiral, and a passive wedge that has to be pushed up and out of the way — and every coefficient in the table is a property of that one drawing.

 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.

 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.

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