Structural Engineering & Statics
where the load goes
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The load must go somewhere — page 5

Essays 97 to 120 of 350 on this thread, in the same order.
An eccentric load is three load cases, and only two of them are checked. A line load of 40 N/mm at 1.5 m from the axis of a 3.0 by 2.0 m box, replaced by the three cases it is equivalent to. Bending is the load on the axis. The torque 60 kNm per metre then splits into a set of edge forces that drives Bredt's shear flow and distorts nothing, and a set with the flange forces reversed — 10.0 kN/m up one web and down the other, 15.0 kN/m across the flanges — which carries no torque at all and squashes the rectangle into the rhombus drawn behind it. Its generalised load is exactly half the torque, so a box girder spends half of an eccentric load's torsion on changing its own shape, and no torsion calculation contains that half. Sections and stress

The section that will not keep its shape

A box girder is closed, so torsion costs it almost nothing. What an eccentric load actually does to it is something a torsion calculation contains no term for — the rectangle becomes a parallelogram, in its own plane, along the whole length of the span.

 Pull it along the girder and it just unfolds. One period of a 30° trapezoidal corrugation, 300 mm of flat and 260 mm of incline, and the same period pulled along the girder's axis. The fold opens by bending the inclined panels out of the web's own plane, so the axial flexibility contains the plate's t³ where a flat web's would contain t — and the effective modulus that comes back from solving the cell as a frame is 222 N/mm², which is 10.6 parts in ten thousand of the steel's 210 GPa. A web with a thousandth of the stiffness carries a thousandth of the stress, which is why the flanges of a corrugated girder carry the whole moment and why the section has 9 per cent less second moment than the flat-webbed girder it replaces. The fold buys freedom from stiffeners and pays for it here. Sections and stress

The web that carries no bending

A corrugated web needs no stiffeners, because the folds give it in one direction a depth it does not have in its thickness. In the other direction the same folds make it an accordion — and a web that cannot be stretched cannot carry a bending stress at all.

 The corner columns take what the middle ones did not. Axial stress in the columns across one flange face of a 30 by 40 m framed tube, at the base. Plane sections says the flat line: every column on the face at the same distance from the neutral axis, therefore at the same stress. The solved distribution is the curve — 69.4 N/mm² at the corner against 18.2 in the middle, a ratio of 3.81. The middle columns lag because the only route the axial force has into them is the in-plane shear of the spandrel frame, bay by bay from the corner. The face is carrying its resultant on an effective width of 51 per cent, and the tube deflects as though its second moment were 72 per cent of the gross. Structural form

The corner columns take more than their share

A framed tube is a hollow cantilever, and a hollow cantilever's flange ought to be uniformly stressed. It is not, and the reason is that the only route the axial force has into a column in the middle of a face is the in-plane shear of the frame — one bay at a time, from the corner inwards.

 The tie is a redundancy, so its stiffness decides the thrust. Thrust and rib bending for a 60 m tied arch of 0.15 rise ratio, against the stiffness of its tie. Cut the tie and the structure is a curved simply supported beam, so the tie force is the one redundant and the force method gives it: with a rigid tie the answer is 2227 kN, within 1.0 per cent of the funicular wL²/8f, and the shortfall is the arch's own axial shortening. A real tie stretches 68 mm and returns 2166 kN — 2.7 per cent of the flexibility is the tie — and whatever thrust the arch does not get, it carries as bending: 760 kNm at 30 m. A tenth of the tie stiffness is not a tenth of the problem; it is a different structure. Structural form

The thrust that never reaches the ground

Every arch on this site has ended at the same sentence — the foundation is where an arch is really decided. A tie changes the sentence without changing the arithmetic: the horizontal force is still there, still the same size, and it now closes on itself through a bar at deck level.

 The windward guy tightens, the leeward one gives way. A 120 m mast on three guy levels, at a wind of 3 N/mm, with the deflection drawn 0.54 times its true size. The guys start at 160 kN each and end at 259 against 95, 299 against 83, 221 against 112 kN. The leeward guys still carry a real force — the lowest keeps 28 per cent of its partner's tension — and supply almost none of the restraint, because their tangent modulus has fallen to 54 per cent of the steel's. The mast top moves 100 mm, its worst bending moment is 554 kNm at 40 m, and it is carrying 801 kN of axial load that nothing but the guys put there. Structural form

Held by something that goes soft

A guy is a cable, so it has no stiffness of its own — what resists a mast's movement is the guy's geometry changing, and how much of that there is depends on the tension already in it. Wind pushes the mast towards the leeward guy, which is the one losing tension.

 Three ways to apply the same force, and one depth to forget the difference. Three end loads on a member 400 mm deep, all with the same resultant and the same moment: a point load, the same force spread over a fifth of the depth, and the same force split in two. What is plotted is the difference between each of them and the beam-theory answer — the self-equilibrating remainder — as a fraction of the mean stress. The point load starts at 20 times it and is under a tenth of it by 0.77 depths; all three are under one per cent by about 1.18. That distance is the licence every figure in this collection is drawn under, and the exact strip eigenvalue agrees with it: 2.106 + 1.125i, whose real part puts one per cent at 1.09 depths and whose imaginary part means the remainder changes sign on the way out, which no statement of the principle mentions. Internal forces

How far a wrong load reaches

Every figure in this collection applies a load as a point, a line or a uniform pressure, and no real load is any of those. The licence is Saint-Venant's, it is usually quoted as a principle, and it is really a statement about a wavelength.

 The same beam, the same load, and one of them has to lift it. Two lower-bound models of one beam, differing only in which chord the 600 kN is applied to. The chord forces are identical in both — the moment diagram does not know where the load arrived — and so are the struts. What is not identical is the vertical at the load: nothing in the top-loaded model, and the whole 600 kN in the hung one, because a bottom-chord panel point touches no strut and the load has nowhere to go but up. That tie is 1200 mm² of steel against the 96 mm² the shear calculation asks for over the same length — a factor of 12.5, in the same place, and additional to it. Internal forces

The load that has to be lifted

Every shear calculation on this site assumes the load arrives on the top of the beam and walks down a diagonal to the support. Hang the same load from the soffit and the diagonal cannot start — the load has to be carried up to the compression zone first, by a tie no sectional calculation contains.

 Half the beam does nearly all of the deflecting. The virtual-work integrand M·m/EI along the member, normalised to its own peak, with the running share of the answer beside it. The integrand is a density: it says how much of the deflection each millimetre of the beam produced. For this case the half nearest the root supplies 87.5 per cent of it, and the rest of the member supplies the remainder. Stiffening the busy 50 per cent by 1.5 times takes the deflection down by 29.2 per cent; the same material spent on the quiet end takes it down by 4.2 — a factor of 7.0 for the same steel. The map of what is contributing is not the map of where the moment is largest, and the second is the one that gets drawn. Deflection

Where a deflection comes from

The unit-load method gives a deflection as an integral, and this collection has treated that integral as a number to evaluate. It is not a number. It is a density, and it says which millimetres of the member produced the answer — which is not the same map as where the moment is largest.

 Strength at an angle, and the straight line that is not it. Compressive strength against the angle between the load and the grain. Hankinson's formula — f₀f₉₀ ÷ (f₀sin²α + f₉₀cos²α) — is an interpolation rather than a failure theory, and what makes it worth having is how far it sits from the straight line anyone would otherwise draw between 21 and 2.5 N/mm². At forty-five degrees it gives 4.5 N/mm² against the line's 11.8: 38 per cent of it, and 21 per cent of the strength along the grain. The curve drops away in the first twenty degrees because the weak direction starts governing as soon as it has any component at all, which is the same arithmetic as a section's weak axis and the reason a skewed bearing detail is a real loss rather than a small one. Materials

The material that has a direction

Every material in this collection so far has had one modulus and one strength. Timber has three of each, differing by more than an order of magnitude, and the consequence is not a correction to steel design — it is a different set of checks with a different one governing.

 The two theorems close on the answer from opposite sides. A pinned-base portal frame under 200 kN at mid-span and 80 kN at the eaves, with a plastic moment of 200 kNm. The three horizontal lines are mechanisms — work equations, each an upper bound: the beam mechanism gives 1.000, sway 1.250 and the combined one 0.714. The curve is the lower bound: for each value of the one redundant, the largest load factor whose moment field stays inside the plastic moment everywhere. Its peak is 0.714, and it touches the lowest mechanism exactly — so the collapse load is known rather than bracketed. Using the beam mechanism instead would have claimed 1.40 times the real capacity, and the sway one 1.75 times, both of them on the wrong side. Equilibrium

Two ways of being wrong

Plastic analysis has two theorems and they point in opposite directions. Any equilibrium field that nowhere exceeds the plastic moment gives a load at or below the collapse load; any mechanism gives one at or above it. Only one of those errors is safe.

 The bond stress is crowded against the loaded end. A 20 mm bar embedded 806 mm, with the force in it and the bond stress on it plotted along the embedment. Uniform bond — the assumption behind every development length ever tabulated — is a flat stress and a straight line of force. An elastic bond of the same peak strength is neither: the slip is largest where the bar is pulled and dies away over 1/α = 471 mm, so the far end of the bar is doing almost nothing. At the design rule's length of 40 diameters the elastic bond is 55 per cent used. The uniform answer is what the bond looks like after it has yielded along the whole length, which is a statement about ductility rather than about strength. Internal forces

The force that arrives along a length

A bolt takes its force at a hole and a weld along a line. A reinforcing bar has no such place — it is a smooth cylinder in a hole of its own shape, and the only thing stopping it sliding out is a stress smeared over its surface. So the force in it is not a number, it is a function of position.

 The chords take the shear the web is credited with. A cantilever of 6 m tapering from 400 to 1200 mm, under a 120 kN tip load, with the shear divided between the web and the two inclined chords. The chord force is M/z and it is not horizontal, so its vertical component is real: V_web = V − (M/z)·dz/dx, which is Résal's result of 1899. At the root the web is left with 33 per cent of the applied shear — exactly d₀/d₁ for a straight taper, with no length, load or material in it. Turn the same member round and the figure becomes 300 per cent: the chords stop helping and start adding, at the section where the moment is largest as well. And the limit is worth having — a member whose depth is proportional to its moment leaves the web nothing to do at all, which is the triangular cantilever every crane jib is. Internal forces

The shear the chords take

Every shear check in this collection has assumed the two chords of a beam are parallel, so that the whole of the shear crosses the web. Taper the member and that stops being true — and the sign of the correction is decided by which end the haunch is at.

 What is left after the first fibre yields, which is a property of shape. The shape factor — plastic modulus over elastic — for six sections, computed by finding each one's equal-area axis and summing ±f_y over it. The numbers contain no dimension, no stress and no material: a rectangle is exactly 3/2 whatever its size, a diamond exactly 2, a circle 16/3π. The spread is the argument. An I-section keeps only 13 per cent in reserve past first yield, because nearly all its material is already at the extreme fibre and there is nothing further in to recruit; a diamond keeps 100 per cent, because most of its material is near the middle and doing very little elastically. So the section shapes that are best at elastic bending are the ones with the least left afterwards, which is exactly backwards from the way the reserve is usually described. Sections and stress

What is left after the first fibre yields

The elastic section modulus stops at the moment the outermost fibre reaches yield. Nothing else in the section has, so it goes on taking load — and how much more it takes turns out to be a property of the shape alone, with no dimension, no stress and no material anywhere in the answer.

 One slit, and the torsional stiffness falls by a factor of hundreds. A 200 by 200 box of 8 mm wall, drawn closed and then slit along its length. Closed, the torque runs round the wall as a shear flow and the torsion constant is 5.66×10⁷ mm⁴; slit, the loop is broken and only each wall's own thickness resists, giving 1.31×10⁵ mm⁴. The ratio is 432 to one, so the same torque twists the slit section 432 times as far and raises a peak shear stress 36 times as high. Nothing about the material changed. Sections and stress

The slit that costs a factor of six hundred

Bending stiffness cares where the material is, and changes by a factor of two or three between sensible sections of the same area. Torsional stiffness cares whether the material forms a closed loop, and the penalty for not doing so is an order of magnitude squared.

 A fan, and where its forces go. Half a cable-stayed bridge: a tower 70 m above a deck, 12 stays reaching out over 200 m, and a uniform 200 kN/m on the deck. Each stay is drawn at a weight proportional to the force in it, from 3427 kN at the innermost to 10090 kN at the outermost — the outer stay carries the same vertical share and is far flatter, so it carries far more. The deck's shading is its own accumulated compression, 61905 kN at the tower, which is 1.55 times the load being lifted and is the horizontal half of every stay force added up. Nothing in this drawing is a catenary: every stay is straight and every one of them is a spring. Structural form

The cable that is a spring

A suspension bridge's cable is a funicular — it takes the shape the load gives it, and the deck's job is to make the load one that shape is right for. A stayed deck is not that at all. Its cables are straight, each reaches the deck at one point, and every one of them behaves as an inclined spring.

 Held up by a pressure nobody can feel. An air-supported roof of 60 m span and 9 m rise. The membrane has no bending stiffness whatever, so the only thing that can hold it in tension is a pressure difference, and the pressure has to exceed the load per unit plan area and nothing else: 0.25 kN/m² of fabric plus 0.6 of snow is 0.85 kN/m², so 1.19 kN/m² does it — 1190 pascals, which is 1.17 per cent of an atmosphere and 121 millimetres of water. Ears do not notice it. A door does: at 2.1 kN on an ordinary leaf, the building needs an airlock rather than a handle. And the whole of it arrives at the foundation as 3365 kN of uplift — 17.9 kN on every metre of perimeter — which is the bill the pressure's smallness conceals. Structural form

Held up by the air inside

A membrane has no bending stiffness at all, so the only thing that can hold it in tension is a pressure difference. The pressure needed to hold up a roof is smaller than the pressure a closed door makes — and the same pressure arrives at the foundation as hundreds of tonnes of uplift.

 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.

 Six ways for one dowel to fail, and the capacity is the smallest. Johansen's single-shear mechanisms for a 12 mm dowel through 40 and 40 mm members, each drawn as the shape it is: the dowel straight and the timber crushing, the dowel rotating rigidly, one plastic hinge, then two. The capacity under each is that mechanism's own, and the joint's strength is the smallest — 5.02 kN by mode c, which is the dowel rotates rigidly and both members crush. That is the kinematic theorem of plasticity: every mechanism gives an upper bound and the true collapse is the lowest of them. The crushed timber is shaded, and the circles are plastic hinges in the steel. Connections

The smallest of six failures

Everything else in this collection that fails does so in one way at a time. A dowel through timber does not — the wood can crush while the steel stays straight, or one plastic hinge can form in it, or two — and the capacity is the smallest of the six, which is the kinematic theorem of plasticity applied to a joint rather than to a frame.

 The worst speed is not the fastest one. Peak deck acceleration against train speed, for a 20 m span at 6.25 Hz under 10 axles 18 m apart. The spikes are not a numerical artefact and they are not about how heavy the axles are: a regularly spaced train is a forcing function with a frequency v/d, and where a multiple of it lands on the bridge's own frequency each coach arrives in step with the motion the last one left. The arithmetic is v = d·f₁/k, which puts peaks at 405, 203, 135, 101 km/h — all of them operating speeds. What fails first is the acceleration rather than any stress: ballast loses its interlock at about 3.5 m/s², and strength does not appear in the equation at all. Here the limit is first passed at 376 km/h. Dynamics

The train that arrives in time with itself

A single load crossing a span is a mild problem. A train is not one load — its axles are regularly spaced, so the forcing has a frequency of its own, and where a multiple of it lands on the bridge's frequency each coach arrives exactly in step with the motion the last one left behind.

 The free body that makes a hoop force a pressure times a radius. Half a ring cut along a diameter, with the pressure drawn normal to the wall wherever the wall is. Vertical equilibrium of the half ring is the whole derivation: the pressure acts over the projected width 2R whatever the shape of the arc, the two cut faces carry N each, so N = pR — 300 kN per metre here at 1 MPa on a 0.3 m radius. The result contains no wall thickness, no second moment, and no length along the pipe, which is why a hoop force is the one internal force in this collection that arrives with no lever arm attached to it. The stress does contain the thickness — 25 MPa at 12 mm — but the force does not, and a thicker wall carries exactly the same force at a lower stress. Internal forces

The force that is only a radius

Every internal force in this collection arrives with a lever arm attached. A hoop force does not. Cut a cylinder along a diameter and the free body settles it in one line — pressure times radius, with no thickness, no second moment and no length in it — which is why a tank wall is thin and why its worst hoop force is not at the bottom.

 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.

 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.

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