Structural Engineering & Statics
where the load goes
The collection

Every essay — page 15

Essays 337 to 360 of 376, in the same order.
Two diagrams for one load, and the second one has no straight-beam ancestor. Bending moment and torsion round a 90° arc of radius 6 m under a uniform load, built in at one end. The bending peaks at 576 and the torsion at 329, 57% of it. Both are zero at the free end and largest at the support, which is where a curved cantilever's bearing has to hold a torque it was probably not asked for. Internal forces

The torque that has nowhere to go

A curved beam on two supports splits its torsion between them, and the two halves cancel at mid-span. A curved cantilever has one end, so every increment of torque accumulates toward it — and the largest action at the root of a curved balcony is one that a straight beam does not have at all.

7 figures · Curved in plan More steel across the crack, until the roughness runs out. Shear resistance of the interface against the reinforcement crossing it. The steel clamps rather than carries, so the resistance is the clamping stress times the interlock coefficient and rises in a straight line — until the asperities crush at 5.50 N/mm², which happens at a reinforcement ratio of 0.79%. Past that the line is flat and every further bar is decoration. The dashed line is what the clamping alone would give if the concrete were unbreakable. Internal forces

Two models of one bracket

A corbel can be designed as a plane that has to be clamped or as a truss that has to be drawn, and the two are not approximations of each other. They describe different failures, they ask for steel in different places, and the honest answer is that both are checked because neither bounds the other.

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

The moment that was shed has to land

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

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

The tendon that can be moved

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

7 figures · Secondary prestress The studs are evenly spaced and the demand is not. The force per unit length the shear connection carries along half of a 12 m composite beam, from Newmark's solution. It is largest at the support — 282 N/mm — falls to nothing at mid-span, and averages 156: the end studs are asked for 1.81 times the mean. Studs are nevertheless placed at a uniform spacing, and the justification is the one the variable-angle truss uses for its stirrups — a ductile connector sheds what it cannot carry to its neighbours, so the uniform distribution is a plastic redistribution and not a description of the elastic state. Internal forces

The connection is busiest where the beam is not

A composite beam's studs are spaced evenly along it and the demand on them is not even at all. It peaks at the supports, where the bending stress is nothing, and falls to zero at mid-span, where the section is working hardest — so the connection is designed from a diagram nobody looks at.

7 figures · Composite action The second moment of area is a function of direction. Second moment of area of an equal angle against the angle of the axis it is taken about, with the product of inertia beneath it. The maximum is 5.943 × 10⁶ mm⁴ and the minimum 1.523 × 10⁶, a ratio of 3.90, and they occur where the product of inertia passes through zero — at 45.0° from the drawn axis. The value the drawing suggests, 3.733 × 10⁶, is neither of them. Sections and stress

The axis a column buckles about

A strut buckles about the axis with the smallest second moment of area, and for a section with no axis of symmetry that axis is neither of the two on the drawing. An angle used as a strut is 2.45 times weaker than the number a designer reads off its own dimensions.

7 figures · Principal axes The shear centre of a channel. A channel of 100 by 250, with the shear flow in its flanges drawn. Those flows form a couple, so the load has to be applied 38.0 outside the web to leave the section untwisted — a point in the air, outside the material entirely. Sections and stress

The eccentricity a purlin cannot avoid

A channel's shear centre is outside the material, so a load applied anywhere on the section misses it. The distance is fixed by the proportions rather than by the detailing, it is 38 mm on an ordinary purlin, and the torque it produces is not an error anybody made.

7 figures · Shear centre The lag follows the shear, so it is worst at the supports. Effective width along the span of a simply supported beam under a uniform load, summed over 25 odd harmonics. Each harmonic has its own half-wavelength L/n and its own, smaller, effective width — 0.815 for the first, 0.400 for the third, 0.269 for the fifth — and near a support the short harmonics carry a bigger share of what little moment there is. So the working fraction is 0.603 at the support against 0.839 at mid-span, a difference of 23.6 percentage points on the same flange. The single sinusoid's answer, 0.815, is drawn as the flat line, and it is only right at mid-span. This is the behaviour a code reproduces by shortening L_e near a support, and the reason it does is that shear lag is driven by w‴, which is the shear force. Sections and stress

The flange works least where the shear is largest

Shear lag is driven by the shear force rather than by the moment, so the effective width of a wide flange is not a property of the beam. It is a function of position along it, worst at the supports, and a single number quoted for a whole span is right at mid-span and nowhere else.

7 figures · Effective width One point, every plane through it, one circle. A point carrying 180 N/mm² across one face, 90 across the other and 40 of shear. As the plane is turned, the pair (σ, τ) runs round a circle of radius 60.2 centred at 135.0 — and it goes round at twice the rate the plane does, which is the part always misremembered and the part that makes the picture work. The principal stresses are 195.2 and 74.8, on planes 20.8° from the face the 180 acts on; the largest shear on any plane is 60.2, exactly the radius, and it sits 45° from those — which is 90° round the circle. The von Mises stress that ranks this state against any other is 170.6. Sections and stress

The circle nobody draws

A plane stress state has three principal stresses and the third is zero. When the two on the drawing share a sign, the largest shear in the state involves the one that is not there — and the circle a designer has drawn is not the circle that governs.

7 figures · Principal stress A short timber beam is a shear problem, and a steel one never is. Utilisation of the bending and shear checks on one beam, against span-to-depth. The two cross where the ratio equals f_m ÷ f_v exactly — no load, no width and no span survives the cancellation — which for this timber is 6.7 and for steel is 1.73. So a timber beam shallower than about six times its depth is governed by shear parallel to the grain, and a steel beam would have to be shorter than twice its own depth before the same thing happened, which is not a beam. The third check is bearing across the grain, which does not move with the span at all: on the beam drawn it is at 0.40, and it is the one that governs. Sections and stress

The shear that decides a timber beam

A steel beam is never governed by shear, because its bending strength is only 1.73 times its shear strength and no beam is that short. Timber's ratio is 6.7 along the grain and 23 across it, so shear governs at proportions people build every day.

7 figures · Shear flow A 10 mm plate, and the width it can be. The elastic critical stress of a plate in compression against its width, with the yield stress drawn across it. Below 462 mm the plate reaches yield before it buckles; above it the plate ripples first, and the fraction of the width still carrying load falls away — at 900 mm only 46 per cent of it is still working. Stability

The coefficient that is not four

A plate's buckling stress carries a coefficient that looks like a constant and is not. It is 4 for an internal element, 0.43 for an outstand and 23.9 for a panel in shear — and the width a 10 mm plate may be runs from 152 mm to 1,130 across that range.

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

Counted, not checked

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

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

The pinned base that is not pinned

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

7 figures · Sway stability Warping stiffens a short member and nothing at all a long one. The stiffening 1/[1 − tanh(κ)/κ] against kL, both axes logarithmic, over kL from 0.05 to 200. At the low end the curve is a straight line of slope −2, because for small kL the bracket is κ²/3 and the stiffening is 3/kL²: it reaches 1201 at kL = 0.05, falls to 1.005 at the top, and every open section ever rolled sits somewhere on it. The same three plates arranged three ways are marked: the 533 by 190 mm I-section at kL 2.76 and ×1.562, the tee at kL 37 and ×1.027, the angle at kL 34 and ×1.030. A tee's warping constant is 288 times smaller than the I-section's and an angle's 236 times, because their plates meet at a point and there is no pair of flanges to bend against each other — so they have no warping resistance to offer at all, and that is the reason an angle is a poor thing to twist. Stability

The restraint that beats the gradient

A moment-gradient factor is worth up to 2.7 on a beam's critical moment and is tabulated everywhere. Holding the ends against warping is worth more, is achieved by a detail rather than by a load case, and appears in no table at all.

8 figures · Moment gradient A channel has three critical loads, not one. The three critical loads of a channel in compression, against its length, with the load it actually buckles at drawn over them. At 3500 mm the flexural loads are 13970 kN about the major axis and 2306 kN about the minor, while twisting about the shear centre takes 1555 kN. The lowest root is 1484 kN, and the column twists. The shear centre sits 109.7 mm from the centroid, so the modes cannot happen separately: the lowest root of the coupled problem is 4.5 per cent below the lowest of the three, and the Wagner coefficient β is 0.60. The governing mode changes at 5813 mm: below that length the column twists, above it, it bends — because the torsional resistance keeps a term that does not grow when the member is shortened, and the flexural loads have none. Stability

The third root of the cubic

A column has three buckling loads and an Euler calculation finds two of them. The third is a twist about the shear centre, and for a section whose shear centre is not at its centroid the three cannot happen separately — so the answer is the lowest root of a cubic and can be a third below anything the two familiar modes report.

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

An influence line is a deflected shape

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

7 figures · Reciprocity Every member's share of the movement, and they are not the members expected. A Pratt truss of eight panels at a depth of 1, carrying 15 kN at each top node, with the movement of the bottom chord at mid-span attributed member by member. The unit-load sum δ = ΣF·f·L/EA gives 2019.41 at EA = 1: 49.4% from six top chords, 30.8% from eight bottom chords, 16.8% from eight diagonals, 3.0% from seven verticals. The single worst member is a top chord at mid-span at 11.9% of the whole. Each member is drawn at the width of its own share. The same deflection from a stiffness solution that shares none of this arithmetic is 2019.41, a relative residual of 1.4e-14. Deflection

The member that is not worth stiffening

A truss's deflection is a sum of one term per member, and a term is zero whenever either force in its product is. A vertical carrying the whole of a panel load can contribute nothing at all to the movement — which a total can never show and a per-member sum shows nothing else.

7 figures · Truss deflection The shear deflection is not a correction. The share of a sandwich panel's deflection that is shear rather than bending, against how slender the panel is. A solid beam at a span-to-depth ratio of 20 spends about a per cent of its deflection on shear; this panel spends 25% at the same ratio, because its core is 2800 times softer in shear than its faces are in tension. At the 37 of the panel drawn it is 9%. The curve falls as the square of the span because bending grows as the fourth power and shear as the second, so the term that is negligible for a long panel is the whole answer for a short one. Deflection

The stiffness that belongs to the span

A section's flexural rigidity is a property of the section, and a member's is not. Once shear deformation is counted the effective stiffness contains the span, so the same panel is a different member at three metres and at six — and for a sandwich the correction is not a correction.

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

Why it converges, and how fast

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

8 figures · Moment distribution The crack length at which the strength stops mattering. Failure stress against crack length for a toughness of 100 MPa√m, with three steel grades drawn. The falling curve is fracture — Kc divided by Y times the root of pi a — and it does not know what the yield stress is. The horizontal lines are the grades. At 275 N/mm² the two cross at a crack 33.5 mm long; At 355 N/mm² the two cross at a crack 20.1 mm long; At 460 N/mm² the two cross at a crack 12.0 mm long. The stronger grade's transition is the shorter one — raising the yield stress does not raise the strength of a cracked member, it only shortens the crack that takes it away. At a working stress of 120 N/mm² the critical crack is 176.0 mm. Materials

Designed to be found in time

A fatigue design that promises a detail will not crack is making a claim about a hundred years of traffic. A damage-tolerant one assumes it will crack, and sets the inspection interval from how long a crack takes to grow from the smallest size anybody can find to the largest the section can survive.

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

The map with three regions

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

7 figures · Shakedown Yielding one way makes it easier to yield the other. Mild steel taken to a strain of 1.20% and then pushed back the other way. The stress falls by 550 N/mm² before it yields again, against a yield stress of 275 — the elastic range is twice the yield stress and not once it, which is the Bauschinger effect and is a consequence of the yield surface sliding rather than growing. Materials

Yielding one way, and then the other

A material that has yielded in tension yields earlier in compression than it did the first time, and by an amount that is exactly what makes its elastic range twice its yield stress rather than once it. That is a property no monotonic test reports and every reversing structure depends on.

8 figures · Ductility Three specimens cannot see the tail. The factor k applied to the sample's own scatter when a characteristic value is estimated from n specimens. With the scatter known in advance it is z·sqrt(1 + 1/n) and barely moves; with the scatter estimated from the same n results it is the Student t quantile instead, and it runs from 7.73 at two specimens to 1.73 at 30. At n = 4 the characteristic strength comes out at 23.8 N/mm² against 27.9 for a population known exactly — 15% lower, for a material that is identical. A small test programme does not report a worse estimate of the strength; it reports a worse strength. Materials

The strength that belongs to the test programme

A characteristic strength is a fractile of a distribution, and a distribution estimated from four specimens is not the same object as one known exactly. The same material tested four times reports a strength 15 per cent below what it reports when its scatter is known — and nothing about the material differs.

7 figures · Characteristic strength The stress that leaks away. A restrained shrinkage strain of 320 microstrain in concrete of modulus 34000 N/mm². Ignoring creep it produces 10.88 N/mm², which is above the tensile strength of 3.8 and predicts that every restrained concrete member ever cast has cracked. Counting creep by the superposition integral leaves 1.72 N/mm² after 55 years, and the one-line age-adjusted shortcut at the usually quoted ageing coefficient of 0.8 leaves 3.05. The two disagree — this creep function implies an ageing coefficient of 1.67, not 0.8 — and both are below the tensile strength, so the conclusion turns on counting creep at all rather than on how it is counted. Materials

The stress that leaks away

Creep makes a load's deflection grow and an imposed strain's stress shrink, and the second is why restrained concrete does not crack as often as an elastic calculation says. The same material property runs both ways, and which way it runs depends on whether the structure was given a force or a movement.

7 figures · Creep