Structural Engineering & Statics
where the load goes
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The load must go somewhere — page 12

Essays 265 to 288 of 450 on this thread, in the same order.
The push that tips it, in every direction. The tipping push in each plan direction, drawn as a distance from the centre, for a 6000 kN body with its weight 15 m up and the push at 18 m. On rigid ground the curve is the hull's: weakest along the axes at 1333 kN. On four equal pads it shrinks, and more toward the corners. With the pads as built — stiffnesses 20000, 20000, 8000, 20000 kN/m — it is no longer symmetric: the weakest direction is 15°, on the soft pad's side, at 1039 kN. Equilibrium

It tips inside its own hull

On rigid ground a body tips when its resultant reaches the edge of its base, and how stiff its supports are has nothing to do with it. On pads that settle, the body leans as it is pushed, the lean moves its weight, and the push that tips it falls by one number a site engineer already has: settlement times the height of the weight, over the square of the half-width. Toward a corner the loss doubles, and one soft pad makes the weakest direction one nobody checks.

 The section, once the joint has supplied the fourth equation. Everything above a cut through panel 3. Four severed members; three equations. Moments about the mid-joint of the horizontal remove both diagonals and leave both legs. Moments about the point where the legs' lines meet, 36.0 m up, remove both legs and leave both diagonals. With the joint's result that the diagonals are equal and opposite, that second equation has one unknown: 20.83 kN in each diagonal. The legs follow: 26.1 T and 86.3 C. Equilibrium

The cut that needs a joint first

The method of sections works because a cut through three members leaves three unknowns and a point about which two of them have no moment. A K-braced tower has no such cut anywhere: every section severs two legs and two diagonals. One joint in the middle of a horizontal supplies the missing equation, and only in that order does each step have one unknown — after which the diagonals turn out to be carrying not the shear but the moment about the point where the legs would meet.

 Capacity against load direction, layout by layout. The same capacities as a function of direction on straight axes, for a load through (150, 150) mm. Three rows of two is weakest at 138°, 135.8 kN. Two rows of three is weakest at 312°, 135.8 kN. A ring of six, r = 70.4 mm is weakest at 132°, 150.3 kN. The curves cross: no layout is strongest in every direction, and the one to choose is the one whose lowest point is highest. Connections

The bracket pushed from the wrong side

A six-bolt bracket checked for a load straight down carries 198.5 kN. Push the same load through the same point at 138 degrees and it carries 135.8 — the direction changes the torque as well as the shear, and the worst direction is one no drawing shows. The capacity of a group whose load can turn is a closed curve, the bolt that governs it changes as the load turns, and what the curve rewards is not a larger polar moment but a smaller distance to the furthest bolt.

 The two methods, direction by direction. Capacity against the direction of the load for three rows of two bolts of 100 kN each, loaded through (150, 150) mm. The elastic method bottoms out at 135.8 kN at 138°; the instantaneous centre at 168.3 kN at 136°, 23.9 per cent higher. Both spike to 600.0 kN where the load aims at the centroid and every bolt carries an equal share. The gap between the curves is not constant: it is widest where the elastic distribution is most uneven, which is the same place the group is weakest. Connections

The better method flatters the worse layout

The instantaneous centre method finds capacity the elastic method cannot see, and how much it finds depends on the layout it is applied to. On a six-bolt rectangle under a load of unknown direction it is worth 24 per cent; on a ring of six with a smaller polar moment it is worth 14. The two methods agree on which layout to choose and disagree by a factor of five about the margin — and the reason is that a layout the elastic method already likes is one with nothing left to redistribute.

 What each missing bolt costs. The weakest-direction capacity of six bolts at 75 by 75 mm, loaded through a point (150, 150) mm from the centroid, with the whole group and with each bolt in turn left out. The full group carries 135.8 kN. Leaving out bolt 3 leaves 90.2 kN, a loss of 34 per cent; leaving out bolt 2 leaves 131.1 kN, a loss of 4. One sixth of the bolts is not one sixth of the capacity, and which sixth it is matters by a factor of 10. The dashed line is the capacity a group that lost a proportional share would have, 113.2 kN. Connections

The bolt that was never fitted

A six-bolt bracket found with five bolts in it has lost a sixth of its fasteners and between four and thirty-four per cent of its capacity, depending which one is missing. The share is the smallest of the three things that changed: the centroid moves away from the gap, which lengthens the load's own lever arm, and the polar moment falls by more than the count does. The bolt whose absence costs most is not the bolt that governed the check.

 A frame bent by nothing at all. A portal of 8.0 m span and 5.0 m columns at EI = 20000.0 kN·m², with its right foot settled 10.0 mm and no load on it anywhere. The moment diagram is drawn on the members: −3.95 kN·m at the left foot, −3.95 at the left knee, −0.00 at the crown, 3.95 and 3.95 on the right. The settlement is drawn hugely magnified; at true scale it is 10.0 mm on an 8.0 m frame. The diagram is antisymmetric, the vertical force the settlement develops is 0.99 kN, and every one of those numbers is proportional to EI. Deflection

The load that is not a load

Settle one foot of a portal frame by ten millimetres and the frame develops moments with nothing applied to it anywhere. In the analogous column the case is simpler than a load case, not harder — the section carries no direct stress at all, and the whole answer is one bending stress. And it scales the wrong way: the moments are proportional to EI, so the stiffer the frame, the more a settlement costs it.

 The modes of five masses joined by springs and held by nothing. Five equal masses joined in a line by equal springs, with nothing holding them to the ground. The first four modes, at zero, 0.62, 1.18, 1.62 times the frequency of one mass on one spring. The first is every mass moving together with no spring stretched at all: a mode at exactly zero frequency, a real solution of the eigenvalue problem, which carries all of the mass and none of the strain. Every other mode has the ends moving against each other, and carries none of the mass under a uniform acceleration. Dynamics

The modes at zero frequency

A structure held by nothing — a span being launched, a segment on a crane, a pontoon — has a mode in which it moves as one body and stretches nothing, at a frequency of exactly zero. That is a real mode, not a glitch in the stiffness matrix. It carries every kilogram of the structure under a uniform acceleration and leaves the flexible modes none at all. A load that is not uniform is a different matter: pushed suddenly from one end, the structure has no static answer to give, only an acceleration with a vibration riding on it.

 Four ways to put six bolts in one plate. Six bolts inside a 150 × 150 mm field of bolt centres, no two closer than 60 mm, loaded through a point (200, 0) mm from the field's centre. The two-column layout carries 193.5 kN complete and 138.8 kN with its worst bolt missing. The ring carries 167.2 kN complete and 117.8 kN with its worst bolt missing. The strongest found carries 234.3 kN complete and 149.7 kN with its worst bolt missing. The most robust found carries 231.6 kN complete and 174.7 kN with its worst bolt missing. The layout found by maximising the complete capacity and the layout found by maximising the worst omission are different layouts, 1 per cent apart when complete and 17 per cent apart with a bolt missing, and both beat the ring — the most evenly spread of the four — on both counts. The dashed line runs from each group's centroid to the load: 200 mm, 200 mm, 180 mm, 188 mm. The ringed bolt is the one each group can least afford to lose. Connections

The strongest layout leans on one bolt

Search a plate for the six bolt positions that carry most and the answer carries 234 kN — and loses 36 per cent of it if one particular bolt is missing. Move that one bolt fifty millimetres, into the corner the optimum had just left, and the group carries 232 kN and loses 25 per cent whichever bolt goes. Robustness here costs one per cent of strength, and a search for strength alone will never find it.

 A tubular K joint with a 12.6 mm gap. A 168.3 × 8.0 mm chord with two 114.3 × 6.3 mm braces at 45°, their toes 12.6 mm apart on the chord's crown. The braces' working lines, carried down through the chord, meet 3.0 mm below the chord's axis — an eccentricity of 0.02 chord diameters, inside the band of −93 to +42 mm in which the joint rules let its moment be neglected. Nobody chose that number. It follows from the gap, the brace diameter and the angle, and the gap was chosen so both toes could be welded. The working lines would meet on the chord's axis at a gap of 6.7 mm, which is a joint no rule permits. Connections

The joint whose lines may not meet

In a welded tubular truss the noding eccentricity is not a fabrication error. It is fixed by the brace diameter, the brace angle and the gap between the toes, and the gap is chosen for the welder. For a third of ordinary proportions the one gap at which the working lines would meet lies in a band no fabrication rule permits — so the concurrent joint the truss was analysed with is the one joint that cannot be built.

 Two flanges, one bolt, one prying force. An end plate 18 mm thick with its bolt 45 mm from the beam web, bolted to a column flange 14 mm thick with the same bolt 30 mm from the column web; their tips are 35 mm beyond the bolt and bear on each other, so the prying force there is one force acting on both. Checked as two separate tee stubs on rigid bases, the end plate carries 109.3 kN per bolt (mode 2, web hinge and bolt) and the column flange 104.4 (mode 1, both hinges), so the component method gives the joint 104.4. The pair carries 92.3 kN, 12 per cent less, with the prying force at 44.7 kN and hinges in the end plate at its web and the column flange at the bolt line — the end plate's web hinge working with the column flange's bolt-line hinge, a mechanism that exists in neither tee stub on its own. Connections

The hinge that forms in the other flange

An end plate bolted to a column flange is two tee stubs sharing one bolt, and the force at their tips is one force pressing on both. Checked separately, as the component method checks them, the end plate carries 109 kN a bolt and the column flange 104. Together they carry 92 — by a mechanism with one hinge in each flange that neither tee stub has on its own.

 The vertical response on layers of different depth. The vertical amplitude of a 150 tonne block 5.0 m across on soil with a shear wave speed of 178 m/s, under a harmonic force of fixed size, as a multiple of its static deflection on a half-space, against frequency. On a half-space the mode is at 12.5 Hz and the peak is 1.10. On 10.0 m the mode is at 14.3 Hz against a cut-off of 7.4 and the peak is 1.03. On 6.0 m the mode is at 15.5 Hz against a cut-off of 12.3 and the peak is 1.74. On 4.5 m the mode is at 16.3 Hz against a cut-off of 16.4 and the peak is 5.84. The layer stiffens the footing and lifts the frequency a little; what changes the picture is the cut-off, which rises as the layer thins and, once it passes the mode, takes the radiation damping away and leaves only the soil's own 5 per cent. Radiation below the cut-off is taken as nothing and its rise above it as linear, which is the usual idealisation of the exact layered solution. Dynamics

The rock that sends the waves back

A machine block on the ground is damped by the waves it launches, and a half-space lets them all escape. Put rock four and a half metres down and none of them can: below the soil layer's own lowest frequency there is no wave that travels, and a vertical mode damped at 47 per cent of critical keeps the soil's own 5. Its resonance grows fivefold, and the heavy block the textbook rule recommends is the one that loses its damping on the deepest ground.

 Thirty years later, two concretes against one. Stress down the composite section after thirty years — a 160 mm slab cast 6 weeks after a 700 mm pretensioned beam — computed twice: with the slab as a second, younger concrete that creeps and shrinks by its own laws, and with it given the beam's concrete and age. With two concretes the slab ends at −0.15 N/mm² at its top and −0.30 at its bottom, the beam at −4.03 at its top and −7.43 at its soffit. With one, the slab is at −0.29 and −1.06, the beam at −2.22 and −8.37. The slab's own shrinkage has taken its compression away and handed it to the top of the beam, and the soffit — the fibre the prestress was designed to keep in compression — has lost 0.94 N/mm² of it. Compression is negative. Materials

The slab that shrinks onto a finished beam

A precast beam with an in-situ slab cast on it is one member made of two concretes, and they do not age together. The slab's shrinkage is nearly the same whenever it is poured; what changes is how much shrinking the beam has left to share it with. Cast the slab at six weeks and it takes a ninth of the soffit's precompression away over thirty years. Cast it at a year and it takes a quarter, and goes into tension itself.

 One volume of steel, divided three ways. A Pratt truss of eight panels at a depth of 1, carrying 10 kN at each top joint, drawn three times with every member as wide as its area, the total volume the same in each. With equal areas the mid-span deflection is 43503; fully stressed, with area in proportion to force, 32702; with area in proportion to the square root of the product of each member's real and virtual forces — the division that makes mid-span as stiff as this steel can make it — 31318. The fully stressed truss is 4 per cent short of the stiffest; the equal-area truss is 39 per cent short. No member is allowed less than 10 per cent of the equal area; the deflections are in units of load × length / (E × volume). Deflection

The truss that is stiff by accident

Give a truss a fixed volume of steel and ask how to divide it among the members. Sized for strength — every member at the same stress — its mid-span deflection comes within four per cent of the stiffest that steel can make, although stiffness was never asked about. The reason is an inequality, and the same inequality says where the accident stops: at the quarter point the strength design is sixty-nine per cent short of the best.

 The toe that runs out of ground. A block 6.0 m square on ground that bears 300 kPa at most — 10800 kN over the whole base — pushed 8.0 m up, weighing 3000 kN, drawn at three stages of the push with the lean exaggerated eight times and the contact pressure under the base; the dashed line is the ground's bearing limit. At a push of 629 kN it touches the ground over 3.9 m of its 6.0 and has not yet brought the ground to its limit anywhere. At a push of 770 kN it touches the ground over 2.5 m of its 6.0 and has brought the ground to its limit under 0.8 m. At a push of 776 kN it touches the ground over 2.3 m of its 6.0 and has brought the ground to its limit under 1.1 m. The last is the tipping push: the heel has lifted, the toe is pressing the ground at its limit over a widening block, and the lever arm of the weight about that block is shrinking as the block grows. Equilibrium

The ballast that helps it over

On rigid ground a heavier block is harder to tip, in exact proportion to its weight — that is the whole of the overturning check. On ground with a bearing capacity it is not. The toe presses the ground to its limit, the weight's lever arm shrinks as the yielded block under the toe grows, and the tipping push peaks when the block weighs half of what the ground under it can bear. Past that, every tonne of ballast added to steady it brings it closer to going over.

 The strength design, and the least steel that is stiffer. A Pratt truss of eight panels at a depth of 1, carrying 10 kN at each top joint, each member drawn as wide as its section from a catalogue whose sections step by 25 per cent in area, the smallest 10 per cent of the largest member's need. Above, every member at the smallest section that carries its force. Below, the least steel that makes mid-span 1.50 times as stiff with every member still strong enough — the discrete optimum — with the 24 members it made larger drawn in the second colour: six of six top chord members, eight of eight bottom chord members, six of eight diagonals, four of seven verticals. The optimum uses 44 per cent more steel than the strength design, and it puts it where a member's real and virtual forces are both large; members either force leaves small are not touched. Deflection

The calculus answer, rounded, is the worst one

A real truss is built from a catalogue, and every member is rounded up to the next section. That rounding costs its stiffness almost nothing: the few per cent that separate the strength design from the stiffest survive it. What does cost is the next step. When a deflection limit governs, the obvious move — take the continuous optimum and round it up — needs more steel than any other way of stiffening the truss, and the exact discrete answer is within one per cent of a bound no catalogue can beat.

 The rotation a bearing sees is made before it arrives. The end rotation of a 12 m pretensioned beam, 300 mm wide and 700 mm deep, with a 160 mm in-situ slab 1.2 m wide, in 60 per cent humidity, from the prestress's release at three days to thirty years, with sagging positive. Release turns each end 4.8 milliradians upward; creep takes it to 7.9 by day 28, when the beam is set on its bearings. The slab brings it back by 0.9, the surfacing by 0.1, and thirty years of creep take it to 8.1 upward. The band after erection is the imposed load (0.9 down) and a night with the top cooler (0.6 down) above the line, and a sunny day with the top warmer (1.0 up) below it. Every value in the band is below zero: under every load it will carry, the beam's ends still point up, and a level bearing is turned the same way for its whole life. Deflection

The angle made in the casting yard

A bridge bearing is designed for the rotation of the beam it carries, and the rotation is listed as a sum of load, temperature, creep and a tolerance. Followed through the life of a pretensioned beam, the largest term is none of those. It is the upward turn the prestress gives the beam's ends in the casting yard, before any bearing exists, and under every load the beam will ever carry its ends still point up.

 The thin plate straightens onto the load line. The centre lines of a 10 mm plate lapped on a 20 mm plate, 100 mm wide, over 60 mm, carrying 60 kN with the grips 1200 mm from the overlap, over 480 mm either side of the overlap, with every vertical distance drawn 14 times the lengthwise scale; dotted, where the centres would be if the joint could not rotate. The load pulls along the straight line between the grips, each of which holds its plate on its own centre, and the moment where a plate enters the overlap is the load times the gap between that line and the plate's centre. The thin arm, flexible under tension, has swung toward the line: its gap is 0.49 of the rigid one, e/2. The overlap has turned with it, and the thick arm, too stiff to follow, is left 1.34 of e/2 from the line. Peak stresses: 193 N/mm² in the thin plate and 121 in the thick one, against 330 and 98 if the joint could not turn. Connections

The thicker plate takes the bend

A lap joint's eccentricity puts a moment into both plates, and the rule is that each takes half of the load times the offset. That is true only while the joint cannot turn. Let it turn, and a thin plate lapped on a thick one straightens under its own tension and hands its share across: the thick plate ends up bending a third more than the rule says, the thin one half as much, and in the limit the thick plate would take the whole of it.

 The forces the girder can hold with nothing on it. The one self-stress state of a K-braced girder of eight 3 m panels, 3 m deep: member forces in equilibrium at every joint with no load and no reaction. It lives entirely in the two middle panels — twelve members: the four diagonals meeting at the middle joint of the central vertical in tension, and the four central chord members and the half-verticals at the outer side of each middle panel in compression, at 0.89 and 0.45 of the diagonal force. It is symmetric about mid-span, so a symmetric load can call on it as freely as any other: symmetry does not supply the missing equation. Every other member of the girder is idle in it, drawn faint. Equilibrium

The redundancy only the sun can find

A K-braced girder that is symmetric about mid-span has one member more than statics can resolve, and it is at the centre, where two K's meet at one joint. Symmetry does not remove it, because the forces it allows are symmetric themselves. Loads barely touch it — a section calculation that ignores it is exact for a load at almost any joint. What finds it is a millimetre of misfit or a sunlit top chord, which put forces into the middle of the girder that no load calculation contains.

 The bow the frames have to hold. A chord 24 m long, EI 2.52 × 10¹² N·mm², held by U-frames every 4 m, carrying 930 kN critical at 1548 kN, drawn with its sideways movement exaggerated; its critical load is 1.66 times the chord force. Dashed, the worst of the bows tried: four half-waves, each 12.0 mm high, a five-hundredth of its own length. Solid, the chord once the compression has grown the bow; each arrow is the force the frame there must supply, the largest 13.6 kN — 1.47 per cent of the chord force. Stability

What the frames hold is the bow

A row of U-frames is sized for a force, and the rules give that force as a percentage of the compression chord's force that nobody derives. Derived, it is not a property of the frames at all. It is the chord's initial bow, grown by how near the chord works to buckling — and it is least, not greatest, near the stiffness at which the chord stops using the frames. Stiff frames tend to a limit that is π²/500 of the chord force: the familiar two per cent, arriving from a bow of one five-hundredth.

 A curved I-beam's flanges curl. The cross-section of an I-beam bent to a radius of 3.0 m, its flanges 300 mm wide and 15 mm thick, with the flanges' radial movement exaggerated 47 times; dashed, where the flanges would be if they did not bend across their width. It is bent the way that opens the curve, so the outer flange is in tension and the inner in compression. A tensioned flange round a curve is pulled toward the centre of curvature and a compressed one pushed away from it, so here both are pressed toward the web, and each outstand bends like a cantilever from the web, its tip moving 0.80 mm. Bent the other way, both would curl away from it by the same amount. The bars are the stress along the beam across each flange: the full value at the web, falling toward the tips to 0.72 of it, so that the flange works as if it were 89 per cent as wide. Sections and stress

The flange that curls away from its stress

Bend an I-beam into a curve and each flange, carrying its stress round the bend, is pressed sideways by that stress. The outstands bend like cantilevers from the web, and in bending they move to a different radius and shed the very stress that pushed them. At a tight radius only a strip beside the web works. But the flange's loss of width arrives slowly, and a stress nobody computes for a straight beam arrives at once: the flange bending across its own width, harder than it is stressed along the beam.

 The same bridge, anchored to the ground and to itself. The deck's bending moment under the live load, for a 500 m suspension span with a 50 m sag, a deck of EI 1.03 × 10¹² N·m², 100 kN/m of dead load and 20 kN/m of live load on half the span. Anchored to the ground, the deck carries at most 56.3 MN·m, because the cable's whole tension works on its deflected shape. Anchored to the deck's own ends, 83.0 MN·m — 1.47 times as much, and a little more than Rankine's elastic theory, 80.1, which leaves the tension out altogether. The deck's compression acts on the same deflected shape as the cable's tension, with the opposite sign, and cancels it. Structural form

The bridge that pays for its own anchorage

A suspension bridge's deck is light because the cable's tension works on its deflected shape and stiffens it. Tie the cable to the ends of the deck instead of to the ground and the deck must carry the same pull as a compression — which acts on the same deflected shape with the opposite sign and cancels the stiffening exactly. A self-anchored bridge is designed by the theory the long suspension bridge was invented to escape, and the price grows as the square of the span.

 The bolt that governs is worst at only one thing. The plate of a bracket of six M20 grade 8.8 bolts in three rows of two at 75 mm, loaded 150 mm to the side of the group and 150 mm in front of the column face, with the load at 60° from straight down, drawn at the group's capacity on the elliptical interaction, 227 kN. Each bolt's arrow is its shear in the plane of the plate and its disc grows with its tension, which comes from the load's distance in front of the column: bottom left 9 kN and 67 kN; middle left 33 kN and 76 kN; top left 74 kN and 84 kN; bottom right 40 kN and 35 kN; middle right 51 kN and 43 kN; top right 84 kN and 51 kN. The most shear is on the top right bolt and the most tension on the top left; the one that reaches its interaction first is the top left, and the next, the top right, is at 0.97 of its own. Connections

The bracket with no worst bolt

A bracket that stands out from its column is pulled off the column by the same load that twists it in its plane, so every bolt carries shear and tension at once. Straight down, one corner bolt is worst at both and the check is that bolt's. Tilt the load and the most sheared bolt and the most pulled bolt come apart; the one that fails is whichever of them its interaction reaches first, and at the angle where they swap the group has two worst bolts and no single one.

 The strain-energy route halves the sag. The midpoint deflection of a cable 10.0 m long, pretensioned to 60 kN, with an axial stiffness EA of 64 MN, loaded across its span at midpoint, against the load, found three ways. Solid: the deflection itself, from the cable's geometry; the dots, the derivative of the complementary energy with respect to the load, lie on it. Dashed: the derivative of the strain energy. Dotted: the pretension alone, as if the cable's tension did not grow. At 10.0 kN the cable sags 212 mm; the strain-energy route gives 107 mm, 0.51 of it, and the pretension alone 417 mm. At 20.0 kN: 294, 128 and 833 mm. The strain-energy route understates the deflection of a member that stiffens by the same mechanism that made it overstate one that softens. Deflection

The sag the strain energy halves

The derivative of the strain energy overstates the deflection of a member that softens. Turn the curve the other way up — a cable that grows stiffer as it sags, a hanger that takes up its play — and the same derivative understates it: by half for a pretensioned cable carrying a modest load, by two thirds for one with no pretension, and by the whole of the slack for a member that has any. The error is one ratio in both directions, and the least-work shortcut built on it hands the load to the member that stiffens.

 The shear centre waits for a thick plate. Where the shear centre of a channel 300 mm deep with 100 mm flanges, all 10 mm thick sits, measured from the web towards the toes, against the thickness of a plate closing the toes as a fraction of the channel's own, on a logarithmic scale. Open, it is 33.3 mm outside the web. Closed by a plate as thick as the rest it is at the middle, 50.0 mm inside the web, towards the toes. With a plate a thousandth as thick it has moved 0.3 per cent of the way; a hundredth, 3 per cent; a tenth, 23 per cent. It is half way at 0.28 of the thickness and crosses the web at 0.22. Sections and stress

The plate that closes the circuit

Close a channel with a plate across its toes and its shear centre is expected to come in from the air beside the web to the middle of the box. It does — once the plate is about a quarter as thick as the channel. A plate a hundredth as thick moves it three per cent of the way while multiplying the torsion constant by eight, so the load keeps its lever, the section stops twisting, and the whole torque goes round the circuit as a shear flow through the thinnest plate in it.

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