Structural Engineering & Statics
where the load goes
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Geometry beats material — page 9

Essays 193 to 216 of 329 on this thread, in the same order.
Where the span moment and the support moment cross. Span moment and support moment against the joint's stiffness, for a 6 m beam under 30 kN/m. They move in opposite directions because they add to a constant — the simple-span 135 kN·m is fixed by statics and the joint only decides how it is split. They cross at 68 kN·m, where the joint is delivering 75 per cent of the fixed-end moment, and neither the crossing nor the fraction depends on the beam, the span or the load: it is the point where f·wL²/12 equals wL²/8 − f·wL²/12, which is f = 0.75 for every beam there has ever been. Connections

The joint that was chosen

A joint's stiffness decides how a beam's moment divides between its span and its supports, and the two add to a constant. So there is a stiffness at which they are equal, the beam is sized by the smaller of two numbers rather than the larger of one, and the design moment is half what a simple connection leaves behind.

 Four things a far end can be doing, and what each is worth. The rotational stiffness of a member at one end, for four conditions at the other, each drawn as the shape the member takes when the near end is rotated through one unit. They are the same expression evaluated four times — M = (2EI/L)(2θ_a + θ_b) — and the only thing that changes is what the far end is known to be doing. A held far end gives 4EI/L and carries over a half; a free one gives 3EI/L and carries over nothing; a far end rotating equally and oppositely gives 2EI/L and carries over minus one, which is what a symmetric structure does to a member crossing its axis; and a far end rotating equally and in the same sense gives 6EI/L and carries over one. None of the four is an approximation. Each removes a freedom that was going to be discovered by iteration. Deflection

Told what the far end is doing

Moment distribution discovers, cycle by cycle, that the pinned end of a beam carries no moment — a fact known before any arithmetic started. Telling it instead changes one stiffness from 4EI/L to 3EI/L and the work from thirty numbers to eight, for the identical answer. Cutting the beam on its own axis of symmetry gets it in two.

 The whole line works, and the axis is not a choice. A weld group 200 mm deep carrying 15 kNm about an axis in its own plane, together with 100 kN of vertical shear. The bending force per unit length runs linearly from 1125 N/mm at one extreme to the same the other way at the other, through zero at the group's own centroid — and the centroid is where the neutral axis is because a weld carries compression across its throat as readily as tension. The 250 N/mm of shear runs along the weld and is uniform over the whole 400 mm, so the two components are perpendicular to one another and are combined on the throat rather than added as vectors in the plane. Connections

The eccentricity at right angles to the drawing

A bracket's load stands off the plane of its welds as well as being offset within it, and the second eccentricity produces a completely different object — bending about an axis through the group rather than torsion about a point in it. The weld line has no compression zone to argue about, so its neutral axis is its own centroid, and the whole length works.

 Two ways for the same plate to fold. An end plate 200 mm wide with a bolt 45 mm from the web face and 55 mm from the edge, and the two families of yield line it can collapse along. The circle closes round the bolt and is 283 mm of hinge — a circle round the bolt. The straight pattern runs out to the plate's free edges and is 249 mm — hinges to the plate edges. The plate folds along whichever is cheaper, which here is the fan, and the 200 mm that comes out is the length of the equivalent tee stub — a dimension that is nowhere on the plate and is 100 per cent of its width. Connections

How much of the plate is bending

A tee stub is an object nobody builds, and the whole component method rests on replacing a real end plate with one. The length of the substitute is not a dimension of the plate — it is the length of the cheapest fold the plate can collapse along, and two families of fold compete for it on a criterion with no strength in it at all.

 Both checks pass, and the contact lets go. A contact pressed together by 1000, with μ = 0.4. Every tangential force the contact can supply lies inside a disc of radius μN = 400.0, because the friction law bounds the length of the force and not its components. The contact is asked for 300.0 one way and 300.0 the other way — 75% and 75% of the radius taken one at a time — and 424.3 together, 106% of it. Each one-direction check passes and the force does not fit: the square those checks describe reaches √2 times further at its corners than the contact can. Equilibrium

Seventy-five per cent each way

A contact asked for friction in two directions at once can supply a force of a certain length pointing any way it likes, so its limit is a disc and not a square. Two checks made one direction at a time, each passing at seventy-five per cent, describe a contact that has already let go.

 The resultant has left the base, and it tips. A body on three supports weighing 60 kN, pushed sideways by 16 kN at a height of 3 m in the plan direction 270°. The push moves the resultant of weight and push 0.80 m from under the weight, and the base — the convex hull of the supports, shaded — lets it go 0.75 m that way before the edge drawn heavy becomes a tipping line: a factor of 0.94, found both along the ray and by moments about that edge. The dashed rosette is the same reach in every direction, from 0.75 m toward the middle of the nearest edge to 1.50 m toward the furthest support. To hold it the support opposite the tipping edge would have to pull 1.3 kN, which a support standing on the ground cannot do, so it lifts and the body turns about that edge. Equilibrium

Half as far between the legs

A body standing on feet, legs or pads has for its base the polygon its supports enclose, and how far its weight can be pushed before it tips depends on which way it is pushed. A three-legged stand pushed toward the gap between two legs has exactly half the reach it has pushed toward one of them.

 Nine-tenths of a tightening torque stretches nothing. Where the torque applied to the nut of an M20 grade 10.9 bolt goes, against the coefficient of friction in its thread and under its nut, taken as equal. The bottom band is the thread's lead — the only part of the work that stretches the bolt — the middle band is friction in the thread and the top band is friction under the nut. At μ = 0.14 the lead takes 11% of the torque, the thread 39% and the nut face 50%. At μ = 0.06 the lead's share is 22% and at 0.24 it is 6%, so a coefficient nobody measured decides how much of a specified torque arrives in the bolt as preload. Connections

The torque that goes into the thread

A preload specified as a torque is a preload specified through two coefficients of friction that nobody measures. Nine-tenths of the torque on a bolt is spent turning against its own thread and the face of its nut, so a change in the grease moves the clamping force by half — and the one method that escapes it does so by yielding the bolt on purpose.

 The compression is under the flange, not at the edge of the plate. A 500 × 400 mm base plate, 20 mm thick, under a 260 mm column carrying 300 kN and 120 kN·m, with holding-down bolts 60 mm from the tension edge. A plate this thick can deliver bearing only 43 mm past the compressed flange, so the compression sits under that flange, 124 mm from the column's centre, and the lever arm to the bolts is 314 mm: the bolts carry 264 kN and the flange zone 564 kN. The dashed block is the rigid-plate answer, 54 mm deep at the plate's far edge with its resultant 223 mm out, which needs only 128 kN in the bolts. Both satisfy equilibrium; only one is a plate that can carry the pressure where it is drawn. Connections

The compression that stays under the flange

Put a moment on a base plate and a pressure-block calculation pushes the compression out to the plate's edge, where the lever arm to the holding-down bolts is longest. A plate that is not rigid cannot put it there. The compression stays within a few tens of millimetres of the compressed flange, the lever arm shrinks, and the bolts carry twice what the pressure block said.

 Each repair reaches the checks its plate touches, and no others. The four checks on a 457 mm beam coped 50 mm deep over 200 mm, carrying a reaction of 300 kN through three bolts, as utilisations, for the end as coped and with three repairs: an 8 mm doubler on the web, a 100 × 10 mm plate along the free edge, and both. As coped the utilisations are flexure 0.46, shear 0.41, tear-out 0.57 and local buckling 0.46. The doubler thickens the web, which is most of what a coped tee is, and lowers all four; the edge plate gives the tee back a flange and lowers only flexure and buckling, leaving shear and tear-out exactly where they were. What governs: as coped, tear-out at 0.57; doubler, tear-out at 0.30; edge stiffener, tear-out at 0.57; both, tear-out at 0.30. Connections

The repair that fixes the wrong check

A coped beam end that fails its checks is usually repaired by welding a plate along the edge the cope left, because the cope removed a flange and the plate puts one back. On ordinary proportions that repair restores the check that was not failing. The check that governs is carried by the web, and only a plate on the web reaches it.

 The whole deflected shape, found from the members' changes of length. A Pratt truss of eight panels at a depth of 1, carrying 15 kN at each top joint, at EA = 1, drawn as built and in its deflected shape, with every movement magnified the same number of times. The shape comes from Williot's construction and Mohr's correction: every joint's movement from the members' extensions alone, less the rigid rotation the supports forbid. The mid-span joint L4 moves down 2019.41 units, and the dots at every joint are a stiffness solution sharing none of that arithmetic, agreeing to 1e-14 of the largest movement. One construction gives all sixteen joints at once. Deflection

The drawing that is right except for a rotation

Williot's construction finds every joint of a truss from its members' changes of length alone, in one drawing — and puts the roller six thousand units off its support. The error is one rigid rotation, Mohr's diagram takes it away, and a drawing started from the member symmetry holds still never makes it.

 Soft in shear, the wall gives up moment to the span. The bending moment along a 300 × 600 rectangle fixed at its left end and propped at its right, under a uniform load, per unit load and span, at span-to-depth ratios of 4, 2, 1, and for the same beam treated as rigid in shear, dashed. Rigid in shear the wall carries 0.125 wL² and the prop 0.375 of the load. Counting the shear it deforms by, the wall moment falls to 0.119 wL² at 4, 0.105 wL² at 2, 0.070 wL² at 1, and the prop's share rises to 0.381, 0.395, 0.430. The load has not changed and the beam is no weaker: its forces have moved, because in a redundant beam they come from how it deforms. Deflection

The shear that moves the moments

In a statically determinate beam, shear deformation adds movement and changes no force. In a redundant one the forces come from how the beam deforms, so a stubby beam soft in shear carries less moment at its wall and more in its span — and the carry-over factor, the one half every hand method passes along, falls to nothing at φ = 2 and then changes sign.

 A room's fire has a peak and an end, and the furnace has neither. Gas temperature in a compartment with 200 MJ/m² of fire load per square metre of its enclosing surface, linings of thermal inertia 1160 J/m²s½K and a medium growth rate, for opening factors of 0.02, 0.04, 0.08, 0.14 m½, against the standard furnace curve, dashed. At 0.02 the fire peaks at 841 °C after 120 minutes and is back at ambient by 435; at 0.04 the fire peaks at 944 °C after 60 minutes and is back at ambient by 171; at 0.08 the fire peaks at 1048 °C after 30 minutes and is back at ambient by 92; at 0.14 the fire peaks at 900 °C after 20 minutes and is back at ambient by 37, having used its fuel before the growth limit, so the fuel rather than the air decided it. A larger opening lets more air in: the fire burns hotter and gets through its fuel sooner. The furnace curve is still rising when every one of them is out. Materials

The fire that goes out

The furnace curve rises for ever. A room's fire has a peak and an end, set by its window, its walls and its fuel, and a bigger window makes it hotter and shorter — which is worse for bare steel and better for protected steel. The protected member is hottest half an hour after the fire has started to die.

 Made from one side, the throat misses the load and its root takes the difference. Two plates 20 mm thick joined by a partial-penetration butt weld with 10 mm of throat, pulled apart by a force on their mid-thickness, with the stress across the throat drawn beside each joint as a multiple of the mean. Above, welded from one side: the throat runs from the top face down to 10 mm, so the force passes 5.0 mm below its centroid. The moment adds tension at the root and removes it at the face: the root carries 4.00 times the mean stress and the face −2.00, which is compression. The unfused 10 mm below the root is a notch, and it is on the tension side. Below, the same 10 mm of throat split between the two faces: the force passes through its centroid and every part of it carries the mean stress. Connections

The throat that misses the load

A partial-penetration butt weld's throat faces the load squarely, so a millimetre of it carries more than a fillet's, and in every grade what limits it is a cap on normal stress rather than the directional criterion. Made from one side, the throat sits off the plates' line of action. At half penetration it keeps a quarter of that capacity, and the stress the eccentricity adds lands on the root, which is a crack.

 The coefficient is an envelope, and its scallops are whole half-waves. The plate buckling coefficient against aspect ratio α = a/b. Each faint branch is one half-wave count m: k = (m/α + α/m)², a curve whose own minimum is exactly 4 at α = m. The plate takes whichever branch is lowest, so the answer is the bold envelope — four touching 4 at α = 1, 2, 3 and 4, with cusps between them at α = √(m(m+1)) = 1.41, 2.45, 3.46, 4.47, where the plate is indifferent between m and m+1 half-waves. The first cusp reaches k = 4.50 and every later one is lower — 4.17, 4.08, 4.05. Past α = 1 the envelope never exceeds 4.49, which is why the length of a plate drops out of a formula that is otherwise entirely geometry. Stability

Four was never a fact about plates

The coefficient every plate calculation starts from is quoted as 4, derived nowhere and remembered by everyone. It is the minimum of a quantity that has nothing to do with plates in it — and what the plate supplies is not the four but the restriction that produces the scallops around it.

 The coefficient a web gets depends on where its neutral axis is. The plate buckling coefficient for an internal element against ψ = σ₂/σ₁, the ratio of the stresses at the two edges of the panel. Uniform compression is ψ = 1 and k = 4; a gradient running from compression to zero is ψ = 0 and k = 7.81; pure bending is ψ = −1 and k = 23.92, six times the value a column's flange gets. The 1800 × 12 mm web drawn here starts at ψ = −1.000 and k = 23.92 and ends at ψ = −0.910 and k = 21.63, because losing width from the compressed half drops the neutral axis and deepens the compression zone. The curve is steepest exactly where a bending web sits, so a small movement of the neutral axis costs 2.29 of coefficient. Stability

Classified by a gradient it does not have

A web in bending is the one plate whose buckling coefficient cannot be looked up. It depends on the stress gradient, the gradient depends on where the neutral axis is, and the neutral axis depends on how much of the web the coefficient has just taken away — so the answer is a fixed point, and the calculation everyone does is its first term.

 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.

 A fixed arch, and the section it is a drawing of. A parabolic arch of 30.0 m span and 6.0 m rise, fixed at both springings, carrying 20.0 kN/m over the span. On the right the analogous column: the arch's own centreline drawn as a section of width ds/EI, so it is narrow where the arch is stiff. Its area is 32.99 and its elastic centre sits 3.86 m above the springings — 2.14 m below the crown and on no part of the arch at all, which is the point of it. The released moments loaded onto that section give a direct stress of 3054.0 kN·m and a bending stress whose gradient is the horizontal thrust, 375.0 kN. Deflection

The centre that hangs in the air

Cross's analogy turns a closed frame into a short column and reads its redundant moments off as that column's stresses. Nothing in it requires the frame to be straight. Run round a fixed arch, the analogous column's centroid is a point two thirds of the way up the rise, on no part of the arch at all, and the bending stress about it is the arch's horizontal thrust. For a parabola under a uniform load the answer it returns is a diagram of nothing, which is exactly right.

 An unsymmetric frame, and a section with a product of inertia. A portal of 8.0 m span whose columns are 5.0 m and 3.0 m, carrying 10.0 kN/m on its beam, and beside it the analogous column — the frame's own centreline at a width of 1/EI. The section is no longer symmetric about a vertical line, so it has a product of inertia of −18.4 against Ix = 31.6 and Iy = 168.0, and its principal axes are tilted 7.54° from the horizontal — the dashed pair through the elastic centre. A section with a product of inertia does not bend about the axis it is loaded about, so the thrust and the redundant shear come out of a two-by-two rather than out of two divisions. Deflection

The axes that have to be turned first

Make one column of a portal shorter than the other and the analogous column stops being symmetric about a vertical line. It acquires a product of inertia, its principal axes tilt seven and a half degrees, and the thrust and the redundant shear stop being two separate divisions. Using the elastic centre and nothing else — which is what the symmetric construction looks like from outside — reports 17.0 kN·m at the left foot where the frame carries 27.8.

 Every shape the analogy reaches, which is every single ring. Four frames of 10.0 kN/m on an 8 m span, each solved by the column analogy and each checked against a stiffness solution of the same frame: a portal, largest moment 40.6 kN·m, agreeing to 0.24 per cent; a pitched portal, largest moment 35.6 kN·m, agreeing to 0.15 per cent; a stepped frame, largest moment 46.6 kN·m, agreeing to 0.16 per cent; splayed legs, largest moment 16.9 kN·m, agreeing to 0.09 per cent. The diagrams are drawn normal to each member. Nothing in the method asks what shape the frame is: it needs an area, a centroid and three second moments of the centreline, and a polyline has all five however it bends. Deflection

Three, and what three is a property of

The column analogy works because a closed ring cut once has three redundants and a plane section has three stress resultants. That match is the whole method, and it is topological rather than geometric — a portal, a pitch, a step, a splay and a polygonised arch are all one ring and all exact. Add a second bay and the frame has six redundants with nothing to be a drawing of, and the outer ring on its own is out by 190 per cent.

 The magnification a crack sees is a ratio, not a depth. The stress magnification at a weld toe against the crack's depth as a fraction of the plate's thickness, on BS 7910's two-branch fit. It is a function of a/t alone, because a weld's own size scales with the plate it is on, so the elevated field is geometrically similar. The dots are the same absolute starting flaw of 0.20 mm in plates of 12, 16, 25, 40, 60, 80, 100 mm: the flaw does not move and its magnification runs from 1.81 to 3.50. A fixed flaw in a thicker plate is a smaller fraction of it, which puts it deeper inside the raised field rather than nearer the edge of it. Materials

The rule that points sideways

Every fatigue code puts the same detail in a thicker plate into a lower category, by a factor of (25/t) to the power 0.2, and explains nothing. It is a strange rule: a detail's strength made to depend on a dimension at right angles to the crack. Integrate a crack through a weld toe's own stress field and the rule falls out — same form, same sign, and an exponent of 0.13 against the design code's 0.2. Remove the toe's magnification and the effect reverses.

 A roof truss and its reciprocal figure. Left, a pitched roof truss of 8 m under three loads of 10.0 kN, its 13 members drawn in the colour of their force — 5 in tension, 6 in compression and 2 carrying nothing — with a letter on every region outside it between one external force and the next and a number on every cell inside it. Right, the force diagram: every lettered or numbered space is a point, every member is the line between the two spaces it separates, drawn parallel to the member and as long as its force, and every joint of the frame is a closed polygon. The eight joints and eleven spaces of the frame have become eight polygons and eight distinct points — fewer points than spaces, because some spaces land on one point, as the two either side of a member carrying nothing always do. Force times length adds to 190.0 kN·m over the tension members and 250.0 over the compression members, and the difference, −60.0, is fixed by the loads and where they act, whatever frame carries them. Each point was placed by crossing one member, and the 8 crossings not used to place anything all close to within 5e-15 kN. Equilibrium

Every space a point, every joint a polygon

A truss's force diagram is a second drawing of the truss in which the joints have become polygons and the spaces between members have become points. Maxwell showed in 1864 that the exchange runs both ways, so a designer can draw the forces first and ask what shape carries them. His theorem also says which frames have such a diagram at all, and the answer is a surprise, because it is about polyhedra.

 A mode that points wherever the asymmetry points. A square floor 24 m on a side, equally stiff in both directions, with a period of 0.80 s. Four copies are drawn on top of each other, each made stiffer by one part in 1,000,000 along a different direction: 0°, 20°, 45°, 70°. The first mode each one returns lies along 90°, 110°, 135°, 160° — at right angles to its stiffening, whatever the size of it — and the two periods differ by one part in 1,000,000. Four structures no instrument could tell apart have first modes pointing four different ways. The floor with no asymmetry at all has no first mode: every direction is one. Dynamics

Two modes that are really a plane

A building equally stiff in both directions has two translational modes with one period, and they are not a pair of shapes but a whole plane of them. The pair an analysis returns is chosen by asymmetries of a millionth, so any result that depends on the pair — a square-root combination, a comparison with measured modes — inherits a choice the building never made. Damping then decides whether the difference can be seen at all.

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