Structural Engineering & Statics
where the load goes
The collection

Every essay — page 19

Essays 433 to 456 of 728, in the same order.
A pin's force does not pass through its centre. Left, a pin of radius 0.15 m in its hole, with a coefficient of friction of 0.15. The reaction at the contact is inclined by φ = 8.5° to the radius through it, because the friction it can develop is that fraction of the force pressing the surfaces together, and the perpendicular distance from the pin's centre to that inclined line is R sin φ = 0.022 m. Every position the contact can take gives a line tangent to the same circle, shaded. Right, a link 0.48 m long pinned at both ends, at the same scale — 3.2 pin radii, which is a stubby linkage rather than a structural tie, drawn that way because at the thirty radii an ordinary tie has, the two circles are smaller than the pencil: its force is a common tangent to the two circles — the two solid lines for the two senses of rotation, the two dashed ones for the senses in which its ends turn oppositely — and the dashed centre line every construction in this collection draws is none of them. A pin of this size carrying 5.0 MN delivers a couple of 111.3 kN·m to whatever it is pinned to, which is the same offset read as a moment rather than as a distance. Equilibrium

The pin that is not a point

Every line of action drawn so far passes exactly through a pin's centre, which is true of a frictionless pin and of nothing else. A real one carries its force tangent to a small circle instead, so a link's line is a band, a construction's answer is a range, and a support drawn as a hinge hands a couple of a hundred kilonewton-metres to whatever it is pinned to.

7 figures · Graphic statics The humidity through a slab's depth, and the strain it asks for. A slab 200 mm thick drying through its top face only into air at 50 per cent, with its underside sealed by a steel deck. Left, the pore humidity against depth at 28 days — 50 per cent at the surface and 100 at the base; 90 days — 50 per cent at the surface and 100 at the base; 1 year — 50 per cent at the surface and 99 at the base; 5 years — 50 per cent at the surface and 74 at the base; 20 years — 50 per cent at the surface and 51 at the base. Right, the free strain each of those profiles asks for, taken as shrinkage at the local humidity. At 28 days the top wants to be 461 microstrain shorter than it was and the bottom 0; at 20 years the two are 461 and 456, and the gradient that produced the curl has gone. Not one of the profiles is straight, and a section that stays plane cannot deliver any of them. Materials

The slab that dries from one face

A creep coefficient is one number for a member, and a member drying through one face does not have one. Give every depth its own humidity and the section a strain profile it cannot deliver, and two things follow that no single coefficient contains — a six-metre slab lifts eleven millimetres at its edges with nothing on it, and its top surface is past cracking before it has been loaded.

7 figures · Creep Three moment diagrams for one pair of beams. Two 15 m spans carrying 12.0 kN/m, drawn sagging downward. As built they are simple spans: 337.5 kN·m at each midspan and nothing over the middle support. Built monolithic they would carry 337.5 kN·m of hogging over the support and 168.8 at midspan. Loaded at 28 days and made continuous at 60, creep takes them 46 per cent of the way from the first to the second: 155.2 kN·m over the support and 259.9 at midspan, after twenty years in which nothing about the loading changed. The support had no moment on the day it was cast and no drawing of the finished structure shows why it has one now. Materials

The support that had no moment when it was cast

Two beams are set on their bearings, carry their own weight for a month, and are then stitched together over the middle support. Nothing about the loading changes afterwards. Twenty years later the stitch is carrying 155 kilonewton-metres, because the concrete went on creeping and the joint would not let it — and how much arrives is decided by a crane schedule.

6 figures · Creep Where a tendon's force goes, over fifty years. The loss of stress in a tendon stressed to 1300 N/mm² and released at 7 days into a member of notional size 300 mm at 70 per cent humidity, with the three causes stacked. At 28 days the total is 61.9 N/mm²; at a year 131.5; at fifty years 190.8, which is 14.7 per cent of what the tendon started with. Creep supplies 107.7 of that, drying and autogenous shrinkage 50.6, and the steel's own relaxation 32.5. Half the loss has happened by 119 d and nine-tenths by 8 y. Materials

The prestress the member takes back

A tendon is stretched, locked off against the concrete, and then has to hold that extension while the concrete underneath it shortens by itself. Fifteen per cent of the force goes, most of it to creep, and how much goes is decided by the shape of the member and the air it stands in rather than by anything about the steel.

6 figures · Creep The column under the first interior support is given a quarter more. Two equal spans of 8.0 m carrying 5.0 kN/m, continuous over three supports. The tributary rule gives each interior support one span's worth of load, 40.0 kN, and each end support half of that. The continuous beam gives 15.0 kN, 50.0 kN, 15.0 kN — ratios of 0.75, 1.25, 0.75 to what the areas say. The reactions still add to the whole load, because they must; what has moved is which support gets it. The end supports are relieved because the span next to them hogs over the first interior support, lifting their end of it. Equilibrium

The column given more than its rectangle

The tributary rule draws a rectangle round each column and hands it whatever stands inside. A floor is continuous over its columns, and a continuous beam does not give each support the load above it — the first interior one takes a quarter more and the end ones a quarter less. On a grid the two directions multiply, and two columns on the same floor differ by a factor of nearly three.

6 figures · Tributary area One bracket, three neutral axes, three sets of bolt forces. A bracket 300.0 mm deep with three rows of two bolts at 75.0 mm pitch, carrying 15.0 kN·m about an axis in the plane of the bolts. Taking the axis at the group's centroid puts the outer rows at 50.0 kN of tension and 50.0 of compression, with the middle row idle. Taking it at the plate's compression edge puts every row in tension — 7.1, 14.3, 21.4 kN from the bottom up — with the top row at 21.4. Solving for it instead, with the plate bearing over 200.0 mm of width and the bolts as areas, puts it 40.2 mm above the edge and the top row at 24.6 kN. All three make the applied moment exactly. The top row differs between them by a factor of 2.33. Connections

The bolt group has no neutral axis

A bolt group carrying a moment about an axis in its own plane has some bolts in tension and something in compression somewhere else. Where that somewhere else is decides the answer, nothing in the group decides it, and the two defensible choices give the worst bolt 50 kilonewtons and 21.

5 figures · Bolt group A portal frame, and the section it is a drawing of. A single-bay portal of 6.0 m span and 5.0 m height with fixed feet, carrying 10.0 kN/m on its beam, with the beam's second moment equal to the columns'. On the right, the analogous column: the frame's own centreline drawn as a section as wide as 1/EI at every point, so it is narrow where the frame is stiff. Its area is 16.000 and its elastic centre sits 1.56 m below the beam — inside the frame, on no member at all. The released moments loaded onto that section give a direct stress of 33.8 kN·m and a bending stress whose gradient is the horizontal thrust, 6.4 kN. Together they give 10.6 kN·m at the feet, −21.2 at the knees and 23.8 at the crown, against a free moment of 45.0. The same frame solved by stiffness gives 10.6, −21.2 and 23.8. Deflection

The elastic centre is not on the frame

Cross's analogy turns a member fixed at both ends into a short column and its end moments into edge stresses. A closed frame's analogous column is the frame's own outline, its elastic centre is a point hanging in mid-air inside it, and the horizontal thrust a gravity load produces is that section's bending stress about the axis through that point.

5 figures · Moment-area 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.

7 figures · Plate buckling 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.

7 figures · Plate buckling Two collapses for the same plate, and only one of them fits. The same three bolt rows at 90 mm pitch, folding two ways. On the left each row folds on its own pattern — 249 mm of hinge line round each bolt — and the patterns overlap, because 249 mm of fold cannot fit in a 90 mm pitch. On the right the plate does what it can actually do: the hinges run straight from one row to the next, and the whole group folds on 429 mm rather than 746. The arithmetic follows the drawing — 377 kN for the group against 657 for the rows added up. Connections

The rows have to share one fold

Each bolt row of an end plate is checked on its own, and the answers are added up. Two rows ninety millimetres apart cannot each fold the plate on a pattern two hundred and fifty millimetres long, because there is only one plate — so the group folds on one shorter pattern, and the sum was never available.

7 figures · Prying The bolt force is the larger of two lines. What an M20 bolt in a 25 mm tee flange actually carries, against the tension applied to the flange. Preloaded to 171 kN it starts there and climbs at Φ = 0.185 — the bolt's own stiffness over the bolt's plus the clamped plates', 857 against 3781 kN/mm — so 18 per cent of every kilonewton applied reaches it and the rest is unloading the contact. At 138 kN the contact runs out and the line joins the one an ordinary bolt has followed from the start, climbing at 2.12. The two lines meet, so the strength is the same either way; what differs is the slope, by a factor of 11.5. The flange's own mechanism is at 172 kN, comfortably past the crossing. Connections

The bolt that was already stretched

Prying is a lever that needs the flange to lift before it can act, and a preloaded bolt does not let it. The bolt carries a fifth of every kilonewton applied until the plates part, and everything after that is the ordinary calculation — which is why preload changes the fatigue answer by a factor of a thousand and the strength answer by nothing at all.

7 figures · Prying A unit load carried by the prop taken away. A beam fixed at its left end and propped at its right, under 4 kN/m over 8 m, asked how far it moves at 4 m. The real moment is the propped cantilever's own, with -32.0 kNm at the wall. The unit load is carried by the prop taken away, whose moment diagram peaks at 4.00. Their product has 98.92 of area on one side and -13.33 on the other, and the net, divided by EI, is 85.33 — the propped cantilever's closed-form deflection, 85.33, which no part of this calculation was given. Deflection

Any structure will carry the unit load

Virtual work has two readings and each is free exactly where the other is bound. A unit load needs only something to stand on in equilibrium, so the deflection of a beam statics cannot solve comes out of a cantilever statics can. A virtual displacement needs only to fit together, so a reaction comes out of pushing a mechanism — and on a redundant beam the unknown cancels out of the equation and nothing is found at all.

6 figures · Virtual work The cross term has the sign of the two contributions. Four responses of the floor, each combined three ways and divided by the complete quadratic combination at ρ = 0.50. Base shear: modal 927 and 839 kN, so the root-sum-square is 0.82 of it and the absolute sum 1.15. Base torque: modal -8640 and 8640 kNm, so the root-sum-square is 1.41 of it and the absolute sum 2.00. Flexible edge: modal 60 and -6 mm, so the root-sum-square is 1.05 of it and the absolute sum 1.15. Stiff edge: modal -5 and 47 mm, so the root-sum-square is 1.05 of it and the absolute sum 1.15. Where the two modes push the same way the root-sum-square is short; where they push opposite ways it is long. Dynamics

The twist the combination rule invents

Two modes close together respond together, and the square root of the sum of squares assumes they do not. The error has the sign of the two modal contributions: where they agree, as they do in base shear, the rule comes up short, and where they oppose, as they always do in torque, it comes up long — by a factor of five for a floor whose stiffness sits twenty centimetres off its mass, a torque the building does not have.

6 figures · Mode shapes The damping that stops helping. The storey drift's white-noise root-mean-square as the bearings' damping is raised from 2 to 50 per cent, each divided by the exact value at 2 per cent. The classical analysis promises that every increment helps, and at 50 per cent predicts 0.20 of the lightly damped drift. The exact analysis flattens: 0.34 at 20 per cent and 0.29 at 50, because the damping force at the bearings is transmitted into the superstructure's own mode, which the classical analysis has decoupled from it. Dynamics

The damping that belongs to no mode

Give every mode its own damping ratio and throw the rest of the damping matrix away, and an isolated building's periods and damping come out right to within a fifth of a per cent. Its storey drift at the superstructure's frequency comes out six times too small. The bearings' dashpot pushes on both modes at once, and the more of it there is the less extra damping buys: the classical analysis promises the drift keeps falling, and it stops.

6 figures · Mode shapes 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.

6 figures · Overturning 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.

6 figures · Method of sections Three routes to one deflection, and the one that is wrong. Two 2500 mm bars of 250 MPa proof stress meeting at a loaded apex, the drop of the apex against the load. The line is geometry: each bar's extension from its own stress, divided by the sine of its slope. The dots are the derivative of the total complementary energy with respect to the load, and lie on it. The dashed line is the derivative of the strain energy — Castigliano's theorem applied to a material that is not linear — which leaves the truth by ten per cent at 99 kN and at 170 kN gives 308 mm for a deflection of 46.0. Deflection

The other area under the curve

Castigliano's theorem says a deflection is the derivative of the strain energy with respect to the load, and it is true only while the material is linear. Past that, the right energy is the area on the other side of the stress–strain curve. On two aluminium bars at their proof stress the strain energy gives a deflection four times too large, and on a redundant truss minimising it picks a set of forces in perfect equilibrium that no deformed shape can produce.

5 figures · Strain energy Settling down or walking away, cycle by cycle. The total plastic hinge rotation of the beam after each cycle of loading — span 1, both, span 2, neither — with the midspan load at 0.98, 1.02, 1.05, 1.10 times the shakedown load of 126.3 kN. At 0.98 it stops at 0.59 mrad. At 1.02 it grows 4.57 mrad a cycle. At 1.05 it grows 11.43 mrad a cycle. At 1.10 it grows 22.86 mrad a cycle. Nothing collapses in any single cycle; above the shakedown load the beam walks. Materials

The load it can carry once

A two-span beam whose loads come and go span by span collapses at 150 kN under any one arrangement, and walks at 127. Between the two it can carry every arrangement once and none of them forever: each cycle leaves a few more milliradians of rotation at the support and a midspan fifteen millimetres lower. Melan's theorem finds the limit as the last residual moment line that fits, Koiter's as a mechanism no single load state can drive, and a cycle-by-cycle calculation walks exactly where both say it will.

5 figures · Shakedown 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.

6 figures · Bolt group 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.

6 figures · Bolt group 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.

6 figures · Bolt group 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.

6 figures · Moment-area 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.

5 figures · Moment-area 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.

5 figures · Moment-area