Structural Engineering & Statics
where the load goes
Theme

The statics of things that do not move — page 2

Essays 25 to 48 of 79 on this thread, in the same order.
The area is the rotation, and its first moment is the movement. A 6 m cantilever under a tip load of 10, with the M/EI diagram beneath it. The shaded area is 180.00, which by the first theorem is the change of slope along the whole member. Its centroid is at 2.000 m, and the first moment about the tip is 720.00 — which by the second theorem is the deviation from the tangent, and for a cantilever that tangent is horizontal, so it is the deflection itself. Integrating the curvature twice instead gives 720.00. Deflection

The area of a diagram is a rotation

A deflection is the double integral of a bending moment, and the two constants of integration are the whole difficulty. Mohr's theorems replace them with two pictures — an area, and where that area's centre of gravity sits.

 Four guesses at one buckling mode. A pin-ended column, with four assumed shapes and the load each of them gives. The reference is a ten-term Ritz expansion solved as an eigenvalue problem, at 9.8696 EI/L² — which is π², as it must be. a half sine gives 9.870, its own sag shape gives 9.882, a mid-span sag gives 10.000, a parabola gives 12.000. Every one of them is high and none of them is low, because an assumed shape is a constraint on the column and a constraint can only stiffen it. Stability

Guessing the shape, and getting the load anyway

A column's buckling load can be had from a shape that is wrong everywhere, because the energy criterion is stationary at the true mode. The error in the load is the square of the error in the shape, and it is always high.

 The plan a straight beam does not have. A beam of radius 12 m turning through 60°, seen from above, with bending drawn outward from the axis in one colour and torsion in the other. The load is vertical and uniform and nothing is applied off the axis. Bending reaches 446 and torsion 155; the two peaks are in different places, which is why the section has to be chosen for a combination rather than for either. Internal forces

Bending that arrives as twist

A straight beam under a vertical load carries no torsion unless something applies one. A beam whose axis curves on plan carries torsion everywhere, from the same load, with nothing applied off the axis — and it cannot be simply supported at all.

 The coefficient is a slope, and that is why it can exceed one. The crack magnified: two rough faces, drawn as a sawtooth at 54° to the plane. Sliding one over the other cannot happen without lifting it, so a shear displacement forces a separation in fixed proportion — the tangent of that angle, which is the number written down as a coefficient of friction and here is 1.40. The bars crossing the plane are stretched by the separation and clamp the faces back together; they are not carrying the shear, they are supplying the normal force that lets the roughness carry it. A bar that is not anchored on both sides supplies nothing. Internal forces

Shear across a crack that is already there

Every shear calculation in this collection starts from an uncracked solid — a principal stress, a shear flow, a diagonal tension. This one starts after the crack, on a plane with no tensile strength at all, and the coefficient it uses is not a coefficient of friction. It is the slope of the roughness.

 A frame with no plane to be drawn in. The tetrahedron, solved: three equations at every free joint, one axial force in every member. Members drawn heavy carry more; tension and compression are separated by the sign that came back from the solve rather than by inspection. The count is m + r = 12 against 3j = 12, which makes it exactly determinate, and joint equilibrium closes to 1.8e-15. Structural form

Three equations at every joint

A plane truss is determinate when m + r = 2j. A space frame needs 3j, and that one changed digit is why a cube of twelve bars is six mechanisms short while looking perfectly solid — and why every three-dimensional frame ever built is made of triangles in several planes at once.

 A basement is a boat. A 20 by 30 m substructure dug 6 m into ground whose water table stands 2 m down. The head on the underside of the base slab is 4.0 m, so the pressure there is 39.2 kN/m² over the whole plan — 23.5 MN of it, pushing upward. Nothing about the structure changes that number. What resists it is weight: 18.7 MN of concrete and whatever is built above, giving a factor of 0.80. The structure floats if the water reaches 2.82 m below the ground, and a base slab alone would have to be 1.64 m thick to hold it down. Equilibrium

A basement is a boat

Every load in this collection presses down and is resisted by strength. Hydrostatic uplift presses up, is resisted by weight, and does not care what is built on it — so the check contains no material property at all. It is a ratio of two weights, and one of them is water.

 Balanced, and four times as heavy on the bearing. A bascule leaf of 900 kN whose centroid is 9 m from the trunnion, balanced by 2700 kN at 3 m on the other side. What balancing achieves is exactly one thing: the moment about the pivot is zero at every opening angle, because both terms carry the same cosine. What it costs is two things that are not zero. The reaction on the trunnion becomes 4.0 times the leaf's own weight, since both weights are still there. And the rotational inertia rises by 33%, so the balanced leaf is the hardest one to start and to stop — which is why the counterweight is put as close to the pivot as it will fit, at the price of being heavy: the same balance at twice the radius weighs 1350 kN and carries 1.25 times the inertia. Equilibrium

Balanced, and four times as heavy

A counterweight cancels a moment about a pivot, and that is the only thing it cancels. The bearing beneath carries both weights, the inertia rises as the square of the radius, and a load that moves cannot be balanced at more than one position at all.

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

Held by something that goes soft

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

 The point the rafter turns about, which is off the frame. A pitched portal of 8 m span and 4.0 m to the eaves, with a 1.5 m rise, collapsing. Each rigid part of the mechanism rotates about some point: the left column about its base hinge, the right about its own. The rafter between them does neither, and its centre is found by one rule — two bodies joined at a hinge share that hinge, so the second body's centre lies on the line through the first body's centre and the hinge, extended. Two hinges give two lines and they cross at (8.0, 11.0) metres, which is 5.5 m above the ridge and outside any drawing of the frame itself. From there the whole collapse is two ratios of lengths and no trigonometry: the load factor is 1.339. Flatten the roof and the centre descends; make the two lines parallel and it goes to infinity, which is the statement that the rafter translates instead of turning. Equilibrium

The point the mechanism turns about

A collapsing frame is a chain of rigid pieces, and every piece is rotating about some point. Find those points and the whole collapse load reads off two ratios of lengths, with no trigonometry anywhere — and for a pitched roof the point in question is well above the top of the drawing.

 At thirty degrees each leg carries the whole load. A 100 kN lift on two legs at 60 degrees to the horizontal. Each leg is a two-force member, so its force is the vertical share it carries divided by the sine of its angle: T = W/(2 sin β) = 57.7 kN, which is 0.58 times the whole load in each. The curve on the right is that division, and it is the reason the rule about sling angles exists rather than a convention: at sixty degrees a leg carries 0.58 W, at forty-five 0.71, at thirty exactly 1.00, and at fifteen 1.93. And the horizontal components do not disappear — they run through the thing being lifted, which here carries 57.7 kN of compression between the two pick points. That is what a spreader beam is for: it takes the compression as a designed strut so the legs above it can stand up. Equilibrium

The angle that doubles the force

A crane hook takes the weight and nothing else. What holds the load is a set of legs running down to it at an angle, and each leg is a two-force member — so at thirty degrees from the horizontal each of two legs carries the entire load, and the difference goes into the thing being lifted as compression.

 Almost all of it is exactly zero. The stiffness matrix of a 2-bay, 3-storey plane frame: 36 freedoms, of which 16.2 per cent of the 36² entries are non-zero. The zeros are not small numbers; they are absences. A member reaches only the two nodes at its ends, so it can contribute nothing to any row belonging to a node it does not touch, and every such entry is zero exactly rather than nearly. The non-zeros therefore sit in a band of width 13 about the diagonal. Before the supports are applied the matrix is singular, and its null space has exactly three dimensions — the three rigid-body motions a plane frame has with respect to the ground, which is the same statement the null vector of the equilibrium matrix makes about a truss that is a mechanism, arrived at from the other end. Deflection

The matrix that replaced the hand methods

Moment distribution passes moments round a frame until they stop moving. Virtual work computes one deflection at a time. Both are exact and both stop scaling in the low tens of members. What replaced them adds no physics at all — the whole of the invention is the bookkeeping.

 A fourth power, and then a cliff. The factor of safety against rolling, against beam length, for one section hung from a roll axis 0.9 m above its centre of gravity. Nothing about the section changes along this axis. z̄ goes as the fourth power of the length — 0.236 m at 30 m becomes 0.747 m at 40 — and the factor of safety is proportional to (y_r − z̄), so it does not decline gently: it falls away and then stops existing. The working factor of 1.5 is lost at about 41 m, and past 42 m there is no hook height at all at which this beam hangs stably. Which is why long girders are lifted with the picks moved inboard, or with the beam braced, or not in one piece. Stability

Hung from above and still unstable

A rigid body hanging from a point above its centre of gravity is a pendulum and cannot fall over. A beam is not rigid, and tilting it puts a component of its own weight sideways — which bows it, which moves its centre of gravity further out. Past a length there is no hook height at which it hangs stably at all, and the length arrives as a fourth power.

 The load that makes itself worse. The amplification of a deflection against the ratio of applied load to buckling load. A structure at half its buckling load deflects twice as far as first-order analysis predicts, and the curve runs away well before the load is reached. Equilibrium

The load that is really a lean

No frame is ever plumb. The columns are out of upright by something like a three-hundredth, and every tonne of gravity load standing on that lean has a horizontal component. The force that represents it is not a safety allowance — it is an exact statics substitution for a geometry nobody drew.

 The length at which a beam stops being a beam. Elastic critical moment against the distance between lateral restraints, with the section's plastic capacity drawn across it. The two cross at 4086 — beyond that length the beam buckles sideways before it reaches the strength its cross-section has, and the capacity is set by the restraints rather than by the steel. Stability

The load that moves with the twist

A beam about to buckle sideways is beginning to rotate, and everything attached to it rotates with it. A load hung from the top flange swings out over the side and drives the rotation on; the same load hung underneath swings back and stops it. Two identical beams, two different capacities, and the only difference is a height.

 The check that everything adds up, and the error it cannot see. Four versions of the same 2-bay, 3-storey frame, with the global equilibrium residual each one produces — the sum of the reactions against the sum of the applied loads, as a fraction of the applied total. It is the first thing every analysis prints and it is worth having: a lost restraint and a load entered in the wrong unit both show up immediately, at 6% and 24%, because both change what the structure is carrying. The fourth bar is the point. A member whose stiffness is wrong by a factor of ten redistributes the internal forces completely — the second bar shows the change in the member forces, 18% — and the global residual is exactly zero, because the wrong answer is still in equilibrium with the same loads. Equilibrium is one equation per degree of freedom of the whole body, and a stiffness error lives entirely in the many equations underneath it. A model can satisfy every equilibrium check ever devised and be a model of a different structure. Stability

The stiffness the load takes away

Buckling is usually taught as an event — a critical load, a bifurcation, a mode. Written as a matrix it stops being an event at all. A compressive load subtracts a stiffness from the structure, the subtraction grows with the load, and the critical load is simply where what is left reaches zero.

 The bearing is one length and the web is loaded over another. A load applied over a stiff bearing of 200 mm on the flange of a girder with a 1200 × 8 mm web. The flange bends under it and the yield lines that form spread the load along the web over 659 mm — 3.3 times the bearing, and 70% of the yield resistance is that spread rather than the bearing. The effective length is not a decision anybody made: it is what the flange's own bending stiffness against the web's own strength works out to. Internal forces

The support that is not a point

A reaction is drawn as a single arrow because the equilibrium equations only need its total. Underneath the arrow is a bearing of some width, delivering a pressure over that width, and almost everything a designer would like to know about the region near a support is a consequence of the width the arrow does not have.

 A force may be moved anywhere, at the price of a couple. A 80 kN force applied 250 mm off the centreline of a body, and the same force applied ON the centreline together with a couple of 20 kNm. The two systems are equivalent: they have the same resultant force and the same moment about every point in space, so no equilibrium equation written about the body can tell them apart. What they are not is the same loading — the stresses inside the body differ, and they differ over a distance of about the body's own depth. The offset is drawn to a scale that keeps the arrow on the body; the number beside it is the real one. Internal forces

The moment the beam left behind

A beam reaction is drawn arriving on a column's centreline. It arrives on a cleat a hundred millimetres out from the face, and the difference is a couple that goes into the column and has to be shared between the lengths above and below it. Nothing about it appears in a frame model whose members meet at nodes.

 The props decide where the stress ends up. Bottom-fibre stress in the steel of a 12 m composite beam carrying 6 kN/m of wet concrete and 9 kN/m afterwards. Unpropped, the bare steel takes the first stage alone and reaches 146 MPa; propped, the finished composite section takes everything and reaches 93 MPa — a ratio of 1.57. 62% of the unpropped beam's final stress was locked in before the slab was structural at all. The deflections differ by 1.73 times for the same reason, and no drawing of the finished beam distinguishes the two. Sections and stress

The section that changed while it was being loaded

A stress is computed from a moment and a section modulus. When part of the moment arrived while the section was a different shape, there is no single section modulus to divide by — the stresses add and the properties do not, and two identical finished beams can differ by half again in stress with nothing on the drawing to say which is which.

 The least reliable number in the material decides the answer, briefly. What a twenty per cent error in the concrete's tensile strength does to a computed deflection, against how far past cracking the beam is. Well past the cracking moment it does almost nothing — at 1.9 times M_cr the spread is 56 per cent — because the section is nearly fully cracked and the interpolation has run out. Just above cracking it does everything: at 1.19 times M_cr the same twenty per cent moves the deflection by 3658 per cent. Tensile strength is the property with the widest scatter and the least direct test, and a beam designed to sit near its cracking moment has put the answer on it. Deflection

The curvature nobody applied

Concrete shrinks as it dries, by about half a millimetre in every metre. In a symmetrically reinforced member that is a shortening and nothing else. In a member with more steel in one face than the other — which is every beam and every slab — the steel holds one side back and the section bends, with no load on it at all.

 One member's stiffness, scattered into the freedoms it touches. A member's own six-by-six stiffness matrix relates the forces at its two ends to the displacements there, and it is written in the member's own axes. Assembly is two operations and no physics: rotate it into the structure's axes, then add each of its thirty-six entries into the row and column of the global freedom that entry belongs to. Every member does the same, and the sum is the structure. The shaded rows and columns are the six freedoms this one member reaches; every other entry it contributes is exactly zero, and that is the whole reason a global stiffness matrix is sparse. Nothing here is an approximation — the result is the same equilibrium and the same compatibility a hand method writes, in an order a machine can follow. Deflection

The answer that depends on how it was divided

Every computed answer in this collection came out of a structure chopped into pieces — elements, strips, stations, trial positions. The chopping is invisible in the result and it is not neutral: some divisions give the exact answer, some give one that is always too stiff, and one of them changes nothing but the cost of getting there.

 What the shape of a load in time is worth, for two load shapes. The peak displacement as a multiple of the static deflection, against the load's duration divided by the structure's natural period, for two load shapes: a load that rises linearly, then stays; a rectangular pulse, then nothing. The lines are closed forms and eight dots are the peak of a complete time integration of an oscillator of 0.300 s period under that load, agreeing with the line to within 0.19% everywhere. Dynamics

The load that is over before it has moved

A blast delivers an enormous pressure for a few milliseconds. Everything else in this field asks what force a structure can carry; a load that has come and gone before the structure has travelled any distance is not asking that question, and the answer turns out to depend on the mass and the ductility with the strength barely in it.

 A broad tank sloshes and a tall one does not. The liquid's division into the part that moves with the wall and the part that sloshes, against the tank's proportion. The convective masses come from the potential-flow solution and the impulsive mass is whatever is left, so the two sum to the liquid's mass exactly at every proportion rather than approximately over part of the range. A tall tank at H/R = 3 is 84% impulsive and behaves almost like a solid; a shallow one at H/R = 0.5 is 72% convective and most of its contents never notice the earthquake. This tank sits at H/R = 1.33, which is 65% impulsive. Dynamics

The liquid has a period of its own

Shake a tank and its contents do not all go with it. Part of the liquid moves as though it were bolted to the wall and part sloshes at a period fixed by gravity and the radius, which the tank's stiffness has no influence over whatever. The split is decided by one proportion, and the two parts then take entirely different amounts of the earthquake.

 The force nobody applied, and the speed it wins at. Lateral force per unit weight for a vehicle on a 400 m curve, against speed. The rising curve is what the free body demands — v²/gR, which is the body's own acceleration written on the other side of the equation — and the flat line is what 6.0° of cant supplies from the weight. They cross at 73 km/h, which is the speed the curve was set out for; below it the deficiency has the other sign and the rail is pushed the other way. The upper line is overturning, at b/2h = 0.399 — and there is no mass in that number, so a loaded vehicle and an empty one go over at the same 160 km/h and only the height of the load decides. At the 108 km/h drawn the deficiency is 0.124 of the weight, which is 49 kN on this 40 tonne vehicle. Equilibrium

The force that is really an acceleration

Every other load in this collection is applied by something. This one is applied by nothing at all — it is the body's own acceleration, written on the other side of the equation so that statics can be used on a problem statics has no business with. The move is legitimate, it is a hundred and eighty years old, and it is exactly half done more often than it is done.

 The worst force in a pile is not at the top of it. A 0.6 m pile 24 m long through ground that is settling, carrying 800 kN at its head. Above the neutral plane the soil moves down past the shaft and the friction acts downward, so the axial force grows with depth; below it the friction acts upward in the ordinary way and the force falls again to the 300 kN the base takes. The maximum is 1282.87 kN at 13.78 m — 1.60 times the load applied, and it is at a depth where nothing is applied, nothing is connected and nothing can be inspected. A pile section chosen for the head load is under-sized by that factor over the middle third of its length. Internal forces

The ground that hangs on instead of holding up

A pile is driven through fill that has not finished settling. The fill goes down past the shaft, the friction along that length turns round, and the pile is now carrying the soil rather than the other way about. The worst force is not at the head and not at the toe, and nothing at that depth can be seen.

All themes