Structural Engineering & Statics
where the load goes
The collection

Every essay — page 10

Essays 217 to 240 of 728, in the same order.
Four ways to make a cell resist being racked, and one that is not one. One cell of a grid shell under the membrane shear it has to carry, by the four mechanisms available for carrying it, with the racking each produces over a 30 m span under 1.2 kN/m² of asymmetric load. A serviceability limit of span/250 is 120 mm. Four pin-jointed bars in a quadrilateral have no in-plane shear stiffness whatever — the cell folds, and the answer is not a large deflection but a mechanism. Rigid nodes carry the shear by bending the members over a cell, which smears to 12EI/s³ and comes to 0.35% of what a continuous sheet of the same stretching stiffness gives: 1205 mm, ten times the limit. One diagonal per cell, or a third member direction, carries it axially instead and lands within a factor of two of the sheet. That is the whole difference between a grid shell and a row of arches. Structural form

A shell only if the grid takes shear

A curved surface carries load in its own plane at a fraction of the material a flat one needs, and every gridshell ever built is an attempt to buy that with members instead of with a surface. The attempt succeeds or fails on one property nobody draws — whether four bars meeting at a corner can resist being racked — and a pinned quadrilateral grid cannot resist it at all.

8 figures · Gridshell 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.

7 figures · Lift stability A straight line, and the comfortable case is already two thirds down it. The capacity of a 215 mm masonry wall as a fraction of its squash load, against the eccentricity of the resultant. Nothing in this figure is a buckling calculation. A material that cannot be pulled bears on a strip of width 3(t/2 − e) under a triangular stress block, so the capacity is exactly 1.5f(t − 2e) — a straight line, zero when the resultant reaches the face, and already at 67% at the edge of the kern. The middle third is treated everywhere as the comfortable case; a wall loaded there has given away a third of its capacity before slenderness has been mentioned. The lower line is the same wall with slenderness in it, which enters as an ADDITIONAL eccentricity of 17.4 mm rather than as a reduced stress — h_ef²/2400t, for h_ef = 3000 mm. Euler's load for this wall is 9.6 times what the eccentricity rule allows, which is why no masonry calculation contains it. Stability

It does not buckle, it runs out of width

Every stability failure in this collection is a member that could have carried tension deciding to go sideways instead. Masonry cannot carry tension, and its failure under an eccentric load is not a bifurcation at all — the bearing area simply shrinks until it runs out. The capacity is exactly linear in the eccentricity, Euler's load is ten times anything allowed, and no material property appears until the very end.

7 figures · Wall slenderness The wind pushes on one face and pulls on three. A 30 × 20 m building in plan, with the measured pressure coefficient on each face and the arrows drawn in the direction the pressure acts. Only the windward face is pushed; the other three are sucked, and the side faces are sucked hardest of all at c_p = -0.7. The horizontal resultant is 842 kN at a velocity pressure of 0.9 kPa, and the arithmetic of it is the whole point: the leeward suction pulls the building downwind, so it ADDS, supplying 38% of the answer, while the two side faces cancel each other exactly and supply none of it. The coefficients are wind tunnel data; what is computed is the free body they are applied to. Equilibrium

Most of it is suction

A wind load is drawn as arrows pressing on the windward face, which is where about three fifths of it comes from. The rest is a pull on the back. The two side faces carry the largest suctions on the building and contribute nothing at all to the answer — and the inside of the building, which nobody draws, decides whether the roof stays on.

8 figures · Wind pressure Two different structures released, and one bending moment diagram. The bending moment in a continuous beam of 8, 10, 8 m under 12 kN/m, solved twice by the force method with different redundants. The first release puts a hinge over each interior support, so the released structure is a row of simple spans and the redundants are moments. The second removes each interior support, so the released structure is one simple span of the whole length and the redundants are reactions. The two released structures have nothing in common — different shapes, different deflections, different everything — and the diagrams they produce lie on top of each other to 9e-15 of the peak moment. Which restraints are released is a choice about the arithmetic and not about the structure, which is a fact worth trusting: it means a hand calculation can pick whichever release makes the sums easiest and be sure of the answer. Deflection

Choose what to take away

The other machine for a redundant structure works by removing restraints until what is left can be solved by statics, then putting back exactly enough force to close the gaps that opened. Which restraints are removed does not change the answer at all, and changes the arithmetic completely — one choice gives a tridiagonal matrix a person can solve on paper, and another gives a full one.

8 figures · Force method The corner is where the plate twists, and it has to be held down. The twisting moment M_xy over a 6 × 6 m simply-supported panel under 10 kN/m², from Navier's double series. It is zero along both centrelines and largest at the corners, which is the opposite of the bending moments and is why no strip reading contains it: 31% of the load crosses this panel in twist, and a strip can only bend. At a free corner the twisting moment is statically equivalent to a downward point force of 2M_xy — 26.7 kN here, 7.4% of the whole load on the panel, applied at a point — so the corner has to be held down. A panel whose corners are free lifts there, deflects more, and cracks diagonally across them. The deflection at the centre is 2.53 mm, against 4.05 mm for the strip that ignores all this — 60% more, which is the size of what the twist is carrying. Deflection

A third of the load crosses sideways

A slab spanning both ways is usually explained as two beams sharing a load by a fourth power, and the explanation is not merely approximate — it is missing a mechanism. A real plate carries load three ways, and the third one has no beam strip in it: it is twisting, it accounts for a third of the load on a square panel, and it is why the corners lift.

8 figures · Plate torsion Three lines through one point, and three different winners. Modulus against density on logarithmic axes, with a guide line for each of three indices drawn through mild steel. A performance index E^(1/n)/ρ is a straight line of slope n on these axes, so ranking materials by it means sliding the line up and to the left and seeing what it leaves behind. The three lines have three different orders, which is why a table of properties cannot answer the question on its own: for a tie the answer is carbon fibre, for a beam it is timber at 4.28 times steel, and for a plate timber wins by more still. The construction is Ashby's; the arithmetic on it is this site's. Materials

The ranking belongs to the load case

Every table of material properties ever printed ranks by strength, and every one of them answers a question nobody asked. What a structure wants is the least mass for a stated performance, and the combination of properties that gives it changes with the shape of the member — so timber beats steel four to one as a beam and loses to it as a tie, without either material changing.

8 figures · Material index Cover enters twice, and the strength of the concrete enters once. How long a 20 mm bar has before the cover over it splits, against the cover, split into the two halves it is always split into. Initiation is the time for the chloride front to reach the bar, which goes as the square of the cover — Fick's law and nothing else — and it is 9.1 years at 35 mm and 36.2 at 70. Propagation is the time from there to a split cover, which is short: 0.9 years, because the cover cracks at a section loss of 0.21% and no strength check in this collection would notice a loss that small. The pressure the cover can take grows with the cover too, so cover appears in both terms and the concrete's own tensile strength appears in one of them, linearly. That asymmetry is why every durability clause in every code is about cover and crack width, and hardly at all about strength. Materials

The load that comes from inside

Every action in this collection has been applied from outside — a weight, a pressure, a movement, a temperature. Corrosion is not applied at all, and the reason it belongs to statics rather than to durability is that what does the damage is a load: rust occupies three times the volume of the steel it came from, and the only place to make room is by pushing the cover apart.

8 figures · Corrosion The width nobody drew. A gusset plate with a brace bolted to it over 240 mm, and the width the profession has agreed to pretend is carrying the force. Everything else on this site arrives with a cross-section; a gusset does not, because it is a piece of steel with something attached somewhere in the middle of it and there is no geometry that says how much of it is working. The answer is the Whitmore section: assume the force spreads at 30° from the first fastener and take the width it has reached at the last, b_eff = w + 2L·tan30° = 367 mm. That is 4.08 times the width anything is actually attached to, and the rule comes from a 1952 master's thesis. It has since been checked against finite element work and holds to about ten per cent, which is fortunate, because moving the assumed angle by ten degrees moves the answer by 34%. On this plate the check that governs is not the stress the rule was written for: it is the Whitmore section buckles, at 721 kN against 1564. Connections

The width nobody drew

Every other member in this collection arrives with a cross-section. A gusset plate does not — it is a piece of steel with a brace bolted to it somewhere in the middle, and no geometry says how much of it is working. The profession's answer is a thirty-degree spread from a 1952 master's thesis, it invents three quarters of the area being checked, and the check it was written for is not the one that governs.

8 figures · Gusset The force falls, the drift rises, and the damping goes the wrong way. What a compliant foundation does to a 0.6 s building on a 8 × 8 m footing, against the stiffness of the ground under it. Three curves, all normalised to the fixed-base answer. The period lengthens — 1.31 times at 200 m/s — because the swaying and rocking of the foundation are flexibilities in series with the structure's own, and the rocking term carries an h², so it is the tall building that feels it. The base shear falls with the period, which is why a fixed base is usually called conservative. The displacement rises, by 51% here, and that is what breaks the cladding, the services and the gap to the building next door. And the effective damping falls rather than rises: the structure's own is divided by the cube of the lengthening — 2.2% of an original 5% — while a slender building's foundation radiates only 0.28% back, because rocking radiates almost nothing at these frequencies. Dynamics

The ground is a spring

Every dynamic result in this collection has assumed a structure rising from something that does not move. Nothing does. A foundation can slide and it can rock, both are flexibilities in series with the structure's own, and the rocking one carries a square of the height — so the period lengthens, the force falls, the drift rises, and the damping goes the wrong way.

8 figures · Soil-structure The snow that left the roof is standing against the wall. A roof in section with a 1.0 m obstruction at its downwind end, drawn with the vertical scale exaggerated 4 times because a metre of snow on twenty metres of roof is thinner than the line it would be drawn with. The balanced layer is 0.48 kN/m² everywhere; the wedge against the wall holds the snow that 30% of the 20 m upwind gave up, so its area is fixed by conservation of mass rather than chosen. That makes it 0.85 m deep and 3.4 m long, with a peak load of 2.60 kN/m² — 5.4 times the balanced value, and still only 27% of the snow on the roof. An intensity several times the design load, made of a quantity nobody would notice had moved. Equilibrium

The load that arrives where the wind stops

Snow is the one load a structure is given rather than subjected to. What falls is spread evenly over a whole region; what a member carries is whatever the wind left above it, and the wind piles it against whatever gets in the way. The heaviest patch on a roof is usually a quarter of the snow on it.

8 figures · Snow drift 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.

8 figures · Notional load Deflection goes as the fourth power of the span. Deflection against span for a constant load intensity and section, with two slower relationships drawn faintly behind it for comparison: the load itself, which grows in proportion to the span, and the bending moment, which grows as its square. Doubling the span multiplies the deflection by sixteen, while the moment only quadruples. Equilibrium

The weight that has to be known before it can be found

Every other load arrives from outside and can be looked up. A structure's own weight depends on how big it is, and how big it is depends on the load — so the first calculation on any project is a fixed point, and the fraction of a member spent carrying itself turns out to be the square of its span as a fraction of a span it can never reach.

7 figures · Dead load 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.

8 figures · Load height A Pratt truss of 8 panels. A Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 14 members came out in tension, 13 in compression and 2 carrying nothing. Stability

The tie that spends an afternoon as a strut

A tension member is chosen by its area and nothing else. A compression member is chosen by how that area is arranged. So a member whose force reverses under some load case is not merely being asked for the same number with the other sign — it is being designed against a different variable, and the same steel can carry seventy times more or less depending on a shape nobody chose for that purpose.

7 figures · Load reversal 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.

7 figures · Geometric stiffness 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.

7 figures · Support width 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.

8 figures · Column eccentricity 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.

8 figures · Staged section The neutral axis is wherever the first moment vanishes. A 300 by 600 section with 1800 mm² of steel at a depth of 540, carrying 250 kNm after the tension side has cracked. The neutral axis is no longer at mid-depth: it has risen to 234.5 mm from the top, which is where the first moment of the compression zone plus the transformed steel vanishes. The compression is 15.4 N/mm² at the top fibre and the steel carries 301 N/mm²; the resulting couple is 541 kN on a lever arm of 462 mm, which multiplies back to the 250 kNm applied. The uncracked section would have had 6673×10⁶ mm⁴ against the cracked 3809×10⁶ — a loss of 43% of the stiffness. Sections and stress

Where the steel is, not how much of it

A section in bending resists a moment with a couple, and a couple is a force times a distance. The force is bought — it is an area of steel at a stress. The distance is free, decided by where the bars were put, and it is the variable almost nobody optimises because it does not appear on an order.

8 figures · Lever arm Three ways to move the same column, and they are not close. The same 2000 kN moved 3 m across 14 m, built three ways and drawn to one scale. The deep beam is 1.14 m of concrete, 24.1 tonnes, and settles 60.6 mm in the long term. The storey-deep truss takes the same moment as a couple at 3.6 m centres, so its chords carry M/h and it weighs 2.6 tonnes — a fifth of the beam — while settling 15.8 mm, and it does not creep. The wall is 72.8 tonnes and hardly moves at all, 1.71 mm, of which 36% is shear rather than bending — which is what a member as deep as it is long always does, and is why beam theory does not describe one. A wall as a deep beam is the stiffest of the three by a factor of 35.4. Structural form

The same span, four ways

A beam, a truss, an arch and a cable can all cross the same gap under the same load, and the choice between them is usually described as a matter of judgement or of taste. It is neither. Each carries the load by a different mechanism, each mechanism has a different exponent, and an exponent decides the ordering at every span rather than at some spans.

8 figures · Form selection The corner columns take what the middle ones did not. Axial stress in the columns across one flange face of a 36 by 36 m framed tube, at the base. Plane sections says the flat line: every column on the face at the same distance from the neutral axis, therefore at the same stress. The solved distribution is the curve — 78.5 N/mm² at the corner against 13.4 in the middle, a ratio of 5.88. The middle columns lag because the only route the axial force has into them is the in-plane shear of the spandrel frame, bay by bay from the corner. The face is carrying its resultant on an effective width of 45 per cent, and the tube deflects as though its second moment were 64 per cent of the gross. Structural form

The columns that lean

A framed tube carries its wind load by bending the spandrel beams between its columns, and it does it badly — the corner columns take nearly six times what the middle ones do. Tilt the columns instead, so the perimeter is triangulated, and the same shear is carried axially. The concentration falls to 1.21 and the tube recovers most of the stiffness the plan said it had.

8 figures · Diagrid 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.

8 figures · Shrinkage curvature 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.

8 figures · Discretisation