Structural Engineering & Statics
where the load goes
The collection

Every essay — page 20

Essays 457 to 480 of 738, in the same order.
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.

5 figures · Moment-area Half the life is spent growing the first half-millimetre. The crack length against time for a detail starting with a 0.50 mm flaw under 800 cycles a day of the same five-band traffic the S-N calculation used, integrated by Paris's law. It reaches 1 mm after 91.2 years, 2 mm after 131.3, 10 mm after 172.9 and its critical length of 125.7 mm after 195.8. The curve is nearly flat and then nearly vertical, because the rate goes as the cube of ΔK and ΔK goes as the square root of the crack: the crack spends most of its life being too small to find and the rest being too large to ignore. The dashed lines are the crack lengths at which each band of traffic starts doing anything at all. Materials

The loops a crack grows on

A Miner sum reduces a hundred years of traffic to a number and throws away everything below the cut-off — on this bridge, a third of the vehicles doing exactly none of the damage. Integrate the same spectrum as a crack instead and that third grows thirty-seven per cent of the crack, because a cut-off is a statement about a constant-amplitude test and a crack's threshold is a length rather than a stress. The two calculations disagree about the life by a third and about which vehicles matter entirely.

5 figures · Fatigue The same cycles, and two different lives. Two sequences made of exactly the same cycles — blocks of 6.00M at 40 N/mm² alternating with blocks of 0.22M at 90, on a detail starting with a 0.50 mm flaw. Taking the small cycles first, the crack reaches its critical 87.7 mm after 12.36M cycles; taking the large ones first, after 6.36M. Miner's sum at the moment of failure is 0.99 for the first and 0.68 for the second, so a rule that predicts failure at a sum of one is right to within a per cent about the first and 48 per cent unconservative about the second. The mechanism is on the axes: ΔK rises with the crack, so a large block met late finds a longer crack and does more with it. Materials

The record played backwards

Miner's rule adds damage, and a sum has no order. A crack does, twice over: a large block met late finds a longer crack and does more with it, and an overload leaves a plastic zone that slows everything after it. The same cycles rearranged fail at 12.4 million or at 6.4, and a Miner sum that is right to one per cent about the first is out by half about the second. One cycle in four million can add fifty-four per cent to a life.

5 figures · Fatigue 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.

5 figures · Fatigue The two knees the environment takes away. The design S-N curve for a category 71 detail and the two shapes a corrosive environment leaves. In air the slope is three to the constant-amplitude limit at 52.3 N/mm², five to the cut-off at 28.7, and nothing below it. With the cut-off removed the second branch continues. Under free corrosion there is one slope of three all the way down and no knee at all. The faint lines are the five bands of the traffic: three of them sit below the cut-off — an ordinary lorry, a van and a car, 90 per cent of the crossings — which is why they do nothing to a detail in air and something to every other curve on the figure. Materials

The cut-off belongs to the water

An S-N curve's cut-off is the most consequential thing on it: on a category 71 detail under ordinary bridge traffic it deletes ninety per cent of the crossings and leaves two bands doing all the damage. It is a property of steel in air. A crack tip that seawater or de-icing salt can reach has no threshold, no endurance limit and no knee, and the same bridge's life runs from 300 years to 45 depending on which of six defensible calculations is asked.

7 figures · Fatigue 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.

7 figures · Graphic statics Two lines and a triangle: a three-hinged arch drawn. A three-hinged arch of 20.0 m span, springings at (0.0, 0.0) and (20.0, 0.0) m and the crown hinge at (10.0, 5.0), under 100.0 kN at 5.0 m. The right half carries no load, so it is a two-force member and its reaction lies along the line from its springing through the crown hinge. That line meets the load's line at K, 7.50 m up, and the left reaction must pass through K too. The triangle of the load and the two reaction directions gives the reactions as 90.1 kN at A and 55.9 kN at B, with a horizontal thrust of 50.0 kN — the values four equilibrium equations return, to 7e-15 kN. The shape of the rib entered nowhere. Equilibrium

The arch that is only its three hinges

A three-hinged arch's reactions come from three points and nothing else, so a parabola, a circle and a portal frame on the same hinges push on their abutments identically. Move a load across and the point where the reactions cross runs along two straight lines through the crown. That is the arch's influence line, drawn with a straightedge — and friction in the hinges it was built around blurs it.

7 figures · Graphic statics Putting a funicular through three points. Four loads — 40.0 kN at 3.0 m, 60.0 kN at 7.0 m, 30.0 kN at 12.0 m, 50.0 kN at 16.0 m — and three points the polygon must pass through: A and B at the springings and C, 5.0 m above their chord at 10.0 m. A trial pole, dashed, draws a polygon from A that ends 6.39 m below B. The ray through the trial pole parallel to its own closing line cuts the load line at Q, 95.0 kN from the top, which is the left reaction of a simple beam on A and B and does not depend on the pole at all. Every pole whose polygon passes through A and B lies on the line through Q parallel to A B; the one whose polygon also reaches C is 98.0 kN from the load line, which is the moment at C of that simple beam, 490.0 kN·m, divided by C's height above the chord. That is the three-hinged arch's thrust — the four equilibrium equations give 98.0 kN. Equilibrium

The polygon that runs out of freedom

A funicular polygon for given loads has exactly three freedoms, so it can be made to pass through three chosen points and no more. Three points is a three-hinged arch, and the drawing solves it. Take the crown hinge away and one freedom is left over. The drawing then offers a whole family of thrust lines and cannot say which one the arch uses — the rib's stiffness decides, and stiffness is not on the paper.

7 figures · Graphic statics Two pencil lines, and where they cross. Two lines drawn with a pencil 0.2 mm wide are two bands, drawn here much wider than a pencil so the shape can be seen, and they cross not at a point but in a parallelogram. At 60° apart the parallelogram's long diagonal is 2.0 pencil widths — 0.40 mm; at 12° apart the parallelogram's long diagonal is 9.6 pencil widths — 1.91 mm. The crossing's uncertainty along the bisector is the width divided by twice the sine of half the angle, so it grows without limit as the lines turn parallel, and a construction that finds a point by crossing two lines inherits it. Equilibrium

How wrong a drawing is

A pencil line is a band, and two bands cross in a parallelogram that grows as they turn parallel. The accuracy of a graphical construction is therefore a property of the angles it makes, not of the hand that made it — and the worst case is the shallow arch, the structure the method was most used on. Measured properly, the drawing's error there is the size of the builder's, and the check draughtsmen relied on cannot see it.

7 figures · Graphic statics Bearing damping that shakes the roof harder. The roof's white-noise root-mean-square acceleration as the bearings' damping rises from 3 per cent to 58 per cent, exact and classical, both divided by the exact value at 3 per cent, for the six-storey building. The classical analysis falls all the way, to 0.33. The exact one falls to a minimum of 0.44 at about 28 per cent and then rises, to 0.50 at 58 per cent: past the minimum, every extra per cent of damping at the bearings shakes the top floor harder, while the classical analysis says it is still helping. Dynamics

The top floor the bearings shake

On a building of several storeys, the damping at the isolation bearings is tied less tightly to each higher structural mode than to the one below. It still does more harm in each, because the ground barely reaches those modes on its own. The error the usual analysis throws away gathers in the upper storeys, and past about a third of critical damping at the bearings the roof shakes harder while the analysis says it is still helping. How much damping is best depends on how tall the building is.

7 figures · Mode shapes 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.

7 figures · Mode shapes The same mode's curvature, which can. The first mode's drift in each storey — the difference between the floors above and below it, which for a shear building plays the part a beam's curvature plays — intact, solid, and with storey 3 10 per cent less stiff, dashed, each scaled to the roof. The damaged storey's drift rises by 9.6 per cent; the storeys above it change by at most 1.8 and below it by at most 1.5. The largest change is in storey 3: the drift finds the damage and says where it is, which the frequency's fall of 0.91 per cent cannot. Dynamics

What a mode shape notices that a frequency does not

Take a tenth of the stiffness out of one storey of a ten-storey building and its first frequency falls by a per cent at most, and by almost nothing if the storey is near the top. The mode shape looks unchanged, its match to the old one 0.99997. But the mode's storey drift — its curvature — rises by about a tenth in the damaged storey and hardly anywhere else, wherever that storey is. The instrument that sees the damage is the one almost nobody installs.

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

The modes at zero frequency

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

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

The strongest layout leans on one bolt

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

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

The joint whose lines may not meet

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

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

The hinge that forms in the other flange

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

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

The rock that sends the waves back

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

6 figures · Vibration isolation Two blows on three foundations. The movement of a 150 tonne block struck by a hammer delivering 18 kN·s, twice, 0.75 s apart. On soil at 12.5 Hz with the 47 per cent of damping a half-space radiates, the block moves 0.86 mm and is still before the second blow. On the same soil over shallow rock, with 5 per cent, it moves 1.42 mm and rings for the whole interval. On springs at 4.0 Hz it moves 4.42 mm, and its ringing has not died when the second blow arrives. The peak force passed to the ground is 1.34, 1.32 and 0.42 MN: the heavily damped block and the lightly damped one pass almost the same force. Dynamics

The blow that has no frequency

A forge hammer does not shake its foundation; it strikes it. The transmissibility curve every isolation design is drawn on has nothing to say about a blow, and the rules it teaches mislead. The force a struck block passes to the ground is least at a quarter of critical damping, not the most; five per cent and forty-seven pass almost the same; and the springs that soften every blow can bring the block down to the hammer's own rate and make a train of blows ring four times as high as one.

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

The slab that shrinks onto a finished beam

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

7 figures · Creep The same restraint, spread and gathered. The buckled shape of a 24.0 m compression chord with the same smeared restraint, 0.35 N/mm per mm, delivered by U-frames at two spacings. With frames every 2.0 m the chord buckles in half-waves of about 5.1 m that ignore the frames — the smeared shape — at 1892 kN, matching the smeared answer. With frames every 4.0 m it buckles between them, with a node at every frame, at 1548 kN: 18 per cent below the smeared 1897 and at the Euler load of one bay. The dots are the frames. Stability

A row of frames is not a foundation

The top chord of a half-through girder is held sideways by U-frames, and the standard calculation smears them into a continuous elastic foundation. That is right while the frames are close and quietly wrong once they are not. The crossover sits at seven-tenths of the buckle's own half-wavelength, whatever the frames' stiffness — and beyond it the chord buckles between frames at a load no stiffening of the frames can raise.

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

The truss that is stiff by accident

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

6 figures · Truss deflection Four ways of being nearly a mechanism. Four measures of the two-bar frame against the angle its bars make with the line between the supports, on logarithmic axes: the bar force as a multiple of the load, the condition number of the equilibrium matrix, the largest relative change of force from a one-millimetre error in any joint's position, and the joint's sag under the load as a share of the rise, for bars of axial stiffness 200 MN over a span of 8.0 m, loaded with 50 kN. The first three grow as one over the angle: at 2° the force is 14.6 times the load, the condition number 54, and a millimetre of error changes the force by 0.7 per cent. The sag grows as the cube: 4.3 per cent of the rise at 8°, 41 per cent at 4°, 3.1 times it at 2°. The measure that ends the analysis is not the arithmetic's. Equilibrium

Rigid by every test, and still folding

Counting the unknowns says whether a frame can be solved and the rank of its equations says whether it can stand, and both answers are yes or no. A frame a few degrees from a critical form passes both and is still nearly a mechanism — and of the ways that shows, the last to arrive is the one the rank test measures. Its own sag under load eats a tenth of its geometry at six degrees; sixteen-figure arithmetic would not notice anything until well below half a degree.

6 figures · Determinacy Four numbers read from a moment diagram. The moment diagram of the fixed-ended beam under a central point load, scaled to a largest value of one, with the four values the quarter-point formula reads: the peak, and the magnitudes at a quarter, a half and three quarters of the span — 1.00, 0.00, 1.00 and 0.00. The formula turns them into a gradient factor of 1.923. Solving the buckling problem for the whole diagram, on a beam 8 m between lateral restraints, fork-supported at its ends, gives 1.723: the formula is 12 per cent too high, on the unsafe side. Stability

The formula that reads four numbers

The factor that credits a beam for the shape of its moment diagram is usually taken from a formula that reads the diagram at four places — its peak and its quarter points — and nowhere else. On the straight-line diagrams it was built around it is safe. On a fixed-ended beam under a central load, whose moment is zero exactly where the formula looks, it is twelve per cent unsafe, and it would give the same answer for a diagram that deserves twenty-two per cent less.

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

The ballast that helps it over

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

6 figures · Overturning