Structural Engineering & Statics
where the load goes
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Drawing as calculation — page 6

Essays 121 to 144 of 162 on this thread, in the same order.
The pole, the strings, and the resultant of any number of forces. Five downward loads on a span of 10 m — 30 kN at 1.5 m, 20 kN at 3.5 m, 45 kN at 5.0 m, 25 kN at 7.0 m, 35 kN at 8.5 m — adding to 155.0 kN. On the right, the loads laid end to end down one line, with a pole 60.0 kN to the left of it and a ray drawn to every division between them. On the left, the funicular polygon: each segment parallel to the ray of the loads it has passed, so the shape is the one a string carrying these loads would hang in. The first and last strings are extended until they cross, at 5.24 m, and that crossing is where the 155.0 kN resultant acts — the same station the moment sum Σ P x / Σ P gives, 5.24 m, reached with no pole in it at all. Equilibrium

The pole decides the drawing, not the answer

Five forces will not pair off the way four do. They need a point that is nowhere on the structure — chosen freely, by whoever is holding the pencil — and the string of lines it generates. Every choice draws a different polygon and finds the same resultant, and the shape it draws turns out to be the beam's bending moment diagram.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 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.

 The forces the girder can hold with nothing on it. The one self-stress state of a K-braced girder of eight 3 m panels, 3 m deep: member forces in equilibrium at every joint with no load and no reaction. It lives entirely in the two middle panels — twelve members: the four diagonals meeting at the middle joint of the central vertical in tension, and the four central chord members and the half-verticals at the outer side of each middle panel in compression, at 0.89 and 0.45 of the diagonal force. It is symmetric about mid-span, so a symmetric load can call on it as freely as any other: symmetry does not supply the missing equation. Every other member of the girder is idle in it, drawn faint. Equilibrium

The redundancy only the sun can find

A K-braced girder that is symmetric about mid-span has one member more than statics can resolve, and it is at the centre, where two K's meet at one joint. Symmetry does not remove it, because the forces it allows are symmetric themselves. Loads barely touch it — a section calculation that ignores it is exact for a load at almost any joint. What finds it is a millimetre of misfit or a sunlit top chord, which put forces into the middle of the girder that no load calculation contains.

 The strain-energy route halves the sag. The midpoint deflection of a cable 10.0 m long, pretensioned to 60 kN, with an axial stiffness EA of 64 MN, loaded across its span at midpoint, against the load, found three ways. Solid: the deflection itself, from the cable's geometry; the dots, the derivative of the complementary energy with respect to the load, lie on it. Dashed: the derivative of the strain energy. Dotted: the pretension alone, as if the cable's tension did not grow. At 10.0 kN the cable sags 212 mm; the strain-energy route gives 107 mm, 0.51 of it, and the pretension alone 417 mm. At 20.0 kN: 294, 128 and 833 mm. The strain-energy route understates the deflection of a member that stiffens by the same mechanism that made it overstate one that softens. Deflection

The sag the strain energy halves

The derivative of the strain energy overstates the deflection of a member that softens. Turn the curve the other way up — a cable that grows stiffer as it sags, a hanger that takes up its play — and the same derivative understates it: by half for a pretensioned cable carrying a modest load, by two thirds for one with no pretension, and by the whole of the slack for a member that has any. The error is one ratio in both directions, and the least-work shortcut built on it hands the load to the member that stiffens.

 A downward pressure that pulls the corners up. The nodal forces that do the same work as a uniform pressure on one eight-node serendipity quadrilateral, as fractions of the whole load, each node's share being the integral of its shape function over the element. The shares are −1/12, −1/12, −1/12, −1/12, 1/3, 1/3, 1/3, 1/3 at its eight nodes, adding up to one. The four corner forces are negative: the pressure pushes down and the work-equivalent forces at the corners pull up. Equilibrium

The corners the pressure pulls up

Replace a uniform pressure on a beam element by its work-equivalent nodal loads and the ends acquire couples the resultant cannot see. Do the same on an eight-node plate element and the corners acquire forces pointing the wrong way: a pressure pushing down is represented by the corners being pulled up, a twelfth of the load each. It is correct, it is the vector that makes the solution the best one available, and it is the reason a printout of nodal forces is not a picture of where a load goes.

 Two quotients from one guessed shape. The critical load of a pin-ended column whose outer quarters keep 10% of the middle's flexural rigidity from four guessed shapes, each worked two ways: Rayleigh's quotient, strain energy of the guess's own curvature over the work of the load, and Timoshenko's, which uses the curvature the guess's bending moment would cause instead. The exact load is 3.225 EI₀/L². a half sine: 8.256 by Rayleigh and 3.745 by Timoshenko; its own sag shape: 8.041 by Rayleigh and 3.712 by Timoshenko; a mid-span sag: 8.875 by Rayleigh and 3.847 by Timoshenko; a parabola: 6.600 by Rayleigh and 3.492 by Timoshenko. Both are upper bounds, and from the same shape the second is never the worse of the two. Stability

The bound from underneath

Rayleigh's quotient turns a guessed shape into a buckling load that is always too high. Divide the same guess differently — use the curvature its bending moment would cause instead of its own — and a parabola that was 21.6 per cent high is 1.3 per cent high. Add one more number that needs no guess at all and the load is caught from below as well, which is the only side of it an amplifier can safely use.

 A butt throat is strongest a little off square. The resultant a millimetre of throat can carry in S355 against the direction of the load, from straight across the weld (0°) to straight along it (90°). The fillet's falls throughout, from 385 to 314 N/mm². The butt's rises from 441 to 478 at 22.7°, because its cap limits only the normal stress and a little shear costs it nothing, and then falls to 314. So the butt's advantage over the fillet is 1.15 straight across, 1.29 at 22.7°, and none straight along. Connections

The shear that helps a butt weld

A partial-penetration butt throat pulled straight across its weld is capped at 0.9fu, which in S355 makes it 15 per cent stronger than a fillet throat. Add a little shear along the weld and the butt throat gets stronger, not weaker, because the cap limits only the normal stress — until the shear is 42 per cent of the pull, where it is 29 per cent stronger than the fillet and carries its largest resultant, 478 N/mm² against 441 straight across. Along a moment connection's web weld the mix of pull and shear sweeps through that peak, and in the direction that is all shear a single-sided throat is as good as two.

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