Drawing as calculation — page 5 · Structural Engineering & Statics

3 continuous spans against 3 simple ones. The bending moment in a continuous beam, solved by the stiffness method, drawn over the moment in the same spans made simply supported. The peak sagging moment falls from 28.1 to 18.0, and a hogging moment of 22.5 appears over the supports where there was none.

Structural form

The chord is a continuous beam

A truss analysis reports one number per member, and for a chord running through eight panels that number is an axial force with no bending in it. The chord is a single piece of steel spanning between its own nodes, and anything landing between them bends it.

A truss drawn inside a solid, and solved as one. A deep member 5000 mm between bearings and 2500 mm deep, carrying 2400 kN at mid-span. The model is two struts and one tie, on a lever arm of 2000 mm, and it is solved by the truss solver rather than by a formula: the tie comes back at 1500 kN and each strut at 1921 kN, at 38.7° to the horizontal. Spread over a strut width of 1031 mm the compression is 3.7 N/mm² against a limit of 18.1 for concrete cracked across its own strut, and the tie needs 3448 mm² of steel. A beam calculation on the same member would have asked the tie for 1404 kN, which is 7% less than the model does.

Internal forces

The node is the part that is checked

A strut-and-tie model is safe if every part of it can carry what the model asks. The struts are wide and lightly stressed, the ties are steel and easy, and the whole of the difficulty collects at the points where they meet — in a volume of concrete a few hundred millimetres across that no drawing shows.

Flattening the truss saves stirrups and crushes the web. Two capacities against the angle of the cracks, for a web 350 mm wide with a lever arm of 630 mm. The rising line is the stirrups: a cut along the crack severs z·cot θ/s of them, so flattening the crack from 45° to cot θ = 2.5 takes the 315 kN they carry to 787 — 2.5 times as much from the same steel. The falling line is the concrete strut, whose stress is V(cot θ + tan θ)/b_w z and therefore least at 45°. They do not cross in this range, so the stirrups govern throughout and the angle is a free choice.

Internal forces

The angle is a choice, not a property

The truss inside a cracked concrete web has a strut angle, and nothing measures it. The designer picks it, the stirrup requirement falls as it flattens, the web stress rises, and every choice in between is a different structure that carries the same load.

More steel across the crack, until the roughness runs out. Shear resistance of the interface against the reinforcement crossing it. The steel clamps rather than carries, so the resistance is the clamping stress times the interlock coefficient and rises in a straight line — until the asperities crush at 5.50 N/mm², which happens at a reinforcement ratio of 0.79%. Past that the line is flat and every further bar is decoration. The dashed line is what the clamping alone would give if the concrete were unbreakable.

Internal forces

Two models of one bracket

A corbel can be designed as a plane that has to be clamped or as a truss that has to be drawn, and the two are not approximations of each other. They describe different failures, they ask for steel in different places, and the honest answer is that both are checked because neither bounds the other.

A reaction with no load, and the moment it bends the beam with. The prestress moments in a 2-span beam. The primary moment is −P·e, the tendon acting on its own section, and it reaches 1440 kNm over the middle support. The secondary moment is what is left when the primary is taken off the total, and it is 720 kNm — 50% of the primary, with the same sign, so it does not cancel anything. It comes from the middle support refusing to let the beam lift: 102.9 kN pressing down there and 51.4 kN lifting at each end, a reaction set that sums to 0e+0 because nothing external was applied. Its diagram is straight between supports to 3.6e-13% of its own peak, which it has to be: reactions are point forces and a point force puts no curvature in a span.

Internal forces

The tendon that can be moved

Lift a continuous beam's tendon at its interior support without changing its drape and nothing about the beam's total moment changes. The primary falls, the secondary rises by exactly as much, and the pressure line stays where it was — which turns a parasitic effect into a quantity a designer can place.

The second moment of area is a function of direction. Second moment of area of an equal angle against the angle of the axis it is taken about, with the product of inertia beneath it. The maximum is 5.943 × 10⁶ mm⁴ and the minimum 1.523 × 10⁶, a ratio of 3.90, and they occur where the product of inertia passes through zero — at 45.0° from the drawn axis. The value the drawing suggests, 3.733 × 10⁶, is neither of them.

Sections and stress

The axis a column buckles about

A strut buckles about the axis with the smallest second moment of area, and for a section with no axis of symmetry that axis is neither of the two on the drawing. An angle used as a strut is 2.45 times weaker than the number a designer reads off its own dimensions.

One point, every plane through it, one circle. A point carrying 180 N/mm² across one face, 90 across the other and 40 of shear. As the plane is turned, the pair (σ, τ) runs round a circle of radius 60.2 centred at 135.0 — and it goes round at twice the rate the plane does, which is the part always misremembered and the part that makes the picture work. The principal stresses are 195.2 and 74.8, on planes 20.8° from the face the 180 acts on; the largest shear on any plane is 60.2, exactly the radius, and it sits 45° from those — which is 90° round the circle. The von Mises stress that ranks this state against any other is 170.6.

Sections and stress

The circle nobody draws

A plane stress state has three principal stresses and the third is zero. When the two on the drawing share a sign, the largest shear in the state involves the one that is not there — and the circle a designer has drawn is not the circle that governs.

Warping stiffens a short member and nothing at all a long one. The stiffening 1/[1 − tanh(κ)/κ] against kL, both axes logarithmic, over kL from 0.05 to 200. At the low end the curve is a straight line of slope −2, because for small kL the bracket is κ²/3 and the stiffening is 3/kL²: it reaches 1201 at kL = 0.05, falls to 1.005 at the top, and every open section ever rolled sits somewhere on it. The same three plates arranged three ways are marked: the 533 by 190 mm I-section at kL 2.76 and ×1.562, the tee at kL 37 and ×1.027, the angle at kL 34 and ×1.030. A tee's warping constant is 288 times smaller than the I-section's and an angle's 236 times, because their plates meet at a point and there is no pair of flanges to bend against each other — so they have no warping resistance to offer at all, and that is the reason an angle is a poor thing to twist.

Stability

The restraint that beats the gradient

A moment-gradient factor is worth up to 2.7 on a beam's critical moment and is tabulated everywhere. Holding the ends against warping is worth more, is achieved by a detail rather than by a load case, and appears in no table at all.

A channel has three critical loads, not one. The three critical loads of a channel in compression, against its length, with the load it actually buckles at drawn over them. At 3500 mm the flexural loads are 13970 kN about the major axis and 2306 kN about the minor, while twisting about the shear centre takes 1555 kN. The lowest root is 1484 kN, and the column twists. The shear centre sits 109.7 mm from the centroid, so the modes cannot happen separately: the lowest root of the coupled problem is 4.5 per cent below the lowest of the three, and the Wagner coefficient β is 0.60. The governing mode changes at 5813 mm: below that length the column twists, above it, it bends — because the torsional resistance keeps a term that does not grow when the member is shortened, and the flexural loads have none.

Stability

The third root of the cubic

A column has three buckling loads and an Euler calculation finds two of them. The third is a twist about the shear centre, and for a section whose shear centre is not at its centroid the three cannot happen separately — so the answer is the lowest root of a cubic and can be a third below anything the two familiar modes report.

Deflection

An influence line is a deflected shape

Finding where a load has to stand to be worst means solving the structure once for every position it could stand in. Reciprocity says the answer is a single deflected shape — release the quantity being asked about, move it by a unit, and the shape the structure takes is the influence line.

Deflection

The member that is not worth stiffening

A truss's deflection is a sum of one term per member, and a term is zero whenever either force in its product is. A vertical carrying the whole of a panel load can contribute nothing at all to the movement — which a total can never show and a per-member sum shows nothing else.

The answer arrives in instalments. The hogging moment at support 1 of a three-span beam, cycle by cycle. It starts at the fixed-end moment of 98.0 kNm — the value with every joint clamped — and settles at 156.9 kNm against an exact 156.9. The error falls by about a factor of four per cycle: 21.03, 5.92, 1.54, 0.60 kNm after one, two, three and four. Two cycles is an engineering answer and nobody had to invert anything.

Deflection

Why it converges, and how fast

Moment distribution is an iteration, and iterations do not always converge. This one always does, at a rate the beam's own proportions fix — about a factor of four per cycle on a regular beam and considerably worse on an irregular one, which is where the method's reputation for two cycles being enough comes from and where it stops being true.

Throat stress round a fillet weld group. A l shape weld group carrying 150 kN at 200 mm from its centroid. The peak throat stress is 2.17 kN per mm of throat, at (139.53, 100); the worst point at maximum radius from the centroid carries 1.99. Checking by radius is wrong here by 9.17%, and points at identical radius differ by a factor of 1.

Connections

The radius rule, and where it fails

A weld group under an eccentric load is checked at the point furthest from its centroid, on the reasoning that the stress from the twist grows with the radius. That reasoning ignores the direction the two stresses point in, and for one common shape it misses the peak by nine per cent.

Cut the throat, and the face carries a moment and a tension at once. A crane hook of trapezoid section, 50 to 120 mm radius and 40 to 15 mm wide, carrying 100 kN, drawn beside the section at the cut and the stress across it. The load hangs on a line through the centre of curvature, so cutting the throat and taking everything below the cut as the free body leaves a face carrying a direct tension of 100 kN and a moment of N·R = 7.970 kN·m about the section's own centroid, which sits a full R = 79.70 mm from the load line. The stress is a hyperbola, zero at r = 75.04 mm rather than at the centroid 4.65 mm outside it, reaching 497.5 N/mm² of tension at the inner fibre and 281.3 of compression at the outer. The straight-beam formula, drawn dashed, reports 323.4 N/mm² for the bending part against the true 445.5, and leaves the 51.9 N/mm² of direct tension out altogether — between them, 54% under the real peak, at the fibre where a hook actually breaks. This is why a hook is trapezoidal: both effects are worst inside, so the material goes there.

Sections and stress

The wide side goes inside

A crane hook's section is a trapezoid with its broad face towards the centre of curvature, and that is not a casting convenience. Turn the same section round — same area, same depth, same moment — and the stress at the fibre that breaks rises by thirty-nine per cent. The shape is doing two things at once, and only one of them is in a straight beam's arithmetic.

Two load sets with the same resultant and different work. The two ways of putting a uniform load of 10 kN/m onto a beam element 6.00 m long. Both put 30.0 kN at each node, so both have the same resultant and the same moment about any point — they are equivalent for a rigid body. The consistent set adds a couple of 30.0 kN·m at each end, in opposite senses, which is what makes it do the same virtual work over the element's shape functions as the real load does. The couples cancel in the resultant, which is exactly why the resultant cannot see them, and they are the whole difference between an exact answer and one that is a third out.

Equilibrium

Equivalent in work, not in resultant

Two force systems with the same resultant and the same moment about every point are interchangeable — for a rigid body. A finite element is not a rigid body, and substituting one for the other on a beam element leaves the tip of a cantilever a third too low with no warning of any kind.

A portal on a stepped base — the two passes added. Bending moments on a single-bay portal with columns of 5 m and 3.5 m under one horizontal beam of 9 m, carrying 10 kN/m down and 60 kN across, with fixed bases. This frame is the two passes added. The corner moments are 5.8 and 99.1 kNm, and the short column's top carries 17.01 times what the tall one does. The diagram is drawn on the tension side of each member.

Deflection

Every joint balanced, and the frame still leaning

Moment distribution enforces one equation per joint, and a frame free to translate has one more equation than it has joints. So a table that balances perfectly can describe a structure held up by a prop nobody built — and finding the prop, then removing it, is a second pass whose unknown is a distance rather than a rotation.

Four things a far end can be doing, and what each is worth. The rotational stiffness of a member at one end, for four conditions at the other, each drawn as the shape the member takes when the near end is rotated through one unit. They are the same expression evaluated four times — M = (2EI/L)(2θ_a + θ_b) — and the only thing that changes is what the far end is known to be doing. A held far end gives 4EI/L and carries over a half; a free one gives 3EI/L and carries over nothing; a far end rotating equally and oppositely gives 2EI/L and carries over minus one, which is what a symmetric structure does to a member crossing its axis; and a far end rotating equally and in the same sense gives 6EI/L and carries over one. None of the four is an approximation. Each removes a freedom that was going to be discovered by iteration.

Deflection

Told what the far end is doing

Moment distribution discovers, cycle by cycle, that the pinned end of a beam carries no moment — a fact known before any arithmetic started. Telling it instead changes one stiffness from 4EI/L to 3EI/L and the work from thirty numbers to eight, for the identical answer. Cutting the beam on its own axis of symmetry gets it in two.

Every number in the table is a rotation, and none of them is a moment. The rotation contributions of a three-span beam of 8, 10, 8 m under 24 kN/m, sweep by sweep. There are six of them, one per member end, and not one is a bending moment: the moment is assembled at the end from M = FEM + 2m′ + m′ of the far end, and until that is done the table holds quantities that mean nothing on their own. That is the trade. A moment distribution stopped after two cycles hands over moments that are wrong by a known amount and can be used; this table stopped after two sweeps hands over nothing that can be read at all — and it gets there in four sweeps against Cross's own count on the same beam, writing six numbers a sweep rather than one per distributed member end plus a carry-over.

Deflection

The table that cannot be read halfway

Kani's method converges at exactly the rate moment distribution does, sweep for sweep and digit for digit, because it is the same iteration. What it changes is what is written in the boxes — rotations rather than moments — and that buys a shorter table that repairs its own mistakes and cannot be stopped early.

The whole line works, and the axis is not a choice. A weld group 200 mm deep carrying 15 kNm about an axis in its own plane, together with 100 kN of vertical shear. The bending force per unit length runs linearly from 1125 N/mm at one extreme to the same the other way at the other, through zero at the group's own centroid — and the centroid is where the neutral axis is because a weld carries compression across its throat as readily as tension. The 250 N/mm of shear runs along the weld and is uniform over the whole 400 mm, so the two components are perpendicular to one another and are combined on the throat rather than added as vectors in the plane.

Connections

The eccentricity at right angles to the drawing

A bracket's load stands off the plane of its welds as well as being offset within it, and the second eccentricity produces a completely different object — bending about an axis through the group rather than torsion about a point in it. The weld line has no compression zone to argue about, so its neutral axis is its own centroid, and the whole length works.

Three fasteners, three completely different clocks. What each of three shear fasteners carries against how far the joint has moved. A 400 kN fillet weld is linear to 0.4 mm and then gone — it is stiff and it is not ductile. A 380 kN bolt in a hole 2 mm larger than itself carries nothing until the hole closes and then rises over several millimetres of hole elongation. A 250 kN preloaded bolt is at its slip resistance in under a tenth of a millimetre and holds it until it slips. Two of these in one joint are at the same displacement, so the one that gets there first carries the load — and the weld gets there 23 times sooner than the bearing bolt does.

Connections

Two fasteners that never arrive together

Every steel code forbids adding a weld's capacity to a bolt's in one shear joint, and states it as a rule rather than deriving it. It is derivable. Two fasteners in parallel are at the same displacement rather than the same force, and a weld has ruptured at four tenths of a millimetre while a bolt in a standard hole has not yet touched the side of it.

Three quantities converging at three different rates. How much of the exact answer a truncated modal analysis of a ten-storey building reaches, against how many modes it keeps. The mass count is the rule — ninety per cent, reached at two modes. The base shear is ahead of it: one mode carries 85 per cent of the mass and 98 per cent of the base shear. The force in the top storey is behind it, at 80 per cent on one mode and 94 on two. The rule is written in the quantity that converges fastest, and it is checked against a quantity nobody plots.

Dynamics

The modes that were left out

Nobody runs every mode a model has, and the rule for how many is a mass count — enough to account for ninety per cent of the structure. The rule is written in the one quantity that converges fastest. On a twenty-storey frame two modes give the base shear to within one per cent and the force in the top storey to within twenty.

The whole deflected shape, found from the members' changes of length. A Pratt truss of eight panels at a depth of 1, carrying 15 kN at each top joint, at EA = 1, drawn as built and in its deflected shape, with every movement magnified the same number of times. The shape comes from Williot's construction and Mohr's correction: every joint's movement from the members' extensions alone, less the rigid rotation the supports forbid. The mid-span joint L4 moves down 2019.41 units, and the dots at every joint are a stiffness solution sharing none of that arithmetic, agreeing to 1e-14 of the largest movement. One construction gives all sixteen joints at once.

Deflection

The drawing that is right except for a rotation

Williot's construction finds every joint of a truss from its members' changes of length alone, in one drawing — and puts the roller six thousand units off its support. The error is one rigid rotation, Mohr's diagram takes it away, and a drawing started from the member symmetry holds still never makes it.

A fixed-ended member's end moments are the edge stresses of a column. A prismatic member, fixed at both ends under a point load 0.333 of the span from its left end, drawn three ways. At the top, the member. In the middle, the column the analogy puts in its place: a strip as long as the member and as wide at each point as 1/EI there, with its centroid, the elastic centre, 0.500 of the span from the left. At the bottom, the simply supported moment diagram, dashed, which is the load on that column; the straight line, which is the stress that load produces, P/A + M(x − x̄)/I; and the member's own moment diagram, which is the difference. The end moments are the column's edge stresses: 0.148 PL at the left and 0.074 at the right, which are Pab²/L² and Pa²b/L², and the largest sagging moment is 0.099 PL.

Deflection

The fixed-end moment is a column stress

A fixed end forbids a change of slope and a deviation, so on a member fixed at both ends Mohr's two theorems both have the answer nothing. Written with 1/EI as a width, those two conditions are the P/A + My/I of a short column under an eccentric load — the end moments are its edge stresses, and a haunched member's are read from where its elastic centre has moved.

Four forces pair off, and the line joining the pairs carries both resultants. A beam 8 m long held by the vertical link at the left end, the strut at 6 m, the horizontal link at the right end, under a load of 10 kN at 3 m inclined at −60.0°. Pairing the load with line A: the two cross at P, (0.00, 5.20) m, and lines B and C cross at Q, (6.00, 0.00). The resultant of the first pair passes through P and that of the second through Q, and since they balance each other both lie on PQ, dashed. The force polygon on the right is the load, then A, B and C, closing where it began, with PQ's direction as the diagonal that splits it into two triangles. The forces are A 4.33 kN along its line, B 6.12 kN along its line, C 0.67 kN against its line, as the three equations of equilibrium give them.

Equilibrium

The line that pairs four forces

Three forces in equilibrium meet at a point; four need not. But they pair off. The resultant of two passes through the point where their lines cross, the resultant of the other two through theirs, and the two resultants must share the line joining those points. Culmann's line turns a four-force body into two triangles — and the method of sections into a drawing.