Structural Engineering & Statics
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The figure library — page 6

Every picture here is generated from code. This page lists the generator families, how many essays call each one, and how many separate drawings each has produced. Page 6 of 7, stiffness-matrix to vortex-lock-in.
Almost all of it is exactly zero. The stiffness matrix of a 2-bay, 3-storey plane frame: 36 freedoms, of which 16.2 per cent of the 36² entries are non-zero. The zeros are not small numbers; they are absences. A member reaches only the two nodes at its ends, so it can contribute nothing to any row belonging to a node it does not touch, and every such entry is zero exactly rather than nearly. The non-zeros therefore sit in a band of width 13 about the diagonal. Before the supports are applied the matrix is singular, and its null space has exactly three dimensions — the three rigid-body motions a plane frame has with respect to the ground, which is the same statement the null vector of the equilibrium matrix makes about a truss that is a mechanism, arrived at from the other end.

stiffness-matrix

6 essays The characteristic strength, which nothing was measured at. A lognormal population of strengths with a mean of 30 N/mm² and a coefficient of variation of 0.15. The characteristic value is the 5% fractile — 23.2 N/mm², which is 77% of the mean, and which need not be the strength of any specimen that was tested. Dividing it by 1.50 gives 15.5, and the shaded sliver below that is the fraction of the population that would fail to reach it: 6.4e-6, or one in 155,818. A factor applied to a fractile is not covering the scatter, because the scatter has already been spent getting to the fractile.

strength-scatter

7 essays Three times the stress, and it does not matter how big the hole is. The hoop stress around a circular hole in a wide plate pulled at 100 N/mm², from Kirsch's exact solution. At the sides of the hole it is 3.0 times the applied stress — 300 N/mm² — and the factor is the same for a hole of any radius, because the radius cancels. At the top and bottom of the hole it is -1.0 times the applied stress, which is compression in a plate that nothing is pushing. The disturbance dies quickly: the stress is within 5% of the applied value by 3.5 hole radii, which is Saint-Venant's principle with a number on it.

stress-concentration

8 essays Three ways to apply the same force, and one depth to forget the difference. Three end loads on a member 400 mm deep, all with the same resultant and the same moment: a point load, the same force spread over a fifth of the depth, and the same force split in two. What is plotted is the difference between each of them and the beam-theory answer — the self-equilibrating remainder — as a fraction of the mean stress. The point load starts at 20 times it and is under a tenth of it by 0.77 depths; all three are under one per cent by about 1.18. That distance is the licence every figure in this collection is drawn under, and the exact strip eigenvalue agrees with it: 2.106 + 1.125i, whose real part puts one per cent at 1.09 depths and whose imaginary part means the remainder changes sign on the way out, which no statement of the principle mentions.

stress-decay

4 essays Three materials pulled until they stop. Three stress-strain curves — mild steel, high-strength steel, aluminium alloy — plotted to a strain of 2.0%. One of them has a plateau, so the stress at which yielding starts is something the specimen does rather than something anyone chooses. The 0.2% offset construction is drawn on the high-strength steel: a line of slope E from a strain of 0.002, cutting the curve at 460 N/mm².

stress-strain

16 essays A truss drawn inside a solid, and solved as one. A deep member 4000 mm between bearings and 2000 mm deep, carrying 1200 kN at mid-span. The model is two struts and one tie, on a lever arm of 1600 mm, and it is solved by the truss solver rather than by a formula: the tie comes back at 750 kN and each strut at 960 kN, at 38.7° to the horizontal. Spread over a strut width of 812 mm the compression is 3.0 N/mm² against a limit of 15.8 for concrete cracked across its own strut, and the tie needs 1724 mm² of steel. A beam calculation on the same member would have asked the tie for 702 kN, which is 7% less than the model does.

strut-tie

11 essays Two curves climbing together, and the one that catches up first. A 6 m member tapering from 200 to 600 mm, with the moment it carries and the moment it can carry drawn on the same scale below it. The demand rises linearly and the capacity as the square of the depth, so the gap between them closes and then opens again. It is narrowest at 3.00 m from the free end, where the member is 400 mm deep and 79% used, against 70% at the root where the moment is largest.

tapered-beam

3 essays A buckled panel is a truss that nobody drew. A 1000 × 1000 panel of 6 mm web, at d/t = 167. It buckles in shear at 63.8 N/mm², which is 383 kN — and it then carries 696 kN, 1.82 times as much, because the tension diagonal takes over from the compression one that has gone. The band runs at 22.5° with a membrane stress of 252 N/mm² over a width of 541 mm, and it pulls on the flange at 221.3 N per millimetre of its length. A web that never buckled at all would have reached 953 kN, so the panel ends at 73% of a stocky web's capacity on a fraction of its steel.

tension-field

4 essays It moves, or it pushes. Never both, and never neither. A 30 m steel member 30 °C warmer than it was built, in three conditions. Free, it grows 10.8 mm and carries nothing. Held, it moves nothing and carries 75.6 MPa in compression — which is E·α·ΔT and contains neither the length nor the area of the member, so the identical stress arises in a two-metre strut. Held by a spring it does some of each: 3.2 mm of movement and 53.2 MPa, and the split is decided by the spring rather than by the member.

thermal-restraint

10 essays Three forces must meet at a point. A body held by two supports and one load. The reaction at the roller is vertical and the load's direction is given, so their lines of action fix a meeting point — and the pin reaction has to point at it.

three-force

15 essays A line of thrust, and the masonry it has to stay inside. An arch ring of 9% of the span in thickness, rising 28% of the span, under its own weight as a uniform load. Any horizontal thrust between 3.85 and 5.23 puts a line of compression entirely inside the masonry, so the arch stands — and which of them it actually takes is not decided by statics. The two extremes are drawn: the minimum-thrust line, which rides high at the crown and low at the haunches, and the maximum-thrust line, which does the opposite.

thrust-line

10 essays One slit, and the torsional stiffness falls by a factor of hundreds. A 200 by 200 box of 8 mm wall, drawn closed and then slit along its length. Closed, the torque runs round the wall as a shear flow and the torsion constant is 5.66×10⁷ mm⁴; slit, the loop is broken and only each wall's own thickness resists, giving 1.31×10⁵ mm⁴. The ratio is 432 to one, so the same torque twists the slit section 432 times as far and raises a peak shear stress 36 times as high. Nothing about the material changed.

torsion-compare

16 essays A torque diagram is a shear diagram about a different axis. A torque of 40 kNm applied 2 m along a member of 6 m held against twist at both ends. The two ends take 26.7 and 13.3 kNm, in inverse proportion to their distances, because the two halves are springs in parallel and torsional stiffness is GJ over length. The diagram steps at the applied torque and closes at the far end, exactly as a shear diagram does — the only difference is which axis the arrows turn about.

torsion-path

6 essays One of these two curves is a stiffness and the other is a statement of statics. The torque a spandrel beam carries, against how much of its torsional stiffness is left. The rising curve is compatibility torsion — a floor beam framing into the side of the spandrel, which shares its fixed-end moment of 197 kNm between the spandrel's torsional stiffness and its own flexural one. Uncracked, the spandrel takes 51% of it, or 100 kNm; at a quarter of that stiffness it takes 21%, or 41 kNm, and the floor beam picks up what was shed. The flat line is equilibrium torsion — a canopy cantilevering 2.2 m off the same spandrel, whose 116 kNm is fixed by statics and contains no stiffness at all. The first can be designed away by accepting a rotation. The second cannot be designed away by anything.

torsion-redistribution

3 essays 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 3000 mm the flexural loads are 19014 kN about the major axis and 3139 kN about the minor, while twisting about the shear centre takes 1962 kN. The lowest root is 1879 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 4.2 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.

torsional-buckling

3 essays The depth is decided by how far it moves, not by what it can carry. A column carrying 6000 kN landing 3 m into a 12 m transfer member. The free body is the member itself, cut under the column: M = P·a(L − a)/L = 13500 kNm, with 4500 kN of shear on one side of the cut and 1500 on the other. At an allowable stress that moment asks for 1.94 m of depth — the dashed outline — and keeping the settlement it causes inside the floors' own bending asks for 2.55 m, which is the member drawn solid. 31% more depth is bought by nothing the strength calculation can see. The depth grows as √(P·a), so four times the load is exactly twice the depth, and depth in a transfer member is a storey nobody occupies.

transfer-structure

3 essays Where a beam's load comes from. A 8 × 6 m panel carrying 5 kN/m², divided at 45° from the corners. The long beams take a trapezoid of 15.0 m² each and the short beams a triangle of 9.0 m²; the four areas sum to 48.0 m², which is the panel, so no load has been invented or lost. The line loads quoted are the uniform equivalents; the real distributions peak at 15.00 kN/m at midspan.

tributary-load

9 essays A Pratt truss of 6 panels. A Pratt truss under equal panel-point loads. The joint equilibrium equations were assembled and solved; 10 members came out in tension, 9 in compression and 2 carrying nothing.

truss

16 essays Every member's share of the movement, and they are not the members expected. A Pratt truss of six panels at a depth of 0.85, carrying 10 kN at each top node, with the movement of the bottom chord at mid-span attributed member by member. The unit-load sum δ = ΣF·f·L/EA gives 631.06 at EA = 1: 47.2% from four top chords, 28.5% from six bottom chords, 22.3% from six diagonals, 2.0% from five verticals. The single worst member is a top chord at mid-span at 14.8% of the whole. Each member is drawn at the width of its own share. The same deflection from a stiffness solution that shares none of this arithmetic is 631.06, a relative residual of 3.6e-15.

truss-deflection

9 essays Chord force against truss depth. The force in a truss chord for a fixed bending moment, against the depth of the truss. The relationship is a reciprocal: the chords form a couple whose lever arm is the depth, so a shallow truss pays for it steeply.

truss-depth

7 essays 3% of the mass, hung on a spring, against the peak it removes. The magnification of a structure with 1.0% damping, with and without a tuned mass damper of 3.0% of its mass, tuned to 0.9709 of its frequency with 10.5% damping of its own. The bare peak is 50; with the absorber the single peak becomes two of 7.34, a reduction to 15% — a factor of 6.8. The marked points at frequency ratios 0.923 and 1.044 are the fixed points: the response there is the same whatever damping the absorber is given, which is what makes the optimum a question with an answer. The absorber's own stroke at the worse of the two peaks is 29.57 times the structure's static deflection, and that stroke is what decides whether it fits.

tuned-mass

6 essays The deflection at x = 3, by virtual work. Three diagrams: the moment from the real load, the moment from a unit load placed where the answer is wanted, and their product. The area under the third, divided by EI, is the deflection — 197.50 here. No standard case was consulted, so the method works for any load pattern at all.

unit-load

13 essays The shear goes round the corner instead of across it. A 6-panel Vierendeel girder, 12 m by 1500 mm, under 100 kN at mid-span. There is no diagonal in it, so each panel's 50 kN of shear is carried as bending in the chords: the curves drawn along them are the chord moments, and every one passes through zero at the middle of its own panel. The local moment is the panel shear times the panel length over four, 25.0 kNm, and it adds to an axial force of 200 kN from the global moment at the same point. The girder deflects 6.20 mm against 3.18 mm for the same members triangulated — 1.95 times — and 68% of that movement is chord bending that a diagonal would have removed entirely.

vierendeel

4 essays Lock-in: the frequency the wind sheds at, and what it does to the chimney. A 1.2 m cylinder at 0.9 Hz. Vortices leave it at St·U/D, so the critical speed at which they match the structure is 6 m/s — a breeze, met many times a year rather than once in fifty. The upper panel shows the shedding frequency locking on to the structure across a band from 5.1 to 7.8 m/s; the lower shows the amplitude that results. At a Scruton number of 11.2 the peak amplitude is 180.94 mm, which is 15% of the diameter.

vortex-lock-in

3 essays