Structural Engineering & Statics
where the load goes
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The figure library — page 3

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 3 of 7, joint-effect to p-delta.
What the joint does to the beam. End moment as a fraction of the fixed-end value wL²/12, against the joint's rotational stiffness, for a beam of EI/L = 14000. At the rigid boundary of 112000 kN·m/rad the joint delivers 80% of it and at the pinned boundary 20%. Everything between the two lines is a redistribution nobody chose and every analysis assumed away.

joint-effect

4 essays The end bolts do the work and the middle ones very nearly nothing. A lap of 8 bolts at 70 mm pitch transferring 800 kN between two plates, with the force each bolt actually carries drawn above it and the flat line a division by the bolt count would have given drawn behind. The end bolts carry 1.09 of their nominal share and the middle ones 0.94. The reason is not in the bolts: at the leading end the first plate is carrying everything and the second nothing, so the two strain at different rates and the slip between them is largest there. In the middle they strain alike, there is almost no slip, and a bolt with no slip across it transfers almost no force. The mean over the worst is 0.918, and the end bolt has to slip 1.36 mm before the rest catch up.

joint-length

3 essays The joints are not pins, and this is what that costs. A 4-panel Pratt truss solved twice on the same stiffness matrix: once with a moment release at every member end, which is the pin-jointed idealisation, and once with the joints continuous, which is what welding them produces. The axial forces are the same to within a per cent; the bending the second solution adds is worst in member 0, where the bending stress reaches 24.3% of the axial stress. Members are shaded by that ratio.

joint-rigidity

5 essays A joint is springs in series. The five components of an end-plate joint, with each bar the flexibility it contributes. The column flange in bending is 39.62% of the total on its own, and doubling its stiffness raises the joint's by a factor of 1.25 — while doubling the stiffest component buys 1.05. The joint's rotational stiffness is 25227.71 kN·m per radian.

joint-springs

2 essays The middle third, computed. The kern of a 400 × 600 mm rectangle, computed by asking, for every direction, how far the resultant can move before the far face would be pulled. It reaches ±100.0 mm vertically and ±66.7 mm horizontally, which are h/6 and b/6 exactly, and the region between is a rhombus rather than the ellipse the two numbers suggest — a resultant on a diagonal has less room than either axis allows.

kern-zone

8 essays A built-up column has a second way to bend. A 12 m column of two chords 300 mm apart, joined by double lacing. On the left it buckles the way a solid column does, by bending; on the right the chords stay straight and the lattice racks, which a solid column cannot do at all. Neither happens alone, and the two flexibilities add rather than the two stiffnesses — so the critical load is 1287 kN against an Euler load of 1341 kN, which is 96% of it, and the column behaves as though its slenderness were 80 rather than 79.

laced-column

3 essays Five millimetres short, and a hundred kilonewtons in every member. An X-braced bay 6 m by 4 m in which one diagonal was fabricated 5 mm short, with the force that leaves in every member. Nothing is applied to this frame. The forces are the self-stress state the frame's one redundancy supports, scaled so that the diagonal is pulled back to the length it should have been: tension in both diagonals at 100 kN, compression in the four members round the outside, and the whole set in equilibrium with nothing. That is 25% of the force the diagonal was sized to carry, and it is there for the life of the structure. Take one diagonal out and the frame becomes determinate: the short member then simply puts the joint somewhere else, and the structure is in the wrong place instead of under stress. Redundancy is bought, and this is the price.

lack-of-fit

1 essay Two shapes that are the wrong way up for each other. Deflected shapes of a 20-storey building under a uniform wind, drawn to the same scale. The wall alone bends: its shape is flattest at the base and steepest at the top, reaching 146 mm. The frame alone shears: it is steepest at the base where the storey shear is largest, reaching 140 mm. Tied together at every floor they reach 58 mm — less than a quarter of either, and less than the 72 mm two springs in parallel would give, because each is stiff exactly where the other is not.

lateral-system

7 essays A fourth power, and then a cliff. The factor of safety against rolling, against beam length, for one section hung from a roll axis 0.9 m above its centre of gravity. Nothing about the section changes along this axis. z̄ goes as the fourth power of the length — 0.236 m at 30 m becomes 0.747 m at 40 — and the factor of safety is proportional to (y_r − z̄), so it does not decline gently: it falls away and then stops existing. The working factor of 1.5 is lost at about 41 m, and past 42 m there is no hook height at all at which this beam hangs stably. Which is why long girders are lifted with the picks moved inboard, or with the beam braced, or not in one piece.

lift-roll

2 essays Ten decades of time for forty per cent of the strength. Strength as a fraction of the five-minute test value, against the length of time the load is held, over 11 decades of seconds. The relation is a straight line on this axis, which is the reason a single duration factor works at all: every decade of time costs about the same amount of strength. A load held for 50 years leaves 59% of the short-term strength, and the member need not have moved — this is not creep, and it is not fatigue, because nothing about the load varies. It is that a static stress slowly breaks fibres. The curve has an asymptote at 18% that nobody quotes: below about a fifth of its short-term strength a member has no time to failure at all, so there is a stress under which duration stops being a question rather than merely becoming a small one.

load-duration

1 essay A triangular load and the force that replaces it. A triangular distributed load with its resultant computed by integration: an area of 24.0 acting at 5.33 from the left. The two moment diagrams below show what the substitution costs — the reactions are identical and the peak moment is not.

load-resultant

3 essays Two beams tied together, and the deeper one takes 89% of the load. Two simply supported beams of 6 m, one twice as deep as the other, tied together at midspan so that they have to move as one. A load of 100 kN stands on the tie. Point stiffness is 48EI/L³, so the deeper beam is 8 times as stiff — depth cubed, nothing else — and the load divides in that ratio: 11.1 kN into the shallow beam and 88.9 kN into the deep one, 11% against 89%. Both midspan points move 5.00 mm, which is the whole of the argument: the geometry of the load never entered it. The deflection is drawn 78 times full size — the real sag is 5.00 mm on a 6 m span, about 1 in 1200.

load-sharing

2 essays The length at which a beam stops being a beam. Elastic critical moment against the distance between lateral restraints, with the section's plastic capacity drawn across it. The two cross at 3803 — beyond that length the beam buckles sideways before it reaches the strength its cross-section has, and the capacity is set by the restraints rather than by the steel.

ltb-curve

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

material-index

4 essays Take that one away and the load finds another route. A 6-panel pratt truss under 20 kN at each top node, before and after member 2 is removed. The load redistributes. The worst-affected survivor now carries 2.03 times what it did, and four members that carried nothing before are now working. Whether that is survival depends on how much spare capacity was there, which is a different question from whether the frame was strong enough.

member-loss

5 essays How much of the mass each mode carries, over eight modes. The effective modal mass of each of the eight modes of a frame of eight storeys, as a percentage of the total. Mode 1 carries 85.6% and mode 2 9.1%; two modes are needed to reach 90% of the mass. The masses sum to exactly the total, which is a property of the eigenvectors rather than a normalisation applied afterwards.

modal-mass

3 essays Three modes of a five-storey frame. The first three mode shapes of a five-storey shear frame, from the eigenvalue problem rather than sketched. Mode 1 has a period of 0.6 s, no node and carries 88.0% of the mass; Mode 2 has a period of 0.21 s, one node and carries 8.7% of the mass; Mode 3 has a period of 0.13 s, two nodes and carries 2.4% of the mass. The nth mode crosses the axis n−1 times, which is a theorem rather than a drawing convention.

mode-shapes

14 essays One point, every plane through it, one circle. A point carrying 140 N/mm² across one face, 0 across the other and 45 of shear. As the plane is turned, the pair (σ, τ) runs round a circle of radius 83.2 centred at 70.0 — 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 153.2 and -13.2, on planes 16.4° from the face the 140 acts on; the largest shear on any plane is 83.2, 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 160.2.

mohr-circle

7 essays The area is the rotation, and its first moment is the movement. A 6 m cantilever under a tip load of 10, with the M/EI diagram beneath it. The shaded area is 180.00, which by the first theorem is the change of slope along the whole member. Its centroid is at 2.000 m, and the first moment about the tip is 720.00 — which by the second theorem is the deviation from the tangent, and for a cantilever that tangent is horizontal, so it is the deflection itself. Integrating the curvature twice instead gives 720.00.

moment-area

9 essays What it costs to reach the plastic moment, for two shapes. Moment against curvature for two cross-sections of identical area (3000 mm²) and identical depth (200 mm), in mild steel, each divided by its own first-yield moment and its own first-yield curvature. The rectangle has a shape factor of 1.50 and reaches 98% of its plastic moment at 4.3 times the curvature at first yield; The I-section has a shape factor of 1.09 and reaches 98% of its plastic moment at 1.2 times the curvature at first yield. The dashed lines are the rigid-plastic moments, computed from the equal-area axis rather than read off the curves, and no curve reaches its own.

moment-curvature

12 essays 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 53.3 kNm — the value with every joint clamped — and settles at 64.0 kNm against an exact 64.0. The error falls by about a factor of four per cycle: 14.00, 3.50, 1.95, 0.59 kNm after one, two, three and four. Two cycles is an engineering answer and nobody had to invert anything.

moment-distribution

9 essays Moment against rotation, for three real joints. Three connections on one plot, with the classification boundaries for a beam of EI/L = 14000 drawn as rays through the origin. web cleats is pinned, flush end plate is semi-rigid, extended end plate is semi-rigid. The boundaries are multiples of EI/L, so the same joint is rigid on a short stiff beam and semi-rigid on a long slender one.

moment-rotation

4 essays Nothing happens, and then everything happens. The moment capacity left to a section already carrying shear, against the shear as a fraction of what the web can take. The web holds 26.1% of this section's plastic modulus and the flanges hold the rest, and only the web's share is reduced — by the factor √(1 − v²) that von Mises leaves it. So the curve is flat for most of its length: the first per cent of moment is not lost until v = 0.27, half the shear capacity costs 3.5%, and 15% is not reached until v = 0.9. The tangent at v = 1 is vertical, which is why the last tenth of the shear range costs more than the first eight.

mv-interaction

5 essays The net section, and the path the tear takes. A 200 mm plate with two holes staggered by 50 mm at a gauge of 60 mm. The straight path through one hole leaves 178 mm; the diagonal path through both leaves 166.42 mm after the s²/4g correction adds 10.42 mm back. The shorter of the two decides, at 83.21% of the gross section.

net-section

10 essays The load that makes itself worse. The amplification of a deflection against the ratio of applied load to buckling load. A structure at half its buckling load deflects twice as far as first-order analysis predicts, and the curve runs away well before the load is reached.

p-delta

12 essays