Structural Engineering & Statics
where the load goes
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Geometry beats material — page 2

Essays 25 to 48 of 243 on this thread, in the same order.
The hour that is really a temperature. The retention factors for carbon steel against temperature: the yield stress and the elastic modulus. The modulus falls away first — at 500°C the steel has kept 78% of its strength and 60% of its stiffness — so a member's failure mode can change during a fire. A member working at 60% of its cold capacity runs out of strength at 558°C, and out of the stiffness for the same ratio at 500°C, 58 degrees earlier. There is nothing about time in any of it: a fire rating is a temperature the member must not reach, converted into the minutes a particular fire takes to get it there. Materials

The hour that is really a temperature

A fire rating is quoted in minutes and there is no time in the physics anywhere. What decides is a temperature, and the stiffness reaches its limit sixty degrees before the strength does — so the way a member fails can change while it is burning.

 Prying action in a tee stub. A tee stub pulled by its web with 100 kN per bolt. The 20 mm flange is in the one-hinge regime, so the prying force at the flange tip is 50.63 kN and the bolt carries 150.63 kN — 1.51 times what was applied. The flange stops prying entirely at 26.97 mm thick, and collapses on its own at 110 kN. Connections

The force the bolt never saw applied

Pull a tee stub with a hundred kilonewtons and its bolt carries a hundred and fifty. The extra comes from the flange bending and pressing its own edge against the thing it is bolted to, and no free body of the connection as a point contains it.

 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. Connections

The tear that goes diagonally, and the correction that has no derivation

Stagger the holes so that no straight line crosses more than one and the plate does not get its strength back. The tear runs at an angle instead, and the arithmetic that makes it come out right is a century-old piece of curve-fitting nobody has improved on.

 Bearing and tear-out against end distance. A 20 mm bolt in a 10 mm plate. Below 165 mm of end distance the bolt tears a channel out to the end and the capacity is proportional to that distance; above it the plate crushes in front of the bolt and the end distance stops mattering. At 40 mm the capacity is 52.12 kN and the mode is tear-out. Connections

The hole that goes oval, and the one that tears to the edge

A bolt pressing on the side of its hole either crushes the plate in front of it or shoves a channel of metal out to the end. Which one happens is decided entirely by a distance that is usually set by a minimum in a table.

 A fillet weld is stronger across than along. Capacity of a 4 mm throat over 100 mm, against the angle between the weld's axis and the load. Loaded along its length it carries 116.83 kN; loaded across it, 143.09 kN. The ratio is 1.22, which is √3/√2 exactly, and it comes out of the failure criterion rather than out of a test. Connections

The weld that is stronger across than along

The same fillet weld, the same size, the same steel, carries twenty-two per cent more when the load runs across it than along it. The factor is exactly the square root of three over the square root of two, and it comes out of the yield criterion rather than out of a test.

 An angle bolted through one leg. A 100 × 75 × 10 angle connected through its 100 mm leg with three bolts at 75 mm pitch. The centroid sits 19.77 mm from the connected face over a connection 150 mm long, so U = 1 − 19.77/150 = 0.87 and 13.18% of the net area is not working. Connections

The angle that uses half of itself

Bolt an angle through one leg and the other leg is not fully working. The correction is one over a length — both halves of it are geometry, neither involves a material, and a two-bolt connection throws away a quarter of the section.

 How a moment crosses a gap. A moment end plate with three bolt rows. The moment is carried as a couple: tension in the rows, compression through bearing at the bottom flange. The plastic distribution reaches 172.8 kN·m and the elastic one 142.29 kN·m, a factor of 1.21 — and the compression at the bottom flange is 540 kN either way, which is the check that gets forgotten because it is not a bolt. Connections

Making a moment cross a gap

A moment is not a thing that can be handed across a joint. It has to be turned into a pair of forces, carried separately, and reassembled — and the whole design is a question of how the tension is shared between bolt rows that are not equally able to take it.

 Rayleigh's method: the frequency read off the deflection that was computed anyway. A simply supported beam of 8 m, sagging 13.08 mm under its own weight, sampled at 21 stations. Rayleigh's quotient over those deflections gives 4.91 Hz against the exact 4.91 Hz — 0.119% high, and high rather than low because an assumed shape is a constraint and a constraint stiffens. The rule of thumb, 18 divided by the square root of the deflection in millimetres, gives 4.98 Hz. Dynamics

The period nobody chose

Every structure has a natural period, it decides the answer to every question in this field, and no drawing anywhere records it. It is a consequence of a mass picked for one reason and a stiffness picked for another — and it has already been computed, by the serviceability check.

 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. Dynamics

The mass that helps by being late

Hang three per cent of a building's mass from a spring in its roof, tune the spring so the mass arrives a quarter-cycle behind the motion, and the peak response falls by a factor of seven. Nothing was strengthened and nothing was stiffened.

 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. Internal forces

The internal force with no diagram

A cut through a member reveals four things, and this collection has drawn diagrams for three of them. The fourth is a torque, it obeys exactly the same rules, and whether it exists at all can depend on a decision the designer is free to make.

 Loaded straight down, and moving sideways. An equal angle with a moment applied about the horizontal axis. Its principal axes lie at 45.0° to the drawn ones, so the neutral axis runs at -30.6° rather than horizontally, and the section moves 59% as far sideways as it moves down. The product of inertia that causes it is -1.066 × 10⁶ mm⁴, and it is zero for every section drawn in this field until now. Sections and stress

Loaded straight down, and it moves sideways

Every section drawn here so far had an axis of symmetry, and that symmetry has been doing silent work. Take it away and a vertical load produces a neutral axis that is not horizontal, a deflection that is not vertical, and on one ordinary section a sideways movement larger than the downward one.

 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. Sections and stress

The middle third

A material that cannot be pulled imposes a condition on where the load may land, and the condition is a region rather than a point. For a rectangle it is the famous middle third; for every other section it is a shape nobody quotes, and one ordinary section's is nearly twice as generous as the rule allows.

 A load with a maximum in it, and nothing bifurcates. Load against apex movement for a two-bar frame of half-span 1000 mm and rise 150 mm. The load rises to 133.4 kN at a movement of 64 mm — well short of the 150 mm that would bring the apex level — and then falls. Past that point the frame can only be held by taking load away, so under a dead weight it goes: 260 mm of movement at constant load, arriving inverted and in tension. The minimum on the path is -133.4 kN, the exact negative of the maximum, because the geometry is symmetric about the flat position and the arithmetic knows it. Stability

The roof that jumps

Every stability failure in this collection so far has been a bifurcation — a straight thing discovering it can be bent. A shallow frame does something else entirely. It stays perfectly symmetric, deforms steadily, and at some point the load it can carry starts to fall while it is still moving in the direction it was pushed.

 A two-way slab is a one-way slab as soon as it is not square. The share of the load carried by the strips spanning the short way, against the ratio of the sides. The two families of strips cross at the centre and must deflect equally there, and a strip's deflection goes as the fourth power of its span — so at a ratio of 1.33 the short strips already take 76% and at 2 they take 94%. The panel drawn here is 6 × 8 m, a ratio of 1.33, and its short strips take 76.0%. Two-way action is worth having at a ratio of one and worth almost nothing by two. Structural form

The slab that spans both ways

A panel supported on four sides sends its load in two directions at once, and the share is decided by a fourth power — so a panel a third longer than it is wide has already stopped being a two-way slab in any useful sense. What it does at collapse is a different calculation with a different answer.

 The reaction lies inside the cone, so the block stands. A block of 100 on a plane at 15°, against a coefficient of friction of 0.35. Resolving across and along the plane gives a normal force of 96.6 and a friction demand of 25.9, against a capacity of μN = 33.8 — a ratio of 0.77. Added together the two make one contact reaction leaning 15.0° from the normal, and the admissible reactions fill a cone of half-angle arctan μ = 19.3°. Equilibrium is possible exactly when the demanded reaction lies inside that cone, which here it does. The weight enters neither the cone nor the lean: a block of any weight on this slope leans its reaction by the same 15.0°, which is why the angle of repose is a material property and the size of a heap of sand is not. Equilibrium

The force that is whatever it needs to be

Every other force in statics has a value the equations produce. Friction has an inequality instead, so it takes whatever value equilibrium demands and the bound only ever says no — which means a problem with friction in it has a range of answers rather than one.

 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. Internal forces

The stiffest path takes the load

When two members share a force the split can be argued about. When they share a displacement it cannot — stiffness settles it, and nothing about the load or the plan drawing gets a vote. The consequence is that stiffening a lightly loaded member raises its stress, and the way to unload something is to soften it.

 The flanges go opposite ways, and the pair of them is the bimoment. A 305 by 165 mm I-section held against warping and twisted by 0.5 kN·m, with the section on the left and the two flanges seen in plan on the right. At the built-in end each flange bends in its own plane, one way at the top and the other at the bottom, through 29.3 mm at the free end — drawn 20 times its true size against the 6 m length. The pair of flange shears is 1.69 kN each, and 1.69 × 295 mm is 0.500 kN·m — the whole torque at that section, carried by two forces neither of which is a torque. The pair of flange moments is 3.04 kN·m each, and 3.04 × 295 mm is 0.897 kN·m², which is the bimoment. It puts 67.0 N/mm² into two diagonally opposite flange tips and takes the same out of the other two, so its net force and its net moment about every axis are zero — which is exactly why no member diagram has a place for it. Sections and stress

The section that cannot stay flat

Twist an I-section and its cross-section dishes out of its own plane. Stop that happening at one end and the member finds a second way to resist — the flanges bend in opposite directions — and the stress resultant that describes it has units nothing else in statics has.

 The flange that is drawn, and the strip of it that is working. A plan of a flange 3 m each side of the web on a 20 m span, with the working strip shaded. The effective width is 2.531 m per side, 84.4% of what is drawn, so 101238 mm² of the 120000 mm² of flange is doing the work and 18762 mm² is not. Widening the flange does not move the shaded strip much, because its width is set by the span: the design code would allow 2.500 m per side here and no elastic flange of any width beats 3.183 m, which is L/2π. The lag exists because longitudinal stress can only enter the flange through shear along the junction with the web, so the free edges arrive late — 76.5% of the web's stress at the edge, in this case. Sections and stress

The flange that is not all there

Stress can only get into a wide flange through shear along its junction with the web, and shear takes distance to do it. So the width that is working is set by the span, and a flange 3 m wide on a 20 m span has 18762 mm² of steel that is there, and paid for, and hardly carrying anything.

 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 50 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 50 kN and a moment of N·R = 3.985 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 248.7 N/mm² of tension at the inner fibre and 140.6 of compression at the outer. The straight-beam formula, drawn dashed, reports 161.7 N/mm² for the bending part against the true 222.8, and leaves the 26.0 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 bar that was bent before it was loaded

In a curved bar plane sections still stay plane, and the bending formula is wrong anyway. The fibres were different lengths before anything was applied, so an equal rotation of two plane faces produces unequal strain — the stress is a hyperbola, the neutral axis has moved inward, and a crane hook carries half as much again as My/I reports.

 A cruciform has three critical loads, not one. The three critical loads of a cruciform in compression, against its length, with the load it actually buckles at drawn over them. At 3000 mm the flexural loads are 921 kN about the major axis and 921 kN about the minor, while twisting about the shear centre takes 700 kN. The lowest root is 700 kN, and the column twists. The shear centre is the centroid, so the three modes are independent and the envelope is simply the lowest of them. The governing mode changes at 3442 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 column that twists instead of bending

Euler's column has one mode. A real column has three, and which of them governs is settled by where the shear centre sits. A cruciform strut buckles by rotating about its own length at a load that does not change no matter how short it is made.

 One restraint, and several times the load. The same portal — the same columns, the same beam, the same steel — buckling with its head held against sway and with its head free to sway. The braced frame's critical load is 16.46 EI/L² and the swaying one's is 5.69 EI/L², a factor of 2.89, and the effective length factor that comes out of each eigenvalue is 0.774 against 1.317. Both are eigenvalues of the assembled frame at a beam-to-column stiffness ratio of G = 1.00; the buckled shapes are the mode vectors themselves, drawn at 18 per cent of the storey height so that the movement can be seen. Stability

Held, and not held

One horizontal restraint at the head of a storey, carrying no vertical load whatever, moves the critical load of the columns beneath it by a factor of 2.89. Effective length is a property of the frame, not of the member.

 The shear part is not a curve at all. A 300 × 600 rectangle spanning 2,400 mm, a span-to-depth ratio of 4, with its bending deflection and its total drawn together. Bending gives 0.0254 mm and shear a further 0.0049 mm, so 16.3% of the movement is the term beam theory drops. The shear part is drawn separately beneath: two straight lines meeting under the load, each limb straight to 2e-16 of the peak. It is the shear diagram integrated once where the moment diagram above it is integrated twice, so it carries one degree of curvature less than the bending shape — none at all here. The real movement is 1 in 79,116 of the span; both panels are drawn at about 11,932 times it. Deflection

The deflection that is not bending

Engineer's beam theory computes a deflection as the second integral of a moment, and that calculation silently drops a term. The beam also shears, and the shear deflection is not a correction to the curve — it is a different shape, and for a deep member it is most of the answer.

 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. Deflection

Which member moved the roof

A beam sags because it curves. A truss has no curvature anywhere — it comes down because every one of its members changes length, and the sum of those changes, weighted member by member, is a ranking that names which ones are worth stiffening. Usually not the ones a designer worries about.

 The further it deflects, the harder it pulls back. Total load against midspan sag for a 30 m cable of 1000 mm² prestressed to 500 kN, carrying 5 kN/m. The cubic H³ − T₀H² − w²L²EA/24 = 0 was bisected at every point of the curve, so the sag at the full 150 kN is 0.740 m rather than the 1.125 m the flat-cable formula WL/8T₀ gives — the straight dashed line, which is the tangent to this curve at the origin and nothing more. Its slope is the initial stiffness 8T₀/L = 133.3 kN/m; at the marked point the tangent has reached 341.2 kN/m, 2.56 times as stiff, and the horizontal component of the tension has risen from 500 kN to 760 kN. Nothing about the steel changed. The geometry got better at the job. Structural form

The stiffness that comes from the shape

A cable has no bending stiffness whatever, and it still holds up a roof. What resists the load is the change of its own geometry, so its stiffness is a function of the tension already in it — and prestress buys stiffness that no change of material could.

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