Structural Engineering & Statics
where the load goes
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The load must go somewhere — page 3

Essays 49 to 72 of 306 on this thread, in the same order.
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.

 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.

 The hoops change their mind at an angle no proportion chose. The two membrane forces of a spherical dome of radius 30 m under 3 kN/m² of surface, taken from the crown to a base at 60°. The free body for the meridional force is the cap above a cone of half-angle φ, and vertical equilibrium of it gives N_φ = −wR/(1 + cos φ) directly: -45.0 kN/m at the crown falling to -60.0 at the base, compression everywhere. Equilibrium normal to the surface then gives the hoop force, which starts at -45.0 kN/m and reaches 15.0 — it changes sign, and the angle at which it does was found here by bisecting N_θ rather than quoted: 51.827292°. Setting N_θ = 0 gives cos²φ + cos φ − 1 = 0, so cos φ is (√5 − 1)/2, the reciprocal of the golden ratio — an identity this site's solver gate checks against the bisection to nine decimals rather than asserting, because it is too pretty to be believed on sight. Below that parallel the hoops are in tension, which masonry has none of, and that is where every old dome is cracked. Structural form

The surface that carries by being curved

A flat plate spends its thickness on a lever arm of a few millimetres. Curve the same sheet and the load is carried in the surface itself, at a thirtieth of the thickness — and the entire bill is presented at the edges.

 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. Structural form

The column that stops

A load path that runs straight to the ground costs almost nothing. Interrupting one costs depth in proportion to the square root of the load times the distance it is moved — and the interruption's own deflection becomes the settlement of everything standing on it.

 A general force system is a screw, not a force. Two forces that are neither parallel, nor concurrent, nor coplanar. Their resultant is (30, 0, -60) and no point in space reduces them to it alone: the moment left on the central axis is (14.40, 0.00, -28.80), which is parallel to the resultant and has a pitch of 0.4800. The simplest object equivalent to the system is therefore a force and a couple about the same line — a wrench — and a planar drawing can never produce one, because a coplanar system has its moment perpendicular to its resultant by construction. Equilibrium

Moving a force, and what it costs

Every free body begins by putting a force somewhere convenient. That move is free along the force's own line, costs a couple across it, and in three dimensions leaves behind something no choice of point can remove.

 Four camber rules, and what each leaves on the finished beam. The same 12 m composite beam, cambered against four different things, followed through its own load history. Positive is a sag and negative a hog, and the point at the left of each line is the shape it was fabricated to. Cambering against the wet concrete leaves 12.7 mm of sag at the end and a flat beam on the day the slab is poured; cambering against the total load leaves the beam dead flat when fully loaded and hogged 37.9 mm — one part in 316 of the span — before anything is on it at all. Deflection

Built to the wrong shape on purpose

A cambered beam is fabricated curved upward so that load bends it down to something like straight. Nothing in the analysis changes, no stress anywhere is altered, and almost every mistake made with it is a bookkeeping mistake about which loads count.

 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. Stability

The panel that carries more after it has failed

Everywhere else in this field a critical load is where the argument ends. A thin web is the exception — it buckles visibly, in waves anybody can see, and then goes on to carry nearly twice as much again by turning itself into a truss nobody drew.

 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. Internal forces

When there is no section to design

Beam theory needs a section, and a section needs the strain to be linear across it. Within about a depth of a support, a load, a corner or a hole it is not — and those are the regions structures actually fail in.

 Two beams, or one beam four times as stiff. Two 200 × 150 planks spanning 4 m under 6 per millimetre. Loose, they have 112.5×10⁶ mm⁴ between them and deflect 16.2 mm, with the two faces at the interface sliding past one another. Bonded, the pair has 450.0×10⁶ — exactly 4 times as much, because doubling a depth cubes — and deflects 4.0 mm at half the extreme-fibre stress. Nothing was added but a restraint on slip. With connectors of stiffness 200 the same beam deflects 5.4 mm, which is 89% of the way from one bound to the other. Internal forces

Two beams, or one beam four times as stiff

Stack two planks and they bend as two beams whose faces slide past one another. Bond the faces and the pair has one neutral axis, four times the second moment and half the stress. Nothing was added but a restraint on slip.

 The same restraint, twice, with opposite signs. A 4 m strip of 200 mm slab whose ends cannot move apart, against deflection measured in its own thicknesses. The flat line is what a yield-line calculation gives, which is what the same strip would carry if its ends were free: 30.0 per unit width. The rising branch is compressive membrane action — the deflected strip is forced into an arch — and it peaks at 116.6, which is 3.89 times the yield-line load, at a deflection of 0.24 of the thickness. Past that the arch runs out of depth and the load falls back to the flexural one; past a deflection of one thickness there is no arch left and the reinforcement starts carrying the strip as a cable. It gets back to the arch's load at 2.17 thicknesses, which is one part in 9 of the span — a sag nobody would design for and exactly what a floor does instead of falling. Internal forces

The force nobody put in the model

A slab strip whose ends cannot move apart is not the strip in the yield-line calculation. Deflecting shortens the chord between its ends, the ends do not come in, and the strip is forced into an arch — worth four times the load it was designed for, at a movement nobody would see.

 The same strain, two moduli, and a width multiplied to say so. A timber section with a steel plate in it, carrying 20.0 kNm. Plane sections stay plane, so the strain at a height is the same in both materials; Hooke's law then puts the stresses in the ratio of the moduli, which here is 19.09. Multiplying the stiffer material's WIDTH by that ratio gives a fictitious section of one material with the same neutral axis and the same forces — 595.2×10⁶ mm⁴ of it, against 351.0 for the same shape with the moduli ignored. The steel plate is 3.8% of the area and carries 43% of the moment, at 96 N/mm² against the timber's 5.0. The transform is not an approximation: it is compatibility and Hooke's law written down. Sections and stress

A section made of two materials, one of them pretended away

Multiplying a material's width by the ratio of the moduli produces a fictitious section of one material with the right neutral axis and the right forces. It is not a trick — it is compatibility and Hooke's law written down — and it says a stiff material takes what its modulus asks for.

 A hole in a web is a Vierendeel panel. A 400 × 300 rectangular opening in a 533 deep beam, 15% along a 9 m span carrying 20 per metre — where the moment is 103 kNm and the shear 63 kN. The moment is a couple on the two tees, 301 kN on a lever arm of 343 mm, which is 70.3 N/mm² of uniform stress. The shear has nowhere to go but through the tees, so each carries 32 kN over the opening and bends in double curvature: a Vierendeel moment of 6.3 kNm and 167.6 N/mm² on top. So 70% of the stress at the corner exists because the hole has a LENGTH, and only 30% of the section's second moment has gone. Sections and stress

The hole that costs nothing, and everything

A service opening removes 30% of a beam's second moment and 0.7% of its deflection. What it costs is not that. Across the opening the shear has nowhere to go but through the two tees, and a tee carrying shear over a length bends.

 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. Structural form

The truss with no diagonals

A rectangle without a diagonal is a mechanism, so a Vierendeel girder makes its corners rigid instead. The shear a diagonal would have carried as axial force now travels as bending in the chords, and bending is a far more expensive way to move a force.

 A cable alone goes to a kink, and a kink is not a road. A point load of 1000 at mid-span of a 900 m suspended deck. The upper shape is the cable with no girder at all: two straight lines meeting under the load, because a cable takes the funicular shape of whatever is on it and the funicular of a point load is a kink — 0.0083 radians of it here. The lower shape is the same cable with the girder present, peaking at 1.125 against the bare cable's 1.873. The girder is not carrying the load — it takes only 17% of it — it is spreading it, over a characteristic length of √(EI/H) = 183 m, and what reaches the cable is spread over that length rather than arriving at a point. Structural form

The deck is not there to carry the load

A cable takes the shape of whatever is on it, which is exactly the problem — under a point load its shape is a kink, and a kink is not a road. The stiffening girder exists to spread the load until what reaches the cable is something the cable's own shape is right for.

 A check made on a perimeter, not on a section. One bay of a flat slab, 7.2 m square, on a 400 × 400 mm column. The heavy closed line is the control perimeter, 2d from the column face with its corners rounded at that radius — 4427 mm long against 1600 mm round the column itself. The shaded area inside it delivers no shear across it and is subtracted from the load; everything outside arrives through the perimeter. At 12 kN/m² that is 604 kN across 4427 × 225 mm, a shear stress of 0.606 N/mm² against a resistance of 0.658. Internal forces

A check made on a perimeter, not on a section

Every shear check in this collection is made on a plane cut through a member. A slab sitting on a column has no such plane, because the shear leaves in every direction at once — so the check is made on a closed line, and a line grows with the column while the load grows with the square of the bay.

 The plan a straight beam does not have. A beam of radius 12 m turning through 60°, seen from above, with bending drawn outward from the axis in one colour and torsion in the other. The load is vertical and uniform and nothing is applied off the axis. Bending reaches 446 and torsion 155; the two peaks are in different places, which is why the section has to be chosen for a combination rather than for either. Internal forces

Bending that arrives as twist

A straight beam under a vertical load carries no torsion unless something applies one. A beam whose axis curves on plan carries torsion everywhere, from the same load, with nothing applied off the axis — and it cannot be simply supported at all.

 The coefficient is a slope, and that is why it can exceed one. The crack magnified: two rough faces, drawn as a sawtooth at 54° to the plane. Sliding one over the other cannot happen without lifting it, so a shear displacement forces a separation in fixed proportion — the tangent of that angle, which is the number written down as a coefficient of friction and here is 1.40. The bars crossing the plane are stretched by the separation and clamp the faces back together; they are not carrying the shear, they are supplying the normal force that lets the roughness carry it. A bar that is not anchored on both sides supplies nothing. Internal forces

Shear across a crack that is already there

Every shear calculation in this collection starts from an uncracked solid — a principal stress, a shear flow, a diagonal tension. This one starts after the crack, on a plane with no tensile strength at all, and the coefficient it uses is not a coefficient of friction. It is the slope of the roughness.

 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. Sections and stress

The section that changes along the span

A prismatic beam is checked where the moment is largest, and everyone knows where that is. A tapered one is not, because the capacity is moving too — and for a cantilever with a load at its tip the governing station is exactly where the depth has doubled, with no length, no load and no material in the answer.

 A frame with no plane to be drawn in. The tetrahedron, solved: three equations at every free joint, one axial force in every member. Members drawn heavy carry more; tension and compression are separated by the sign that came back from the solve rather than by inspection. The count is m + r = 12 against 3j = 12, which makes it exactly determinate, and joint equilibrium closes to 1.8e-15. Structural form

Three equations at every joint

A plane truss is determinate when m + r = 2j. A space frame needs 3j, and that one changed digit is why a cube of twelve bars is six mechanisms short while looking perfectly solid — and why every three-dimensional frame ever built is made of triangles in several planes at once.

 Two centres, and the distance between them is a torque. A storey 30 by 18 m with its walls drawn heavy, pushed in one direction by 1000 kN. The force acts through the centre of mass and the storey turns about the centre of rigidity — the stiffness-weighted centroid of the walls, at x = 15.0 m — and the distance between the two is an eccentricity of 0.00 m before the 5% that has to be assumed anyway. The table below the plan splits each wall's force into its direct share and its torsional one. Torsion relieves the walls near the centre of rigidity and loads the far ones, so the wall in trouble is not the wall carrying the most: west wall is asked for 8% more than its direct share, and the walls at right angles to the push carry 19 kN each with nothing applied along them at all. Structural form

The corner that moves most

A lateral force is shared out in proportion to stiffness only if it passes through the centre of rigidity, which is not the centre of the plan and not the centre of mass. The distance between the two is a torque, and the wall that pays for it is the one furthest away and carrying least.

 Whether the floor shares the load out by stiffness or by area. The share of a uniform storey force taken by each of three equally stiff walls, against the stiffness of the floor plate that spans between them. A plate far stiffer than the walls translates almost rigidly, every wall deflects the same and the share is the ratio of stiffnesses — 60% to the middle wall here, the same as everyone else. A plate far softer than the walls behaves as a continuous beam over them and the middle wall takes 33%. Neither end is the tributary-area answer of 50%, which assumes a plate that is both soft and discontinuous over the wall, and which nothing here ever reaches. Structural form

The floor is a beam lying down

A floor plate spans horizontally between the walls that resist a lateral load, carries a distributed inertia load, and has chords, a web and a span-to-depth ratio like any other beam. Its stiffness decides whether the walls share the load by their stiffness or by the area of floor nearest them — and the familiar tributary answer turns out to be neither limit.

 The bearing is one length and the web is loaded over another. A load applied over a stiff bearing of 100 mm on the flange of a girder with a 1200 × 8 mm web. The flange bends under it and the yield lines that form spread the load along the web over 559 mm — 5.6 times the bearing, and 82% of the yield resistance is that spread rather than the bearing. The effective length is not a decision anybody made: it is what the flange's own bending stiffness against the web's own strength works out to. Stability

The load that chooses its own length

Every other load in this collection arrives over a length somebody decided. A wheel on a crane girder does not — the flange bends under it and spreads it along the web, and how far it spreads is an output of the flange's own stiffness against the web's own strength. The effective length is 5.6 times the bearing that produced it.

 How stiff a brace has to be before the frame stops swaying. The effective length factor of a swaying portal against the stiffness of a horizontal spring at its head. The curve starts at k = 1.317, the unbraced value, and falls to 0.774 — the factor for the same frame with its head held — at a brace stiffness of 23.2 EI/L³. Past that point the frame buckles in the non-sway mode, which the brace does not restrain, and further stiffness buys nothing at all. The threshold is worth stating as 1.41 N꜀ᵣ/L, which is the form the number is memorable in: for a storey carrying a thousand kilonewtons over four metres it is about 0.35 kN per millimetre of sway. Against the frame's own lateral stiffness of 12.0 EI/L³ it is a factor of 1.93. Stability

The most dangerous day is before it is finished

A structure is analysed once, complete, with every restraint present. It spends weeks in states nobody drew — a beam landed with no deck on it holds 17% of the moment its section is worth, a frame not yet braced buckles at a third of the load it will, and a bolt not yet tightened is a pin where the analysis assumed a fixity.

 Two differences up the same building, peaking in different places. Differential shortening between a perimeter column and the core of a 40-storey building, plotted up the height. The part driven by load peaks at level 20 — exactly half way up, because a floor near the top has almost nothing built above it to shorten what is beneath, and a floor near the bottom has almost nothing beneath it to shorten. The part driven by shrinkage does not care what is above it at all and accumulates all the way to the roof. Their sum is worst at level 40, at 43 mm, which across a 9 m bay is a floor out of level by one in 208. Deflection

The columns are shorter than the core

Every column in a tall building gets shorter as the building is built on top of it, and the core beside it gets shorter by a different amount. The floors between them tilt by the difference — and the difference is largest exactly half way up, because a floor near the top has almost nothing built above it and a floor near the bottom has almost nothing beneath it.

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