Structural Engineering & Statics
where the load goes
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Drawing as calculation — page 2

Essays 25 to 48 of 147 on this thread, in the same order.
The moment does not stop at the end of the beam. A portal frame of 8 m by 4 m with fixed bases, carrying 20 kN/m on the beam. The bending moment is drawn on the tension side of every member, and it runs round the corner without a break: 65.2 kNm arrives at the end of the beam and 65.2 kNm leaves down the column, which is the same number, since joint rotational equilibrium is one of the equations the frame solve satisfied. Midspan carries 94.8 kNm, and the two add to 160.0 — the 160.0 kNm of a simply supported span, to 0.0e+0 kNm. The corner takes 61% of the wL²/12 a fully built-in beam would have carried, because the columns are springs rather than walls: the beam-to-column stiffness ratio is 1.27. The beam's moment crosses zero 0.92 m from the corner and the column's 1.33 m above its base. Internal forces

The moment that goes round the corner

At a rigid knee the bending moment does not stop at the end of the beam. It turns and runs down the column, and in the same instant the beam's shear becomes the column's axial force — while the block of steel that has to carry the turn appears on no member diagram anywhere.

 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.

 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.

 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.

 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.

 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. Deflection

The area of a diagram is a rotation

A deflection is the double integral of a bending moment, and the two constants of integration are the whole difficulty. Mohr's theorems replace them with two pictures — an area, and where that area's centre of gravity sits.

 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.

 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. Sections and stress

The worst stress is not where the worst bending is

Every stress this collection has quoted is a stress on a particular plane, and neither the bending stress nor the shear stress is a property of the point. Turn the plane and both change; one pair of numbers does not, and on a short beam it peaks where neither of them does.

 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 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 540 kNm over the middle support. The secondary moment is what is left when the primary is taken off the total, and it is 306 kNm — 57% 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: 51.0 kN pressing down there and 25.5 kN lifting at each end, a reaction set that sums to -2e-13 because nothing external was applied. Its diagram is straight between supports to 2.0e-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 prestress that pushes back

On a simply supported beam a tendon is an internal matter and changes no reaction. Put the same beam on three supports and the tendon lifts it off the middle one, the support refuses, and the force it takes to hold the beam down is a reaction produced with no load applied at all.

 The cut that severs the stirrups is the cut that counts them. A cracked web drawn as the truss it has become: a tension chord along the bottom, a compression chord along the top, concrete struts between the cracks and stirrups crossing them. The free body is a cut parallel to the cracks, which over a lever arm of 495 mm severs z·cot θ/s = 8.3 stirrups at cot θ = 2.5. Every one of them is at yield, so the shear is their number times their strength, and flattening the crack raises the count rather than the strength. The same cut passes through the bottom chord, which is where the chord force the moment diagram does not contain comes from. Internal forces

The beam that becomes a truss

Once a web has cracked in shear there is no shear stress field in it any more. There are concrete struts, two chords and whatever crosses the cracks, and the angle of those cracks is not a property of the material — it is something the designer chooses, and every quantity in the beam moves when it changes.

 Two null spaces of one matrix, and the count is their difference. Two pin-jointed frames, each with the forces it can carry with nothing applied to it drawn on its bars — tension one colour, compression the other, thickness in proportion. That force set is the null space of the equilibrium matrix; a mechanism is the null space of its transpose; and Maxwell's count b + r − 2j is the difference of their dimensions and knows neither of them separately. A square with both diagonals has s = 1 and m = 0. Two bars in a straight line has s = 1 and m = 1 with a count of 0, so the count is satisfied by a frame that both folds and can be prestressed — and the prestress stiffness is positive, which is why a tensioned pair of collinear bars is stiff at all. Structural form

The forces that are there with nothing applied

Maxwell's count is the difference between two dimensions, and it knows neither of them separately. A frame can satisfy it exactly and still both fold and be prestressable — and when it does, the second of those is what stops the first.

 The force runs along the chord whatever route the member takes. A member pinned at two points 10 m apart, loaded only at those two points, and bowed 1.2 m off the line between them. Equilibrium leaves the two end forces no choice: equal, opposite, and along the chord. So at every section the axial force is P cos α, the shear is P sin α, and the bending moment is the force times the perpendicular offset — 240 kNm at the crown for 200 kN at 1.2 m, which is a multiplication rather than an analysis. Straighten the member and the moment diagram is identically zero; that is the case a truss member is in, and the reason it carries one number. Equilibrium

The member with only one direction

If a body is in equilibrium under forces applied at exactly two points, those forces are equal, opposite and along the line joining the points. It is three conclusions from two equations, it is the shortest real theorem in statics, and nearly everything that follows depends on it without saying so.

 A derivative taken with a ruler, and the step that makes it worst. Castigliano's theorem says the deflection is ∂U/∂P, and the derivative here is taken numerically — two solves at ±dQ and a central difference. Against the unit-load answer of 1.720635e-2 it agrees to 1.6e-13, which for a linear structure it must: ∂N/∂P is exactly the force a unit load produces, so the two expressions are the same sum written twice. The error against step size is the classic pair of straight lines — truncation falling as the step shrinks, round-off rising as the difference of two nearly equal energies loses its digits — meeting near dQ = 1.2e+1. For a linear structure the truncation term is exactly zero, so what is drawn here is round-off alone. Deflection

The deflection that is a derivative

A structure's strain energy is one number. Differentiate it with respect to a load and out comes the displacement under that load — and the trick that makes it a method rather than an identity is that the load does not have to be there.

 The force spreads, and the spreading needs a tie. The end block behind an anchorage of 1200 kN on a 200 mm plate, in a section 700 mm deep. Half the force enters at the quarter point of the plate and leaves at the quarter point of the section, so a strut between the two rises 125 mm and needs a transverse tie to turn it. Placing the tie 0.5 depths from the face makes that tie force 214 kN — and at exactly half a depth this reproduces Guyon's 0.25P(1 − a/h) to the digit, which makes that famous coefficient a lever arm somebody chose rather than a property of concrete. The bearing stress under the plate is 20.0 N/mm² against 5.7 once the force has spread. Internal forces

The force that splits what it pushes on

A prestressing tendon delivers its whole force through a plate a fraction of the section deep. One depth further along the stress is uniform, and the spreading in between requires a transverse tension nobody applied — the force that splits end blocks, and the only number in the design that no equilibrium equation on the member can see.

 Four inequalities, and the wedge between them. The Magnel diagram: every limit on a prestressed section, plotted as a bound on 1/P against the eccentricity. Two of the four come from transfer, when the force is largest and the only moment is the beam's own weight, and two from service, when 20% of the force has been lost and the moment is 640 kNm. Each is linear in 1/P, which is the substitution that makes the problem a picture rather than a search. The shaded region is every force-and-eccentricity pair the section will accept: it is a wedge opening to the right, so the cheapest prestress is always at the largest eccentricity the cover allows — 1029 kN at e = 400 mm here. The section's kern is 241 mm, and every useful answer is outside it. Sections and stress

Four inequalities and a wedge

A prestressed section has to satisfy two stress limits when the force is largest and the load smallest, and two more when the force has relaxed and the load has arrived. Each is linear in one over the force — which turns a search for a prestress into a region on a page, and turns an impossible section into an empty one.

 The neutral axis obeys neither the load nor the moment. A 305 × 102 mm I-section carrying a moment 5° out of the plane of its web. The moment vector is the short arrow; the neutral axis is the long line, at 69.7° to the strong axis. They do not line up, and the reason is that the neutral axis follows the moment ratio scaled by the stiffness ratio: tan α = (M_z/M_y)(I_y/I_z), and I_y ÷ I_z is 30.8 here. So a 5° tilt of the load puts the neutral axis 70° over, the corner that ends up furthest from it carries 489 N/mm² against the 258 the straight-down case would give, and the section has lost 47 per cent of its capacity to a misalignment nobody would draw on a detail. Sections and stress

Two moments and a neutral axis that obeys neither

Tilt the load on a rolled beam by five degrees and the neutral axis swings by seventy. The section is doubly symmetric, its product of inertia is exactly zero, and none of that helps — because what decides the axis is the moment ratio multiplied by a stiffness ratio of thirty.

 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. Internal forces

How far a wrong load reaches

Every figure in this collection applies a load as a point, a line or a uniform pressure, and no real load is any of those. The licence is Saint-Venant's, it is usually quoted as a principle, and it is really a statement about a wavelength.

 The same beam, the same load, and one of them has to lift it. Two lower-bound models of one beam, differing only in which chord the 600 kN is applied to. The chord forces are identical in both — the moment diagram does not know where the load arrived — and so are the struts. What is not identical is the vertical at the load: nothing in the top-loaded model, and the whole 600 kN in the hung one, because a bottom-chord panel point touches no strut and the load has nowhere to go but up. That tie is 1200 mm² of steel against the 96 mm² the shear calculation asks for over the same length — a factor of 12.5, in the same place, and additional to it. Internal forces

The load that has to be lifted

Every shear calculation on this site assumes the load arrives on the top of the beam and walks down a diagonal to the support. Hang the same load from the soffit and the diagonal cannot start — the load has to be carried up to the compression zone first, by a tie no sectional calculation contains.

 Half the beam does nearly all of the deflecting. The virtual-work integrand M·m/EI along the member, normalised to its own peak, with the running share of the answer beside it. The integrand is a density: it says how much of the deflection each millimetre of the beam produced. For this case the half nearest the root supplies 87.5 per cent of it, and the rest of the member supplies the remainder. Stiffening the busy 50 per cent by 1.5 times takes the deflection down by 29.2 per cent; the same material spent on the quiet end takes it down by 4.2 — a factor of 7.0 for the same steel. The map of what is contributing is not the map of where the moment is largest, and the second is the one that gets drawn. Deflection

Where a deflection comes from

The unit-load method gives a deflection as an integral, and this collection has treated that integral as a number to evaluate. It is not a number. It is a density, and it says which millimetres of the member produced the answer — which is not the same map as where the moment is largest.

 One criterion inside the other, touching at six points. The two yield criteria in principal stress space with the third principal stress zero, both normalised by the yield stress. Von Mises is the ellipse — σ₁² − σ₁σ₂ + σ₂² = f_y², which is a circle seen at an angle — and Tresca is the hexagon inscribed in it, touching at the six points where one principal stress is zero or the two are equal. Everywhere else Tresca is the smaller, by up to 15.5 per cent, and the widest gap is at pure shear, where σ₁ = −σ₂ and the two answers are 205 and 178 N/mm². The ratio there is exactly 2/√3, computed rather than quoted, and it is the whole reason a web is checked against f_y over root three. Materials

The shear strength nobody measured

Every web on this site is checked against the yield stress divided by the square root of three, and no test produced that number. It is a consequence of a decision about what makes a metal yield, and the alternative decision gives a different answer by fifteen per cent.

 The two theorems close on the answer from opposite sides. A pinned-base portal frame under 200 kN at mid-span and 80 kN at the eaves, with a plastic moment of 200 kNm. The three horizontal lines are mechanisms — work equations, each an upper bound: the beam mechanism gives 1.000, sway 1.250 and the combined one 0.714. The curve is the lower bound: for each value of the one redundant, the largest load factor whose moment field stays inside the plastic moment everywhere. Its peak is 0.714, and it touches the lowest mechanism exactly — so the collapse load is known rather than bracketed. Using the beam mechanism instead would have claimed 1.40 times the real capacity, and the sway one 1.75 times, both of them on the wrong side. Equilibrium

Two ways of being wrong

Plastic analysis has two theorems and they point in opposite directions. Any equilibrium field that nowhere exceeds the plastic moment gives a load at or below the collapse load; any mechanism gives one at or above it. Only one of those errors is safe.

 The chords take the shear the web is credited with. A cantilever of 6 m tapering from 400 to 1200 mm, under a 120 kN tip load, with the shear divided between the web and the two inclined chords. The chord force is M/z and it is not horizontal, so its vertical component is real: V_web = V − (M/z)·dz/dx, which is Résal's result of 1899. At the root the web is left with 33 per cent of the applied shear — exactly d₀/d₁ for a straight taper, with no length, load or material in it. Turn the same member round and the figure becomes 300 per cent: the chords stop helping and start adding, at the section where the moment is largest as well. And the limit is worth having — a member whose depth is proportional to its moment leaves the web nothing to do at all, which is the triangular cantilever every crane jib is. Internal forces

The shear the chords take

Every shear check in this collection has assumed the two chords of a beam are parallel, so that the whole of the shear crosses the web. Taper the member and that stops being true — and the sign of the correction is decided by which end the haunch is at.

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