Structural Engineering & Statics
where the load goes
Generator

The deflected shape is the moment, integrated twice

Rendered here at the parameters it defaults to, with every essay that calls it — which is the same list as the blast radius of changing it.
The deflected shape is the moment, integrated twice. A loaded beam with its deflected shape above and its bending-moment diagram below. The shape was obtained by integrating the moment twice and fitting the constants to the supports; the vertical scale is exaggerated enormously.

The deflected shape is the moment, integrated twice. A loaded beam with its deflected shape above and its bending-moment diagram below. The shape was obtained by integrating the moment twice and fitting the constants to the supports; the vertical scale is exaggerated enormously.

18 essays call deflected-shape. The drawing above is what it returns with no arguments at all; every call below passes it something, because a placement that passes nothing draws whichever member of the family the generator happens to default to rather than the one its essay argues about.

Where it is called

Changing this generator changes every one of these figures.

Load, shear and moment — a simple span. The applied load, the shear force it produces and the bending moment that follows, drawn one above another to the same horizontal scale. Shear is the integral of the load and moment is the integral of shear. Internal forces

The diagram is an integral, and that is why it can be drawn by eye

Load, shear and moment are one function and its two integrals. Once that is seen, the diagrams stop being things to calculate and become things to sketch.

 3 continuous spans against 3 simple ones. The bending moment in a continuous beam, solved by the stiffness method, drawn over the moment in the same spans made simply supported. The peak sagging moment falls from 30.6 to 19.6, and a hogging moment of 24.5 appears over the supports where there was none. Internal forces

The moment over the support, and what it buys

Run a beam over its supports instead of stopping at each one, and the mid-span moment falls by a third while a new moment appears where there was none. Nothing was added but continuity.

 Where the peak moment goes when a restraint is added. The same uniformly loaded beam with three sets of restraints, and the bending moment in each. Adding restraint moves moment from mid-span to the supports and lowers the peak — but only the first case can be solved by statics. Deflection

One support too many, and what it costs to know

Add a redundant restraint and the load has two routes to the ground. Equilibrium cannot say how it splits, and the answer turns out to depend on stiffness — which is a different kind of question.

 The deflection at x = 4, 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 — 213.33 here. No standard case was consulted, so the method works for any load pattern at all. Deflection

One deflection, without solving everything

To find how far one point of a structure moves, put an imaginary force of one unit there, multiply two moment diagrams together, and integrate. The answer arrives without ever solving for the deflected shape.

 Two triangles that cross zero, and a block that does not. Stress across a 300 × 700 mm section at each stage, compression positive. The prestress alone gives -6.33 MPa at the top and 20.61 at the bottom; at transfer, with only self-weight on it, the top is at -2.47 MPa and in service the section runs from 7.61 to 3.82 MPa — compression everywhere. The same beam with no prestress reaches -12.67 MPa at the bottom fibre, which is 4.2 times what the concrete can hold. Internal forces

The load put on backwards

Every other structure in this collection waits for its load and then resists it. A prestressed one is given a load first — chosen, permanent, and pointing the wrong way — so that when the real one arrives the two nearly cancel and the material never has to do the thing it is bad at.

 The props decide where the stress ends up. Bottom-fibre stress in the steel of a 12 m composite beam carrying 12 kN/m of wet concrete and 18 kN/m afterwards. Unpropped, the bare steel takes the first stage alone and reaches 292 MPa; propped, the finished composite section takes everything and reaches 186 MPa — a ratio of 1.57. 62% of the unpropped beam's final stress was locked in before the slab was structural at all. The deflections differ by 1.73 times for the same reason, and no drawing of the finished beam distinguishes the two. Structural form

The structure that was never complete

Every analysis in this collection is of a finished structure loaded once. Real ones are built in pieces, and each piece carries whatever was present at the moment it became structural — so the stress in a member depends on when it arrived, which appears nowhere on any drawing.

 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.

 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.

 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.

 The building does not care how far it went down; it cares how much it tilted. Five footings on soil that is 35% as stiff under one of them, carrying 60 kN/m. They settle between 12 and 54 mm, and the number that matters is neither of those: it is the angular distortion between neighbours, 4.32 per thousand, or one in 231 — against a limit of one in 500 for cracking in finishes, which this does not. A building that went down half a metre uniformly would be undamaged and would need a new front step; this one has moved a twentieth as far and has cracked. Deflection

The settlement that matters is the difference

A building that goes down half a metre uniformly is undamaged and needs a new front step. One that goes down a twentieth as far, unevenly, has cracked. The superstructure can even the difference out — and the only way it can do so is by carrying the difference itself, as a force.

 One coefficient, and nothing else in it. The deflected shapes of one beam under four load cases, each scaled so that its mid-span deflection is the same, with the tangent at the left-hand support drawn on each. The end rotation is that deflection times a coefficient that depends only on the shape of the load: 3.20 for a uniform load, 3.00 for a load at mid-span, 2.99 for a triangular load, 3.60 for a load on half the span. Every material property, every second moment and the span itself cancel out of the ratio θL/δ, so a beam at any deflection limit has an end rotation that is known before anything about it is: at L/360 it is 8.89 milliradians, or 0.51 of a degree. Deflection

The angle nobody limits

Every serviceability rule in this collection limits a displacement. What a bearing, a joint and a cladding gap actually have to accommodate is an angle — and the angle is locked to the displacement by a coefficient that contains no material, no section and no span.

 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.

 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.

 A section has two areas and the tables give one of them. Peak shear stress divided by the mean, for four sections of exactly the same gross area and depth. The mean is V/A and is the number a first calculation uses; the peak is what the material actually sees, and the ratio between them is a property of shape alone. A rectangle's is 1.5 — the parabola's peak over its average — and it is one of the few numbers in this subject that is exactly derivable and universally ignored. An I-section's is near 1.98, and the reason is on the second bar: 97% of the shear is inside a web that is 56% of the area. So the flanges carry the moment and almost none of the shear, and the web carries the shear and almost none of the moment — which is why a shear check on an I-section uses the web area and a moment check uses the whole section, and why the two checks are about two different pieces of steel. Sections and stress

The section has two areas

A shear force divided by the area of the section is not the shear stress anywhere in it. A rectangle's peak is exactly one and a half times that number and an I-section's web carries nearly all of the shear over a fifth of the area, which is why a moment check and a shear check on the same member are checks on two different pieces of steel.

 The check that everything adds up, and the error it cannot see. Four versions of the same 3-bay, 4-storey frame, with the global equilibrium residual each one produces — the sum of the reactions against the sum of the applied loads, as a fraction of the applied total. It is the first thing every analysis prints and it is worth having: a lost restraint and a load entered in the wrong unit both show up immediately, at 8% and 32%, because both change what the structure is carrying. The fourth bar is the point. A member whose stiffness is wrong by a factor of ten redistributes the internal forces completely — the second bar shows the change in the member forces, 24% — and the global residual is exactly zero, because the wrong answer is still in equilibrium with the same loads. Equilibrium is one equation per degree of freedom of the whole body, and a stiffness error lives entirely in the many equations underneath it. A model can satisfy every equilibrium check ever devised and be a model of a different structure. Equilibrium

The check that cannot see the error

Every analysis prints a global equilibrium residual, and it is the first thing anybody looks at. It catches a lost restraint and a load entered in the wrong unit immediately. It is structurally incapable of catching a member whose stiffness is wrong by a factor of ten, because the wrong answer is still in equilibrium with the same loads.

 The same curve, computed twice and from opposite ends. A cantilever of 6 m, with its deflected shape drawn from a double integration of M/EI and the bending moment of its conjugate beam drawn on top of it. The conjugate is the same span carrying M/EI as a load, with its supports transformed — a real fixed end becomes a free end and a real free end becomes a fixed one, because a fixed end has no slope and no deflection and the conjugate therefore needs no shear and no moment there. The two curves agree to 4.7e-6 of the largest deflection, which is the trapezium rule and not the method. The largest deflection is 36.00 mm against the closed form's 36.00. The reaction of the conjugate beam is -9.000 milliradians, which is the real beam's rotation at that support — so the whole of a slope calculation is one reaction. Deflection

The beam whose moment is a deflection

A beam's bending moment is the second integral of its load. Its deflection is the second integral of M/EI. They are the same problem, so a deflection can be found by loading a fictitious beam with M/EI and asking a statics question — and the only thing to remember is the supports, which are not remembered but derived, one boundary condition at a time.

 Two answers added, and the answer to the two together, drawn on top of each other. A 8 m beam under a 60 kN point load at mid-span (152.38 mm), under 12 kN/m of uniform load (152.38 mm), and under both at once (304.76 mm). The sum of the first two is 304.76 mm, and the residual between it and the third is zero — not small, zero, to the last bit of the arithmetic. That exactness is not a numerical accident: the governing equation is linear in the load, so the response is a linear operator applied to it, and a linear operator distributes over addition by definition. Every calculation that adds one load case to another is standing on that one line. Deflection

The addition everything else rests on

Influence lines add, the unit-load method adds, moment distribution adds, load combinations add, and a stiffness matrix is linear by construction. All of it stands on one sentence with three hypotheses in it — and when they fail, two of the failures point in opposite directions.

 Soft in shear, the wall gives up moment to the span. The bending moment along a 300 × 600 rectangle fixed at its left end and propped at its right, under a uniform load, per unit load and span, at span-to-depth ratios of 4, 2, 1, and for the same beam treated as rigid in shear, dashed. Rigid in shear the wall carries 0.125 wL² and the prop 0.375 of the load. Counting the shear it deforms by, the wall moment falls to 0.119 wL² at 4, 0.105 wL² at 2, 0.070 wL² at 1, and the prop's share rises to 0.381, 0.395, 0.430. The load has not changed and the beam is no weaker: its forces have moved, because in a redundant beam they come from how it deforms. Deflection

The shear that moves the moments

In a statically determinate beam, shear deformation adds movement and changes no force. In a redundant one the forces come from how the beam deforms, so a stubby beam soft in shear carries less moment at its wall and more in its span — and the carry-over factor, the one half every hand method passes along, falls to nothing at φ = 2 and then changes sign.

The library, page 1 of 7 — where deflected-shape sits