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.

Assumes One deflection, without solving everything, The deflection that is not bending and The section that changes along the span.

The unit-load method says that to find how far one point of a structure moves, put an imaginary unit force there, multiply two moment diagrams together, and integrate:

δ=MmEIdx\delta = \int \frac{M\,m}{EI}\,dx

Every use of that expression in this collection has treated it as a machine for producing a number. It is also a machine for producing a map, and nobody looks at the map.

The integrand is a density. It says how much of the answer each millimetre of the member contributed, and the distribution is nothing like uniform.

Half the beam does nearly all of the deflectingThe 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.00.20.40.60.8100.20.40.60.81along the memberdensity ÷ its peak, and running sharewhere it comes fromrunning totalhalf nearest the root: 87.5% of it · worth 7.0× stiffening the other half
Fig. 1 The integrand along a cantilever with a load at its tip, and the running share of the answer beside it. Both diagrams are linear in the distance from the tip, so the density goes as the square of it — and the half of the beam nearest the root supplies exactly seven eighths of the tip deflection.

Which free body produced the number

The cantilever case has a closed form and no arithmetic in it worth hiding.

A tip load PP gives M(x)=P(Lx)M(x) = -P(L-x), measuring xx from the root. The unit load at the tip gives m(x)=(Lx)m(x) = -(L-x). Their product is P(Lx)2P(L-x)^2, so the density goes as the square of the distance from the tip, and the share supplied by the outer portion of length aa is (a/L)3(a/L)^3.

The half nearest the tip therefore supplies (1/2)3=1/8(1/2)^3 = 1/8, and the half nearest the root supplies 7/8. No length, no load, no material, no section: seven eighths, for every tip-loaded cantilever there has ever been.

Under a uniform load the real moment goes as (Lx)2(L-x)^2 while the virtual one is still linear, so the density goes as the cube and the root half supplies 1(1/2)4=15/161 - (1/2)^4 = 15/16.

The simply supported case, which is the one everybody has

For a mid-span deflection under a uniform load the two diagrams are different shapes — a parabola and a triangle — and the density is their product, wx2(Lx)/4wx^2(L-x)/4 on the left half.

Integrating it returns 5wL4/384EI5wL^4/384EI, which is the answer everybody knows, and integrating it piecewise returns something nobody quotes: the middle half of the span supplies 67/80 of the deflection, or 83.75 per cent, and the two outer quarters supply the remaining sixth between them.

Half the beam does nearly all of the deflectingThe 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 middle half supplies 83.7 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 27.9 per cent; the same material spent on the quiet end takes it down by 5.4 — a factor of 5.2 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.00.20.40.60.8100.20.40.60.81along the memberdensity ÷ its peak, and running sharewhere it comes fromrunning totalmiddle half: 83.7% of it · worth 5.2× stiffening the other half
Fig. 2 The same density for a uniformly loaded simple span. The middle half supplies 83.7 per cent of its own deflection; stiffening that half by half again takes 27.9 per cent off the answer, and spending the same material on the outer quarters takes off 5.4 — a factor of 5.2 for the same steel.

Which is the whole argument for a haunch, and against one

A haunch is material added where the moment is largest, and it is usually justified on strength: the section is deepened where it has to carry the most.

The map says something stronger and more specific. Deepening the busy region buys stiffness at five times the rate that deepening the quiet region does — so a haunch is not a marginally better place to put material, it is very nearly the only place worth putting it.

And the same map says where not to bother. The outer quarters of a simply supported span contribute a sixth of the deflection between them; a designer worrying about the depth available at the bearing is worrying about a region that could be made half as stiff for an eight per cent penalty.

This is where the argument meets the tapered member from the other side. That essay asks where a tapered beam is checked — a strength question, whose answer is the station where the demand-to-capacity ratio peaks. This one asks where a tapered beam is stiff, and the two stations are not the same. For a tip-loaded cantilever the governing section is where the depth has doubled, and the stiffness is coming from the root.

The fully stressed shape is a parabola, and the taper is a straight line through itThe member as built — a straight taper from 300 to 700 mm — with the profile of constant utilisation drawn behind it. The fully stressed depth follows the square root of the moment, which for this load case is a parabola, and a straight line drawn to touch it at the governing station lies outside it everywhere else. The straight taper carries 7.6% more web than the shape that would be exactly used up at every section, which is the price of a member that can be cut from a plate with one straight line.they touch hereas built: a straight taperbehind it: constant utilisation7.6%more web300 mm at the free end · 700 mm at the root · governing at 4.50 m
Fig. 3 The member the map is an argument about. Its profile is usually chosen from the moment diagram, which is the strength answer. The deflection answer weights the same beam differently, and a profile optimised for one is not optimal for the other.

The map depends on which deflection is asked for

The density contains the virtual moment diagram, and the virtual diagram depends entirely on where the deflection is being asked about.

Ask for the mid-span deflection of a uniformly loaded span and the virtual diagram is a triangle peaking at mid-span, so the map is concentrated there. Ask for the deflection at the quarter point and the virtual diagram peaks at the quarter point, and the map moves — the region supplying the answer shifts towards the end of the beam even though the real moment diagram has not changed at all.

Half of the density belongs to the question rather than to the structure. That is easy to say and easy to forget, and it is why “where the beam is working hardest” is not a well-formed phrase: the beam is working hardest at mid-span for every question about strength and for only some questions about movement.

The clean statement of it is reciprocity: the influence of a load at AA on the deflection at BB equals the influence of a load at BB on the deflection at AA. The density is that theorem written out as a distribution rather than as a pair of numbers.

Maxwell's reciprocal theoremA load at one point and the deflection it causes at another, against the same load moved to the second point and the deflection read at the first. Both integrals return 251.9998, and neither calculation was told about the other. The two deflected shapes are entirely different; the two readings are identical.24 at 3δ at B = 252.00024 at 6δ at A = 252.000the shapes have nothing in commonand the two readings agree to 9e-14which is why an influence line can be measured by pushing the structure where it is easy to push
Fig. 4 The theorem the map is a picture of. Swapping the real and virtual loads leaves the integral unchanged, so the density is symmetric in the two questions — which is why a map drawn for one deflection answers a question about a load somewhere else.

In an indeterminate structure part of the map is negative

Everything so far has had both diagrams of the same sign over the whole member, so every millimetre contributed positively and the running total climbed. That is a property of determinate structures with a single load, not of structures in general.

Take a propped cantilever under a uniform load. Its real moment diagram is hogging over the prop and sagging in the span, so it changes sign; the virtual diagram for a deflection somewhere in the span changes sign too, at a different place. Their product is therefore negative over part of the member, and that part of the beam is reducing the deflection being asked about.

Two consequences follow, and the second is the useful one.

The running total is no longer monotonic, so “the middle half supplies eighty per cent” stops being a well-formed sentence — there are regions supplying more than a hundred per cent and regions supplying less than nothing.

And stiffening a region with a negative density makes the structure deflect more. That is not a paradox: stiffening it attracts moment to it, which is a redistribution, and the redistribution moves the answer the other way. It is the clean statement of something every designer of continuous structures learns by accident — that adding material somewhere can be counterproductive — and the density says exactly where the boundary is.

One support too manyThe 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.simply supportedstatics alonesag 202.5propped at one endneeds stiffnesssag 113.9hog 202.5built in at both endsneeds stiffnesssag 67.5hog 135.0the load never changes; only what is holding the endsthe built-in case peaks at two-thirds of the simple span's moment
Fig. 5 The structure where the map changes sign. Its moment diagram has a zero in it, so the product of two diagrams has two zeros in general — and between them is a region of the beam whose stiffness is working against the deflection being computed.

The shear map is the other half of the beam

Shear deflection has a density of its own, Vv/GAvV\,v/GA_v, and it is the product of two shear diagrams rather than two moment diagrams.

For a uniformly loaded simple span the real shear is largest at the supports and zero at mid-span; the virtual shear for a mid-span deflection is a constant ±12\pm\tfrac12. So the shear density is largest at the supports and smallest in the middle — exactly the opposite of the bending map.

On a rolled steel beam that hardly matters, because the shear term is 1.4 per cent of the total. On a sandwich panel it is 9 per cent at ordinary proportions and 60 at short ones, on a timber joist it is several times a steel beam’s, and on a deep beam it is most of the answer.

For those members the two maps have to be read together, and they point in opposite directions: the bending stiffness is wanted in the middle and the shear stiffness at the ends. A haunched sandwich panel would be a strange object; a panel with a stiffer core at the supports is exactly what a well-detailed one has.

At one span-to-depth ratio the section still decidesThree sections at a span-to-depth ratio of 8 under a central point load, with the share of the deflection each carries in shear. The two rectangles are 200 × 600 and 100 × 1200 — different in every dimension — and both give Q = 3.111 and 4.64% of the deflection in shear, because I/As is d²/10 for every rectangle there is. The I-section shears on its web alone, so κ falls from 0.833 to 0.513, Q rises to 9.618 — 3.09 times — and the share is 13.06%. The rectangle reaches a tenth of its deflection in shear at L/d = 5.29; the I-section is still there at L/d = 9.30, which is a beam nobody would call deep.every section drawn to one scale, every span at L/d = 8shear, as a share of that section's deflectionrectangle 200 × 600κ = 0.833 — five sixths of the gross areaQ = 3.111a tenth of the deflection at L/d = 5.294.64%rectangle 100 × 1200κ = 0.833 — five sixths of the gross areaQ = 3.111a tenth of the deflection at L/d = 5.294.64%I-section 600 × 200κ = 0.513 — the web aloneQ = 9.618a tenth of the deflection at L/d = 9.3013.06%the shape decides and the size does not: Q is 3.111 for every rectangle and 9.618 for this I
Fig. 6 Where the shear term stops being a correction. Its map is the reverse of the bending one, so a member for which both matter is being asked for stiffness in two different places, and the section that answers both is not the section that answers either.

And a truss has the same map, one dimension down

A truss’s deflection is a sum with one term per member rather than an integral, and the terms are FfL/EAF f L/EA — the same product of a real quantity and a virtual one, weighted by a flexibility.

The map is therefore a ranking of members rather than a distribution along a length, and it says the same kind of thing: a handful of members supply most of the deflection, and they are not the members carrying the largest forces. A member with a large force and a small virtual force contributes nothing, and the classic case is a member carrying a full panel load in a truss where the deflection being asked about puts no virtual force through it at all.

The continuum and the discrete versions are the same theorem, and reading them together is what makes the density worth naming: the question “which part of this structure produced the movement” always has an answer, it is always available from a calculation already being done, and it is almost never plotted.

Every member's share of the movement, and they are not the members expectedA Pratt truss of six panels at a depth of 0.9, carrying 40 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 2298.84 at EA = 1: 46.2% from four top chords, 27.9% from six bottom chords, 23.5% from six diagonals, 2.3% from five verticals. The single worst member is a top chord at mid-span at 14.5% 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 2298.84, a relative residual of 2.2e-15.δ = 2298.84 read here40 kN at every top node — each member drawn at the width of its own sharetop chord (four members)46.2%bottom chord (six members)27.9%diagonal (six members)23.5%vertical (five members)2.3%the members that moved the roof, ranked — a symmetric pair is two members and appears twicetop chord, at mid-span14.50%F -200.0 × f -1.667top chord, at mid-span14.50%F -200.0 × f -1.667bottom chord, at mid-span8.59%F 177.8 × f 1.111bottom chord, at mid-span8.59%F 177.8 × f 1.111top chord, near the left support8.59%F -177.8 × f -1.111top chord, near the right support8.59%F -177.8 × f -1.111diagonal, near the right support6.54%F -149.5 × f -0.747diagonal, near the left support6.54%F -149.5 × f -0.747virtual work and a stiffness solution agree to 2.2e-15 — two methods sharing no arithmetic
Fig. 7 The same map for a truss, where it is a bar chart instead of a curve. The members are ranked by FfL/EAFfL/EA, and the ranking names which ones are worth stiffening — usually not the ones a designer worries about.

The map is also a sensitivity, which is what makes it a tool

There is a second reading of the density that is more useful than the first, and it follows from reciprocity.

The contribution of a length dxdx to the deflection is Mmdx/EIM m\,dx/EI, so the sensitivity of the deflection to a change in EIEI there is that contribution divided by EIEI — the density again. A designer asking how much the answer moves when the stiffness at a station changes is asking for exactly the curve that has been plotted.

That converts a plot into a procedure. Compute the deflection once, plot the density, and read off where a change is worth making — no second analysis, no trial section, and no iteration. It is the same trick strain energy differentiated with respect to a load plays for a different variable, and it is available from the first analysis rather than from a sweep.

And it is the honest reply to a common request. Asked to make a beam stiffer, the reflex is to make the whole beam deeper. The density says which part of it is doing the work, and on an ordinary simply supported span the answer — five to one in favour of the middle half — is large enough to change the detail rather than merely to reassure.

A derivative taken with a ruler, and the step that makes it worstCastigliano'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.10^-610^-410^-210^-1610^-1410^-1210^-1010^-8size of the dummy loadrelative errorround-offand nothing elsebest 2e-16∂U/∂P = 1.720635e-2 · unit load = 1.720635e-2
Fig. 8 The same integral read as an energy. Strain energy is M2/2EI\int M^2/2EI and the deflection is its derivative with respect to a load; the density in this essay is what that derivative looks like before it is integrated, which is why it doubles as a sensitivity.

And it says where a splice may go

The map has a use nobody teaches, because it answers a question that is usually settled by convention: where to put a joint.

A splice is a discontinuity in stiffness — a bolted cover plate is stiffer than the section, a site weld with a backing bar is not quite, and a bolted end plate under-develops the flange until it is fully tightened. Wherever it goes it perturbs EIEI locally, and the density says what that perturbation costs: nothing at all in the outer quarters of a simply supported span, and several per cent of the deflection at mid-span.

The received rule is to splice near a point of contraflexure, which is a strength rule about putting the joint where the moment is small. The density says the same thing for a different reason and does not always agree: for a deflection asked at mid-span of a continuous beam, the contraflexure point is where the real moment vanishes and the product need not, because the virtual diagram is not zero there. The two rules coincide often enough that nobody has noticed they are different rules.

The same diagram, from two entirely different arithmeticsBending moments on a two-span beam under 24 kN/m. The curve is the stiffness solution; the marked points are what moment distribution reached after eight cycles of hand arithmetic — 0.0, 192.0, 0.0 kNm at the supports against an exact 0.0, 192.0, 0.0. The two share no code and no equations, which is the only arrangement in which agreement is evidence of anything.0.0192.0the largest disagreement is 2.9e-3 kNm
Fig. 9 Where the moment vanishes on a continuous beam, which is the station the splice rule names. The deflection density has its own zeros, and they are in different places, because one diagram belongs to the load and the other to the question.

Where the model stops

The density assumes linear elastic behaviour and superposition. A member that has yielded, cracked or slipped has a curvature that is not proportional to its moment, and the integrand is then MκM\kappa rather than Mm/EIM m/EI — still a density, no longer a product of two diagrams, and no longer symmetric under reciprocity.

The comparison between stiffening the middle and stiffening the ends holds EIEI constant elsewhere. A real haunch changes the moment diagram too, in an indeterminate structure, by attracting moment towards the stiffened region — so the gain is smaller than the map predicts and the redistribution is another calculation.

Nothing here is about axial deformation. A frame’s sway comes partly from its columns shortening, and that density is a third map with its own shape.

And the map is drawn for one load case. A structure is designed for several, and the region that supplies the deflection under a uniform load is not the region that supplies it under the pattern loads that govern a continuous beam.

What the pictures cannot show

The density is normalised to its own peak in every figure here, so nothing on these axes says how large the deflection is. Two beams with identical maps can differ by a factor of a hundred in how far they move, and the map has no opinion about it.

Nor can it show the practical constraint that decides most real profiles, which is that a beam has to be made. A haunch is a fabrication operation, a tapered member is a cutting and welding sequence, and a variable-depth precast unit needs a variable mould. The map says where the material is worth putting; the price list says whether the saving covers the cost of putting it there, and for spans under about fifteen metres it usually does not.

The assumption the figure rests on

The whole essay assumes the deflection worth knowing is the one at a single named point.

Serviceability is not usually written that way, and where it is, the point is chosen by convention rather than by consequence. What a cladding panel cares about is the differential movement between its two fixings; what a partition cares about is the change in deflection after it was installed; what a crane rail cares about is the slope. Each of those is a different virtual load — a pair of opposed unit forces, a unit force applied only to the loads that arrive later, a unit moment — and each therefore has a different density.

Which means the map is not a property of the beam at all. It belongs to the pair of a structure and a question, and changing the question moves it. That is a discipline rather than a difficulty: it says that “stiffen the middle” is an answer to a specific question, and that the specific question is the one worth writing down first.

The area is the rotation, and its first moment is the movementA 8 m simply supported span under a central load of 10, with the M/EI diagram beneath it. The shaded area is 853.33, which by the first theorem is the change of slope along the whole member. Its centroid is at 4.000 m, and the first moment about mid-span is 3413.3 — 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 1066.7.centroid at 4.00 mM/EIarea = 853.33 · first moment = 3413.3by double integration: 1066.7
Fig. 10 The other reading of the same integral. The area of a curvature diagram is a rotation and its moment is a deflection, which is this essay’s density with the weighting supplied by geometry rather than by a virtual load — two ways of saying that a deflection is an accumulation along a member rather than a property of a section.

Named alongside this one

Essays reaching for the same objects. Nobody chose these; they are what the concept index makes visible.

The objects this essay names

Each one links to every other essay that touches it.

CurvatureDeflectionHaunchInfluenceMoment diagramOptimisationReciprocitySecond moment of areaServiceabilityShear deflectionStiffnessStrain energyTapered memberUnit load methodVirtual work