The beam whose moment is a deflection

Two facts sit next to each other in every course on structures and are almost never introduced.

The first: a beam’s bending moment is the second integral of the load along it, with the two constants of integration fixed by the supports. The diagram is an integral is that statement, and it is why a moment diagram can be drawn by eye.

The second: a beam’s deflection is the second integral of $M/EI$, with the two constants fixed by the same supports.

They are the same problem. So the deflection of a beam can be found by loading a fictitious beam with $M/EI$ and asking what its bending moment is — which converts an integration into a statics question, and a statics question is the one kind that can be answered by drawing.

Which free body produced the number #

The conjugate beam, cut at a section, with the $M/EI$ load to one side of the cut and the transformed supports holding it up.

That is an ordinary free body and the arithmetic on it is ordinary statics: the shear at the cut is the area of the $M/EI$ diagram to one side, and the moment at the cut is the first moment of that area about the cut. Both are numbers a draughtsman can get from a diagram with a planimeter, which is precisely why the method exists.

What has to be got right is the supports, and the derivation is the whole of the method’s content.

A real simple end has $M = 0$ and $\delta = 0$, and a non-zero rotation. The conjugate’s moment is the deflection and its shear is the rotation, so it needs $M = 0$ and a non-zero shear — a simple end.

A real fixed end has $\theta = 0$ and $\delta = 0$. The conjugate needs zero shear and zero moment — a free end.

A real free end has both a rotation and a deflection. The conjugate needs both a shear and a moment — a fixed end.

A real interior support has $\delta = 0$ and a kink in the slope. The conjugate needs $M = 0$ and a jump in shear — an internal hinge.

And a real internal hinge, where the slope jumps and the deflection is continuous, needs a jump in the conjugate’s shear with its moment continuous — an interior support.

Five lines, each one boundary condition read through the correspondence, and not one of them has to be remembered.

The cantilever, which is the memorable one #

A cantilever fixed at the left and free at the right has a conjugate that is free at the left and fixed at the right.

As a structure it is absurd. It is a beam held only at the end that used to be free, loaded along its whole length, with nothing under it anywhere else. Anybody drawing it as a structure would object that it would fall over — and it would, if it were one.

It is not one. A conjugate beam is an integration with supports drawn on it. The “fixed end” at the tip is not a support; it is the statement that the real beam’s slope and deflection are unknown there and therefore that the conjugate’s shear and moment are non-zero. The picture is a piece of bookkeeping, and its absurdity as a structure is the clearest possible sign that it is not being used as one.

Once past that, the arithmetic is a pleasure. For a tip load $P$ the $M/EI$ diagram is a triangle of height $PL/EI$ at the fixed end. Its area is $PL^2/2EI$, which is the tip rotation. Its first moment about the tip is $PL^2/2EI \times 2L/3 = PL^3/3EI$, which is the tip deflection. Two lines and no calculus, and both of the standard cantilever formulae have fallen out of the geometry of a triangle.

Where the method earns its keep #

It is fair to ask what a graphical device is for on a site whose figures are generated by solving equations.

It answers a deflection at a point without the whole curve. One deflection, without solving everything is the virtual-work version of the same economy; the conjugate beam gets there by taking a moment about the point rather than by integrating a product, and for a hand calculation the moment is quicker.

It handles a stepped or varying EI without difficulty. The load on the conjugate beam is $M/EI$, so a change of section is a step in the load — an ordinary thing to have on a beam — where in the direct integration it is a discontinuity in the differential equation. A haunched beam, a cracked concrete beam with a different $I$ over the supports, a composite beam with a different section in the hogging region: all are conjugate beams with awkward-shaped loads and no awkwardness at all.

It answers a rotation as easily as a deflection, which matters more often than it looks. A bearing has a rotation capacity, a movement joint has a rotation the detail must accommodate, and a cladding panel cracks on a rotation rather than on a deflection — the angle nobody limits is the case for that being the criterion that is left out. The conjugate beam gives the rotation as a shear, which is one step less work than the deflection rather than one step more.

And it gives a check. The conjugate’s reaction is the real beam’s rotation at that support, so a slope and a deflection come out of the same calculation and can be verified against each other. The reaction of the simple-span conjugate is $wL^3/24EI$, which is the standard end rotation, and finding it as a by-product is the sort of redundancy a hand method should have.

Two routes, and why agreeing matters #

The figure at the top of this essay draws the same curve twice, and it is worth being clear that the two are genuinely different computations rather than the same one printed twice.

The direct route integrates the curvature twice along the beam and fixes the constants from the boundary conditions — for a cantilever, by starting both integrals at zero at the fixed end; for a simple span, by subtracting the chord through the two ends.

The conjugate route computes the transformed beam’s reactions by statics — moments about one end for the far reaction, then vertical equilibrium — and gets the moment at each section from the load to one side of it.

Neither uses any part of the other. They agree here to five parts in a million, which is the trapezium rule at four hundred stations rather than the method, and both agree with $PL^3/3EI$ to the same order.

That is worth having as a check on the implementation as well as on the mathematics, and it is the habit this site runs on: a claim that two things are the same is worth making only if there is a computation that could have said otherwise.

Reading a real deflection off a diagram #

The method is worth one worked case at the scale a designer meets, because the arithmetic is short enough to do standing at a drawing.

A simply supported span of 6 m under 12 kN/m, with $EI = 20{,}000$ kNm². The free bending moment diagram is a parabola of height $wL^2/8 = 54$ kNm. Divide by $EI$ and the conjugate load is a parabola of height $2.70 \times 10^{-3}$ per metre.

The area of a parabola is two thirds of base times height: $\tfrac{2}{3}\times 6 \times 2.70\times10^{-3} = 10.8\times10^{-3}$. Half of that is each end reaction of the conjugate, $5.40\times10^{-3}$ radians — which is the real beam’s end rotation, and the standard formula $wL^3/24EI$ gives exactly the same.

The moment at mid-span is the reaction times 3 m less the first moment of half the load about mid-span: $5.40\times10^{-3}\times 3 - 5.40\times10^{-3}\times \tfrac{3}{8}\times 6 = 16.2\times10^{-3} - 12.15\times10^{-3} = 4.05\times10^{-3}$ m, which is 4.05 mm. And $5wL^4/384EI$ is 4.05 mm.

Two standard results from the geometry of a parabola, with no integration performed and no formula recalled. That is what the method was for, and it is still the fastest way to get a deflection on a beam whose $EI$ or whose loading is not in anybody’s table.

The duality, and where it stops #

The support table has a pattern in it that is more than a mnemonic: the conjugate of a restraint is a release, and the conjugate of a release is a restraint.

Count them. A real beam with $r$ redundants has $r$ more restraints than statics needs. Transform every one of them and the conjugate has $r$ fewer than it needs — it is a mechanism, short by exactly the number of equations the real beam was short of.

So a propped cantilever, one degree indeterminate, has a conjugate that is free at one end and free at the other and loaded: not a structure. A fixed-ended beam, two degrees indeterminate, has a conjugate free at both ends. A two-span continuous beam has a conjugate with a hinge where a support was and one support where there were three.

That is why the method belongs to determinate beams, and it is not a convention that could have been chosen differently. The missing equations in the real problem are the missing restraints in the fictitious one, and no amount of care with the table will produce them. The moment-area theorems have the same limit for the same reason, and the standard workaround is the same: release the real beam until it is determinate, solve it, and put the redundant back as an unknown force — which is choose what to take away, and is the force method.

Where it came from, and why it looks like that #

Otto Mohr published the two area theorems in 1868 and Christian Otto Mohr’s students turned them into the conjugate beam over the following decades; the elastic-weights version and the column analogy are the same idea pushed further, until it solves indeterminate frames by treating the $M/EI$ field as a load on a fictitious cross-section.

The reason it exists in that form is worth knowing. Mohr was working on the graphical analysis of structures, where a truss’s displacements come out of a Williot diagram and a force system out of a funicular polygon, and the whole intellectual project was to replace calculation with drawing — because drawing was faster, checkable by eye, and available to people who could not integrate. The polygon that finds the shape is that project’s other great success.

Seen that way, the conjugate beam is not a trick for remembering formulae. It is the last step of a programme: turn an integration into a statics problem, because statics can be drawn. The fact that it survives into a period when nobody draws anything is a statement about how good the reduction was.

Where the model stops #

Shear deformation was ignored. The whole correspondence rests on curvature being $M/EI$, and for a deep beam it is not — there is an additional curvature from shear which the conjugate load does not contain. The deflection that is not bending is the missing term, and adding it means adding a second load to the conjugate beam proportional to the shear.

The beam was elastic. A cracked concrete beam has an $EI$ that varies with the moment, so the conjugate load is not a scaled copy of the moment diagram and has to be computed section by section — which is fine, and is exactly how a hand deflection calculation on a concrete beam is done.

Axial force was absent. A beam-column’s curvature includes a $P\delta$ term, which makes the equation non-linear in the unknown, and no fictitious beam loaded with a known distribution can represent it. The load that makes itself worse is that term.

Support settlement was not in it. A support that moves adds a rigid-body component to the deflected shape which the curvature does not contain, so it has to be superposed afterwards rather than found from the conjugate beam. The support that moved is the field it adds, and on an indeterminate beam it also changes the moment diagram the conjugate is loaded with.

And the beam was straight. The correspondence generalises to a curved member only through a different pair of theorems, because the relation between curvature change and displacement is no longer a double integral along a straight axis.

The generalisation #

The idea worth carrying is that two problems with the same differential equation and the same kind of boundary conditions are the same problem, and that recognising it converts one into whichever is easier to solve.

That is a much larger idea than a beam. A beam on an elastic foundation and a cylinder’s edge disturbance are the same equation, so a table of one solves the other — the length a structure was never given uses that in both directions. Torsion of a section and the deflection of a soap film over the same outline are the same equation, which is Prandtl’s membrane analogy and is how torsion constants were measured before they were computed. A steady heat flow and a seepage flow are the same equation. The stress function of a plate and the deflection of a membrane are the same equation.

The conjugate beam is the smallest and most domestic member of that family: a beam and a beam. Which is why it is the one to learn the habit on, and why the habit is worth more than the method. When a calculation is awkward, the useful question is not how to do it better but what else obeys the same equation — because somewhere there is a version of it whose answer is already drawn. The conjugate beam’s answer had been drawn for four hundred years before anybody wanted a deflection out of it: it is a bending moment diagram, and bending moment diagrams were the first thing this subject learned to draw.