One deflection, without solving everything

Integrating the moment diagram twice gives the deflected shape of a beam, and the shape gives the deflection anywhere. It also gives the deflection everywhere, which is a great deal of work when one number is wanted.

There is a method that produces the one number directly. Apply an imaginary force of one unit at the point of interest, in the direction of interest, and draw the moment diagram it produces. Then multiply that diagram by the real one, point by point, and integrate:

$$\delta = \int \frac{M m}{EI}\,dx.$$

Why an imaginary force gives a real answer #

The method looks like a trick and is a theorem, and the theorem is about energy.

Consider a structure carrying a set of forces, in equilibrium, and separately consider a set of displacements that are geometrically possible for it. The principle of virtual work says the work done by the first acting through the second equals the internal work done by the corresponding stresses acting through the corresponding strains.

Crucially, the forces and the displacements do not have to belong to the same problem. One can be real and the other imagined; the principle only requires the forces to be in equilibrium and the displacements to be compatible.

So: take the virtual force system to be a single unit load at the point of interest, with the internal moments $m$ it requires. Take the real displacement system to be the actual deformation of the structure under the actual load, whose curvature at each station is $M/EI$. Then

$$1 \times \delta = \int m \times \frac{M}{EI}\,dx,$$

and the left-hand side is the deflection wanted, because a unit force times a displacement is that displacement. The imaginary force was a device for extracting one number from an equation about work.

Two things follow that are worth stating plainly. The unit load never has to be applied to anything — no structure is ever loaded with it — so it can be placed anywhere and in any direction, including directions the structure is not loaded in. And the answer’s sign tells the direction: positive means the point moved the way the unit load pointed.

Reading the product integral #

The integrand is a product of two diagrams, and looking at it is more informative than evaluating it.

Where the real moment is large and the unit moment is small, the product is small — that part of the structure is heavily loaded and contributes little to this deflection. Where both are large, the product is large, and that region is doing most of the moving.

So the third diagram in the hero figure is a map of where the deflection comes from — which, for a member sized by how far it moves, is the more useful of the two things a moment diagram could tell a designer. For a mid-span deflection under uniform load it is concentrated in the middle half of the beam; for a deflection at a quarter point, it is skewed toward that quarter. A designer wanting to stiffen a structure at one point can read off which regions are worth stiffening, and it is not always where the moment is largest.

That is the practical advantage over integrating the moment diagram twice. Double integration produces a curve and buries the reasoning; the product integral produces a picture of the contribution.

One, done in full #

The method is short enough to run through completely, and doing so recovers a coefficient that otherwise has to be taken on trust.

Take a simply supported beam of span $L$ under a uniform load $w$, and ask for the deflection at mid-span.

The real diagram. $M(x) = \dfrac{wx(L-x)}{2}$, a parabola peaking at $wL^2/8$.

The unit diagram. A single unit load at mid-span gives reactions of a half each, so $m(x) = x/2$ on the left half, rising linearly to $L/4$ at the centre and falling back by symmetry.

The product, integrated. By symmetry, twice the left half:

$$\delta = \frac{2}{EI}\int_0^{L/2}\frac{wx(L-x)}{2}\cdot\frac{x}{2}\,dx = \frac{w}{2EI}\int_0^{L/2}\left(Lx^2 - x^3\right)dx.$$

The bracket integrates to $\dfrac{L^4}{24} - \dfrac{L^4}{64} = \dfrac{5L^4}{192}$, so

$$\delta = \frac{5wL^4}{384EI}.$$

There is the famous $5/384$, and the five in it arrived from $8 - 3$ over a common denominator of $192$. It is not a constant handed down from anywhere; it is what a parabola multiplied by a triangle integrates to.

The generator behind these figures runs the same integral numerically on six hundred stations and returns $213.33$ for a span of eight under a load of four, against the closed form’s $213.33$. Two routes to a number that only one of them was told.

Where along the beam the answer comes from #

The product diagram is a map, and putting numbers on it is worth doing once because the answer is more lopsided than it looks.

For that same beam and that same question, integrating the product over the middle half of the span gives $83.7$ per cent of the total. The two outer eighths — a quarter of the beam’s length — contribute $2.3$ per cent between them.

So a mid-span deflection is very nearly a fact about the middle of the beam. That has two practical consequences worth carrying.

Stiffening works where the product is, not where the moment is. Adding material near the supports of a simply supported beam does almost nothing for its mid-span deflection, however heavily loaded that region is in shear. Adding it at mid-span does nearly everything. A haunch designed to reduce a moment and a cover plate designed to reduce a deflection go in different places.

Which is why continuity is so effective. Hogging over a support subtracts from the real moment diagram exactly in the region where the unit diagram is largest, so it attacks the deflection where the deflection is made. A fixed-ended beam deflects a fifth as far as a simply supported one, which is a much larger ratio than the change in peak moment, and this is why.

The same integral, asking a different question #

Nothing in the derivation required the virtual force to be a force. It required only that the virtual system be in equilibrium and that its work through the real displacement isolate the quantity wanted — so choosing a different virtual system asks a different question.

A unit couple applied at a station gives the rotation there, because a unit moment times a rotation is that rotation. The procedure is identical: draw the moment diagram produced by a unit couple, multiply it by the real one, integrate. It is how the slope at a beam’s end is obtained, which is the quantity that matters when a beam frames into something that has to rotate with it.

A pair of opposed unit loads at two points gives the relative movement between them — the closing of a gap, the convergence of two faces of a cut, the change in a diagonal’s length. That is the version used in the force method, where the released redundant is an internal force and the compatibility condition is about a relative displacement rather than an absolute one.

A unit load at a support gives the deflection there, which is zero if the support does not move — and that apparent triviality is the whole of how support settlement is handled as a load case: set the left-hand side to the settlement instead of to zero, and the same equation returns the forces it generates.

The generality is the reason the principle is stated in terms of work rather than in terms of deflections. Work is a scalar, it pairs any equilibrium force system with any compatible displacement system, and the pairing is what selects the answer. Choosing a virtual system is therefore not a computational trick but the act of asking the question, and the arithmetic afterwards does not know or care which question it was.

Products, not integrals #

In practice nobody integrates. Both diagrams are made of straight lines and parabolas, so the integral of their product has closed forms, and the calculation collapses into looking up a handful of standard cases.

The most-used one: two straight lines over a length $L$, one running from $a_1$ to $a_2$ and the other from $b_1$ to $b_2$, give

$$\int_0^L ab\,dx = \frac{L}{6}\left(2a_1b_1 + 2a_2b_2 + a_1b_2 + a_2b_1\right).$$

A parabola against a straight line, a triangle against a triangle, a trapezium against a parabola — each has an entry in a table that fits on one page. The method reduces to splitting both diagrams at every discontinuity, looking up each segment, and adding.

This is why the technique survived so long as a hand method. A deflection that would take a page of double integration takes four table lookups, and the arithmetic is multiplication rather than calculus.

It has a checkable property, too, which is worth using. The product integral is symmetric in its two diagrams: swapping which is called $M$ and which $m$ cannot change the answer. Anyone who has miscopied an ordinate will usually find it by evaluating the product the other way round.

Trusses, where it is even simpler #

For a pin-jointed frame the same theorem gives an expression with a sum instead of an integral, because each member has a constant force along it:

$$\delta = \sum \frac{F f L}{AE},$$

with $F$ the member force under the real load and $f$ the member force under the unit load.

Both come from the same joint-equilibrium solver, so the calculation is: solve the truss under the real loads, solve it again under a single unit load at the joint of interest, multiply the two force lists together member by member, weight by $L/AE$, and add.

The sum has the same diagnostic value as the product integral. Members with a large $Ff$ product are the ones controlling that particular deflection, and they are often not the most heavily loaded members — a lightly loaded member that is long and thin can contribute more to a specific movement than a short heavily loaded one. Stiffening a truss at the wrong members is a common and expensive way to achieve nothing.

It solves indeterminate structures too #

The method’s second use is larger than its first, and it is the reason it appears in every structural course.

To solve a redundant structure, release the redundant restraint to leave a determinate one, compute the displacement at the released point under the real load, then compute the displacement produced there by a unit value of the redundant force. The redundant force is whatever makes the two cancel:

$$X = -\frac{\delta_{\text{load}}}{\delta_{\text{unit}}}.$$

Both displacements come from the unit-load method, and both are computed on a determinate structure. That is the force method, and it converts an indeterminate problem into two determinate deflection calculations.

For a propped cantilever the two are $wL^4/8EI$ and $L^3/3EI$, giving a prop reaction of $3wL/8$ — the standard result, obtained without ever writing a differential equation or a stiffness matrix.

The method scales badly — $n$ redundancies means an $n \times n$ system of flexibility coefficients, every one of which is a product integral — which is why the displacement method displaced it for computer work. For one or two redundancies by hand it remains the shortest route there is.

Where the model stops #

Linear elastic behaviour. The whole apparatus rests on superposition: the response to two loads is the sum of the responses to each. Once anything is nonlinear — material yielding, geometry changing, a support lifting off — the imaginary load and the real one can no longer be considered separately.

Bending only, unless the other terms are added. The integral above accounts for bending. Shear, axial and torsional deformation each have their own term, and for a slender beam they are negligible while for a deep one, a truss or a member in torsion they are not. Omitting them is a choice rather than a property of the method.

Known stiffness. $EI$ appears in the denominator and has to be supplied, and it is the second moment of area that carries nearly all the variation in it. For a cracked concrete member that is the least certain quantity in the calculation, and the answer is only as good as it.

One point, one direction. That is the method’s advantage and its limit. Getting a whole deflected shape means repeating it at many points, which is exactly the work double integration does in one pass.

Determinate releases. The force method requires a released structure that is stable and determinate, and for a heavily redundant frame choosing the releases well is a skill in itself — a poor choice produces flexibility coefficients that are nearly dependent, and a nearly singular system behaves exactly like a badly arranged one.

Small deflections. The virtual displacements are assumed small enough that the geometry is unchanged, so the unit-load moment diagram is drawn on the undeformed structure. The same first-order assumption as everywhere else on this site, and it fails in the same places.

The figures share a distortion worth naming. The three diagrams are drawn at three unrelated vertical scales chosen to fit the canvas, so the visual size of the product diagram says nothing about the magnitude of the deflection. Only its shape — where the contribution is concentrated — is being communicated, and comparing its height with the height of either parent diagram means nothing at all.

The ladder from here #

Later rungs on this anchor: the principle of virtual work stated and proved. Virtual displacements against virtual forces, and which problems each suits. The product-integral tables. Shear, axial and torsional terms. Deflections of trusses and frames. The force method for one and several redundancies. Maxwell’s reciprocal theorem, which falls out of the same integral. Castigliano’s theorems, which reach the same answers through strain energy. Influence lines by the Müller-Breslau principle. And the unit-load method as the ancestor of the finite-element method’s weak form, which is the same idea with the virtual displacement chosen from a finite set of shape functions.

Maxwell set out the method in 1864 and Mohr independently in 1874, which is why it is often called the Maxwell–Mohr method. Both were looking for a way to analyse redundant frames, and the deflection calculation that is now taught first was, historically, a step on the way to something else.