The deflection that is a derivative

A loaded structure holds energy. Every member that has stretched or shortened has stored $N^2L/2EA$, every length of beam that has curved has stored $\int M^2/2EI$, and the total is one number for the whole structure.

Castigliano’s second theorem says that differentiating that number with respect to one of the applied loads gives the displacement of the point that load acts at, in the direction it acts in.

$$\delta = \frac{\partial U}{\partial P}$$

It is a peculiar-looking statement — a scalar differentiated with respect to a force, giving a length — and it is one of the two standard routes to a deflection on this site. The other is the unit load, and the relationship between them is closer than “two methods” suggests.

Which free body produced the number #

None. That is the point of the method and worth saying plainly: there is no cut, no free body, no equilibrium statement anywhere in the derivation. What there is instead is a statement about work.

Load a linear structure gradually to a set of forces $P_i$. The work done is stored as strain energy:

$$U = \tfrac12\sum_i P_i\delta_i$$

Now add a small increment $dP_k$ to one of them. Approach it two ways.

Apply the whole load and then $dP_k$: the extra work is $dP_k\,\delta_k$ to first order, since the point has already moved $\delta_k$ by the time the increment arrives.

Apply $dP_k$ first and then the whole load: the extra work is again $dP_k\,\delta_k$, because the small force is present throughout the movement caused by everything else.

Either way $dU = \delta_k\,dP_k$, and dividing gives the theorem. The only thing used is that the structure is linear — that superposition holds and the order of loading does not matter.

Why it is the same sum as the unit load #

Take the truss form. $U = \sum N^2L/2EA$, so

$$\frac{\partial U}{\partial P} = \sum \frac{N L}{EA}\frac{\partial N}{\partial P}$$

Now ask what $\partial N/\partial P$ is. The structure is linear, so every member force is a linear function of every applied load: $N_e = \sum_k c_{ek}P_k$. The derivative $\partial N_e/\partial P$ is $c_{eP}$ — the member force per unit of $P$, which is exactly the force a unit load applied at $P$’s position would produce.

So

$$\frac{\partial U}{\partial P} = \sum \frac{\bar{N} N L}{EA}$$

which is the unit-load expression, term for term.

They are not two methods that agree. They are one sum written twice, and the reason to have both is that they are entered differently: one asks for a virtual structure, the other asks for a derivative.

Computed both ways, and the agreement is a measurement #

The theorem is provable, so agreement is not news. Computing it numerically is, because a numerical derivative of a physically meaningful quantity is a thing that can go wrong in interesting ways.

Take a nine-member truss under two 120 kN loads. Its strain energy is 2.13333 units. Solve it twice more with a dummy load $\pm dQ$ at the joint of interest, take a central difference, and compare with the unit-load answer:

$$\frac{\partial U}{\partial P} = 1.720635\times10^{-2}, \qquad \sum\frac{\bar{N}NL}{EA} = 1.720635\times10^{-2}$$

agreeing to $1.6\times10^{-13}$.

And the horizontal movement of a joint that carries no load at all — the dummy-load case, which is the trick the method is really for — comes out at $5.66667\times10^{-3}$ by both routes, sign included.

The dummy load, which is the whole reason to bother #

$\partial U/\partial P$ requires a $P$. If the deflection wanted is at a point with no load on it, there is nothing to differentiate with respect to.

The device is to put one there, of magnitude $Q$, carry it through the algebra, differentiate, and then set $Q = 0$. The member forces become $N = N_0 + \bar{N}Q$; the derivative is $\bar{N}$; and at $Q = 0$ the expression collapses to $\sum \bar{N}N_0L/EA$ — the unit load again.

That is the same manoeuvre a physicist calls a generating function and a statistician calls a moment-generating trick: introduce a variable nobody wants, differentiate with respect to it, and set it to zero. The structure never carries the dummy load; it exists so that a derivative exists.

Done numerically, as here, the dummy load is genuinely applied — twice, at $\pm dQ$ — and the answer is the slope. That is why the round-off curve above matters: too large a $dQ$ and nonlinearity would intrude (though for a linear structure it does not), too small and the two energies differ in their last digits only.

Where it earns its keep #

Three places, and they are not the places a textbook usually starts.

Redundant structures. The theorem of least work — the redundant takes the value that minimises $U$ — is Castigliano applied to a redundant force with the compatibility condition that its point does not move: $\partial U/\partial R = 0$. That is a minimisation rather than a compatibility equation, and it is the same equation: the propped cantilever solved in one support too many by setting the support deflection to zero can be solved instead by differentiating the strain energy with respect to the prop force, and the redundant that comes out is the same number.

Curved and tapered members. A member whose properties vary along its length is awkward for the unit-load method, because both moment diagrams have to be integrated against each other; it is no more awkward for the energy method, because the energy integral was going to be numerical anyway.

Machine and mechanism deflections. A crank, a bracket, a clamp — anything where the load path is a series of segments in bending, torsion and axial force at once. The energy adds up over all four actions and the derivative takes them all at once, whereas the unit-load method needs a virtual diagram of each kind.

What none of the three is, is a different answer. There are now three routes to a deflection on this site — the area of a diagram, the unit load, and this derivative — and they differ entirely in what they ask the user to draw rather than in what they compute. The moment-area route asks for one diagram and a geometric reading of it; the unit load asks for two diagrams and their product; the energy method asks for no diagram at all and a differentiation instead. The figures above and below are all the third route, which is why they are all the same three panels or the same integrand.

The sign is half the answer, and it is easy to get wrong #

A deflection has a direction, and the theorem supplies it — provided the dummy load is applied in the direction the answer is wanted.

That sounds like bookkeeping and it is not. Computing the same truss joint’s horizontal movement with the dummy load applied along the positive axis gives $+5.667\times10^{-3}$; applying it the other way gives the same magnitude with the opposite sign, and both look equally plausible on the page. The convention has to be fixed once and honoured in both routes, or the two methods agree in magnitude and disagree about which way the structure went.

This site’s convention is the one that makes the answer read naturally: the unit load acts in the direction the answer is wanted, so a positive result means the joint moved that way. For a roof that is downward; for a horizontal freedom it is along the positive axis. The two are not the same sign in the arithmetic, and a check that only compares magnitudes cannot see the difference — which is why the gate for this family asserts the horizontal case as well as the vertical one.

Energy against stiffness, which is the same object twice #

There is a second derivative in all of this and it is worth taking.

Differentiate $U$ once with respect to a load and get a displacement. Differentiate the displacement with respect to another load and get an influence coefficient — the flexibility $f_{ij}$, which is how far point $i$ moves per unit of load at $j$. So

$$f_{ij} = \frac{\partial^2 U}{\partial P_i \partial P_j}$$

and since mixed partials commute, $f_{ij} = f_{ji}$. Maxwell’s reciprocal theorem falls out of the symmetry of second derivatives, which is a considerably shorter proof than the usual one and says exactly why it is true rather than merely that it is. The measured version of that statement — two experiments on one beam returning the same number to every digit — is the theorem that swaps the question, and the figure at the top of this page is its second derivative read once instead of twice.

Invert the flexibility matrix and it is the stiffness matrix. So the strain energy, the flexibility method, reciprocity and the stiffness method are four readings of one scalar function of the applied loads — and every method on this site for an indeterminate structure is somewhere in that list.

What it cannot do #

It is a linear theorem. The derivation used superposition twice. For a nonlinear elastic material the correct statement uses complementary energy rather than strain energy — Crotti and Engesser’s theorem — and the two coincide only when the load–deflection relationship is a straight line.

It gives one displacement per load. $\partial U/\partial P_k$ gives the displacement at $k$ in the direction of $P_k$ and nothing else. A joint’s full displacement needs two derivatives in the plane and three in space, each with its own dummy load.

And it needs the whole structure solved. $U$ is a sum over every member, so nothing about the method is local: computing one deflection costs a full analysis, which is exactly what the unit-load method costs too.

Where the energy actually is #

The decomposition is worth reading as a design tool rather than as bookkeeping, because it answers a question no other method on this site answers directly: which member is responsible for the deflection?

On the truss here, the two end diagonals hold 22.3% of the strain energy each, the four chords hold 11.4% each, and two of the interior diagonals hold 4.8% — with one member holding nothing at all. Since every term of the energy is also a term of the derivative, that ordering is the ordering of the members’ contributions to the joint’s movement.

Which means stiffening the structure is a matter of finding the large terms. Doubling the area of a member holding 22% of the energy removes 11% of the deflection; doubling one holding 5% removes 2.5%; doubling the zero-force member removes nothing whatever and adds weight.

That is not the ordering of the forces. A member carrying a large force over a short length holds less energy than a lightly loaded long one, because the term is $N^2L/2EA$ and the length is in it linearly. Stiffening the most heavily loaded member is not the same as stiffening the structure, and the energy decomposition is the thing that tells them apart.

The best place to put the next kilogram, derived #

The energy decomposition says which members matter. One further step says how much area each should have, and the answer is not the one a strength argument gives.

Ask for the areas that minimise a chosen deflection $\delta = \sum \bar{N}NL/EA$ subject to a fixed total volume $V = \sum AL$. That is a Lagrange problem, and differentiating with respect to one member’s area gives

$$-\frac{\bar{N}_e N_e L_e}{E A_e^2} + \lambda L_e = 0 \quad\Longrightarrow\quad A_e \propto \sqrt{\bar{N}_e N_e}$$

The optimal area is proportional to the square root of the product of the real and virtual forces, not to the force itself.

That is worth sitting with, because the alternative is what most designers do. A fully stressed design sizes each member so that its stress reaches the allowable — $A \propto N$ — and it is the right answer when strength governs. A stiffness optimum sizes each member as $\sqrt{\bar{N}N}$, and the two are different distributions of the same tonnage.

The difference has a direction. A square root compresses the range: the heavily loaded members get relatively less area than a fully stressed design would give them, and the lightly loaded ones relatively more. A stiffness-optimal truss is therefore more uniform than a strength-optimal one, which is the opposite of the instinct that says to put the material where the forces are.

Two further readings fall out of the same expression.

A member with $\bar{N}N \le 0$ wants no area at all. Its term in the sum is negative — it reduces the deflection being minimised — so the optimality condition pushes its area up rather than down, bounded only by whatever else constrains it. And a member with $\bar{N}N = 0$ exactly, which includes every zero-force member, has an optimal area of zero: it should not be there. That is a topology statement rather than a sizing one, and it is the reason optimisation programs delete members rather than merely thinning them.

And the optimum is deflection-specific. $\bar{N}$ depends on where the unit load was put, so a truss optimised for the deflection of one joint is not optimised for another. There is no such thing as the stiffest truss of a given weight; there is only the stiffest truss for a given question, which is why real optimisation problems carry several deflection constraints at once and produce something that is best at none of them.

There is a check hiding in the result that is worth using, because the derivation is short enough to distrust. At the optimum, substituting $A_e \propto \sqrt{\bar{N}_e N_e}$ back into the deflection sum makes every member’s contribution proportional to $\sqrt{\bar{N}_e N_e}\,L_e$ — so each member’s share of the deflection becomes proportional to its own share of the volume. A structure at its stiffness optimum is one in which every kilogram is doing equally well, which is the form every optimality criterion in engineering eventually takes and a good sign that the algebra was done right.

Which sharpens the section above rather than replacing it. The energy decomposition says where the deflection comes from; this says what to do about it — and the answer is a square root, not a proportion, and it is different for every deflection anybody might care about.

The member-level reading has a structure-level twin. Span to the fourth makes the same argument one level up: a deflection is a sum, one term of it usually dominates, and the engineering is in finding that term rather than in evaluating the rest carefully. The difference is that a prismatic beam’s answer is known in advance from the exponent on the span, and a truss’s, or a tapered member’s, has to be computed from the integrand itself.

What the pictures cannot show #

The strain energy is a scalar. There is nothing to draw, and the bar chart above is a decomposition chosen because a total is unreadable — it is not a picture of anything the structure is doing.

Nor can the figures show the derivative. What is drawn is the error curve of a finite difference, which is an artefact of the arithmetic rather than a property of the structure; the theorem’s derivative is exact and has no step size at all.

The assumption the figure rests on #

The truss is statically determinate, so its member forces are a linear function of the loads and $\partial N/\partial P$ is a constant. That is what makes the numerical derivative exact rather than approximate: $U$ is a quadratic in $Q$, a central difference of a quadratic has zero truncation error, and the entire error curve above is round-off.

Put a redundant frame in instead and the forces are still linear in the loads — so the property survives, and it survives for the same reason. The linearity is the whole method, and everything else here is arithmetic.

The history worth having #

Alberto Castigliano published this in his 1873 dissertation at Turin, at twenty-six, and died at forty-eight. What is worth knowing is what it replaced.

Before it, a deflection was computed by integrating the differential equation of the elastic line — twice, with constants of integration fixed by boundary conditions — for each member and each case. That is entirely workable for a prismatic beam and hopeless for a frame with thirty members. Castigliano’s theorem turns a boundary-value problem into a differentiation, and Maxwell’s and Mohr’s virtual-work method, arriving at almost the same moment from a different direction, does the same thing.

Between them they made indeterminate structures analysable by hand, which is what made the next fifty years of long-span steel possible.

Shear energy, and the bracket that is all of it #

The truss above stores energy in axial force alone. A beam stores it in bending and in shear, and the ratio between them is a span-to-depth question with a decisive answer.

For a simply supported rectangular beam under a central load, the bending term goes as $P^2L^3/EI$ and the shear term as $P^2L/GA$. Their ratio is proportional to $(L/d)^2$: at a span-to-depth ratio of 20 the shear term is under a per cent, at 5 it is about a tenth, and at 2 it is comparable.

So the usual practice of ignoring shear deflection is right for a beam and completely wrong for a bracket, a corbel, a deep transfer beam or a short link. The energy method makes that visible because both terms are written down before either is dropped, whereas a moment-area or unit-load calculation typically never mentions shear at all.

The ladder from here #

Later rungs on this anchor: the theorem of least work, and redundants found by minimisation rather than by compatibility. Complementary energy and Crotti–Engesser, which is what the theorem becomes when the material is not linear. Castigliano’s first theorem — the derivative of strain energy with respect to a displacement gives the force — which is a different statement and the one finite elements are built on. Strain energy in bending, shear, torsion and axial force together, and the surprise that shear energy is negligible in a beam and dominant in a short bracket. The energy method for a rotation rather than a displacement, using a dummy moment. And the connection to stiffness: the second derivative of $U$ is the stiffness matrix, which is where this page’s method and the whole of matrix analysis turn out to be the same object.