The other area under the curve

Castigliano’s second theorem is one of the most satisfying results in structural analysis. Write down the strain energy stored in a structure as a function of the loads on it, differentiate with respect to one load, and the answer is the displacement at that load in its direction. The essay on it computed a truss’s deflection that way and by the unit-load method and found the two agree to the precision of the arithmetic — because, for a linear structure, they are the same sum written twice.

That essay also said, in a sentence near its end, that the theorem is a linear one: that for a nonlinear elastic material the correct statement uses complementary energy rather than strain energy, and that the two coincide only when the load–deflection relationship is a straight line. The sentence is true and it undersells the difference. On a real metal a little past its limit of proportionality, the two theorems are not slightly different; one of them is wrong by a factor that approaches the exponent of the material’s stress–strain curve.

Two areas, and the one that is not usually drawn #

A member’s stress–strain curve divides the rectangle under any point on it into two regions.

The area between the curve and the strain axis is the strain energy density: $\int \sigma\,d\varepsilon$, the work done on a unit volume of material in straining it. It is the area that is usually shaded, because it is the energy the material stores.

The area between the curve and the stress axis is the complementary energy density: $\int \varepsilon\,d\sigma$. It is not the energy of anything, and it has no obvious physical meaning — which is why it is not usually drawn. Its value is that its derivative with respect to stress is the strain, in exactly the way the strain energy’s derivative with respect to strain is the stress.

The two areas add up to the rectangle $\sigma\varepsilon$. On a straight line through the origin they are equal, each half of it. On anything else they are not.

The curve in the figure is the aluminium alloy that has no yield point: its proof stress is defined by an offset because it has no corner at which to define anything else. The Ramberg–Osgood form describes it as $\varepsilon = \sigma/E + 0.002(\sigma/\sigma_{0.2})^n$ — a linear strain plus a nonlinear one that is exactly 0.2 per cent at the proof stress and grows as the $n$th power of stress. With $n = 10$, the nonlinear part is 0.0075 per cent at 180 MPa — a thirty-fourth of the linear strain — and then it arrives quickly: 0.06 per cent at 220 MPa, 0.2 per cent at 250.

At 240 MPa the curve has bent enough that the strain energy is 60 per cent larger than the complementary energy. At 150 MPa it is still nearly straight and the two differ by under one per cent. The difference is a direct measure of how far the material has left the line, and it is the thing that decides which theorem is right.

Two bars and three routes to one deflection #

Take two aluminium bars, each 2,500 mm long with a 500 mm² section, pinned at supports 4 m apart and meeting at an apex 1.5 m below the line of the supports. A load $P$ hangs from the apex. Each bar makes an angle with the horizontal whose sine is 0.6, so each carries $N = P/1.2$ in tension whatever the material does: the truss is determinate and the forces come from statics alone.

The drop of the apex can be found three ways.

By geometry. Each bar’s strain follows from its stress by the Ramberg–Osgood law; its extension is that strain times 2,500 mm; and a bar sloping at sine 0.6 that extends by $e$ lets the apex drop by $e/0.6$. At $P = 150$ kN the bars are at 250 MPa, exactly the proof stress, and the strain is $250/70{,}000 + 0.002 = 0.005571$, the extension 13.93 mm and the drop 23.21 mm. This route involves no energy at all and there is nothing to argue with.

By Castigliano. The strain energy of the two bars is twice the bar volume times the strain energy density. Differentiate it with respect to $P$.

By Crotti and Engesser. The same, with the complementary energy.

The figure is the hero drawn again, and the two derivatives are taken numerically from the energy sums rather than from formulas, so their agreement or disagreement with geometry is a measurement. The complementary route lies on geometry everywhere. The strain energy route lies on it to about 90 kN, where the bars are at 150 MPa, and then leaves: at 150 kN it gives 98.2 mm, 4.2 times the real drop.

It is worth seeing why by hand, because the reason is three lines.

The complementary energy of a bar is $V\int_0^\sigma \varepsilon\,d\sigma$, so its derivative with respect to the bar force is $V\varepsilon\,d\sigma/dN = \varepsilon L = e$, the extension. Differentiating through $N = P/1.2$ and summing both bars gives $2e/1.2 = e/0.6$ — the geometric answer, exactly.

The strain energy is $V(\sigma\varepsilon - \int\varepsilon\,d\sigma)$. Its derivative with respect to the bar force is $V\sigma\,d\varepsilon/dN = N\,de/dN$. That is the extension only if $N\,de/dN = e$, which is the definition of a straight line through the origin. For the Ramberg–Osgood law, $N\,de/dN$ is the linear extension plus $n$ times the nonlinear one:

$$N\frac{de}{dN} = \left(\frac{\sigma}{E} + n \times 0.002\left(\frac{\sigma}{\sigma_{0.2}}\right)^{n}\right)L.$$

At the proof stress that is $(0.003571 + 10 \times 0.002) \times 2{,}500 = 58.93$ mm of what Castigliano treats as extension, against 13.93 mm of real extension. Times $2/1.2$ gives the 98.2 mm.

The strain energy route weights the nonlinear part of the strain by the exponent. Wherever the curve is straight it is right; wherever the curve is bending it overstates, and the overstatement is largest exactly where the member is most heavily loaded.

The error is the exponent #

The ratio of the two answers has a limit that the formula above makes plain: as the stress rises and the nonlinear term dominates the linear one, $N\,de/dN$ approaches $n$ times the extension. The strain energy route’s overstatement approaches the exponent.

The three curves say something the formula does not make obvious. A sharper knee holds the error back and then releases more of it. An alloy with $n = 20$ has a nonlinear strain under one per cent of its linear strain up to 80 per cent of its proof stress, where an $n = 5$ alloy’s is already nearly a quarter, and then the $n = 20$ curve crosses the others and at the proof stress is nearly twice as wrong. The materials whose stress–strain curves look most like an elastic line followed by a plateau — the ones an analyst is most tempted to treat as linear up to a point — are the ones for which the strain-energy form of the theorem fails most violently past it.

That is also a statement about stainless steels and cold-worked sections, whose exponents are low, against high-strength aluminium alloys, whose exponents are high. The low-exponent material strays early and gently; the high-exponent one holds and then goes.

Least work, and a truss the wrong energy cannot build #

Castigliano’s theorem has a sibling that is used more often in practice: the theorem of least work, which finds the redundant forces in an indeterminate structure as the values that make its energy stationary. For a linear structure it is equivalent to the compatibility condition of the force method, and it is often preferred because a minimum can be found without deciding what the compatibility condition is.

The same question arises. Which energy?

Add a third bar to the truss: a vertical one, 1,500 mm long, from the apex straight up to a support midway between the other two. Make it aluminium, the same 500 mm² alloy, and make the two inclined bars steel of the same area. The truss now has one redundant — say the force $X$ in the vertical bar — and the inclined bars carry $(P - X)/1.2$.

Compatibility says the vertical bar’s extension must equal the apex drop that the inclined bars’ extensions allow. At $P = 260$ kN that fixes $X = 107.7$ kN. The aluminium bar is at 215 MPa and stretches 5.29 mm; the steel bars are at 254 MPa, stretch 3.17 mm each, and let the apex drop $3.17/0.6 = 5.29$ mm. Everything fits.

Minimum complementary energy gives the same $X$, to every digit the search returns, and the reason is the three-line argument above run backwards: the derivative of the total complementary energy with respect to $X$ is the vertical bar’s extension minus the apex drop the others permit, and a minimum is where that is zero.

Minimum strain energy gives $X = 94.0$ kN.

The forces the strain-energy minimum proposes are in perfect equilibrium: 94.0 kN up the vertical bar and 138.3 kN in each inclined bar resolve exactly to the 260 kN load, and nothing in statics can object to them. They are also impossible. At 94.0 kN the aluminium bar would stretch 4.20 mm, and at 138.3 kN the steel bars would let the apex drop 5.76 mm. The vertical bar would have to be 1.56 mm longer than it is to reach.

The wrong energy has found a statically admissible set of forces that no deformed shape of the truss can produce. It has quietly relaxed the one condition that makes a redundant structure’s forces what they are, and it has done so in the direction that puts less force into the yielding bar — which is the unconservative direction for the steel.

This is the same asymmetry the unit-load method’s two readings turned on. The complementary energy is the functional whose stationary point imposes compatibility on a set of forces already in equilibrium; the strain energy is the functional whose stationary point imposes equilibrium on a set of displacements already compatible. Minimising strain energy over forces rather than over displacements mixes the two, and a linear material is the one case where the mixture happens to give the right answer.

A stainless tie at its working stress #

The examples above push aluminium to its proof stress, which is further than a designer normally takes a member in service. The low-exponent materials make the point at ordinary working stresses, and a single stainless steel tie is enough to show it.

Take a 3 m tie in an austenitic stainless steel, with a modulus of 200,000 MPa, a 0.2 per cent proof stress of 240 MPa and a Ramberg–Osgood exponent of 5, which is typical of the annealed grades. At a service stress of 180 MPa — three-quarters of the proof stress, and an ordinary utilisation for a tension member — its linear strain is 0.090 per cent and its nonlinear strain $0.002 \times 0.75^5 = 0.047$ per cent. The nonlinear part is already more than half the linear part, at a stress no designer would call yielding.

The tie’s real extension is $(0.00090 + 0.00047) \times 3{,}000 = 4.12$ mm. The strain-energy route weights the nonlinear part by five and returns $(0.00090 + 5 \times 0.00047) \times 3{,}000 = 9.82$ mm, 2.4 times too much. A linear analysis that ignores the curve altogether returns 2.70 mm, a third too little.

So on the same member at the same load, the three calculations an engineer might reasonably make give 2.70, 4.12 and 9.82 mm. The first ignores the material’s curve; the third uses the curve and the wrong energy; only the middle one is right. And the wrong-energy answer is the one that looks most careful, because it is the only one of the two wrong answers that took the nonlinear material into account.

That is the practical form of the argument. For carbon steel, which is linear to near its yield point, the choice of energy almost never matters in service. For stainless steel and for many aluminium alloys it matters at working stresses, which is why stainless steel design rules compute deflections with a secant modulus evaluated at the working stress — geometry, in effect — rather than from any energy expression.

Why the linear case hides the whole distinction #

For a linear material every one of these distinctions collapses, and it is worth being exact about how.

On a straight line the two areas are equal, so $U = U^*$ at every load. Their derivatives with respect to load are then equal too, and Castigliano’s second theorem and Crotti–Engesser’s are the same theorem. Least work in strain energy and least work in complementary energy are the same minimisation. And the numerical Castigliano calculation that agreed with the unit-load method to within round-off agreed because it was computed on a linear truss, where the strain energy is the complementary energy wearing a different name.

That is why the theorem is taught in its strain-energy form, and why the form survives in so many design aids: for the linear-elastic analysis that governs most serviceability checks, it cannot be distinguished from the correct one. The distinction matters only where a member has left the straight line, and those are exactly the members where a designer is most interested in the deflection — the ones approaching their proof stress, or losing their strength to a weld, or cracking and softening.

Which one a program uses #

A modern nonlinear analysis does not minimise either energy directly. It solves equilibrium incrementally, updating each member’s tangent stiffness at each step, and the result is the geometric answer — the solid line in every figure above — because the incremental method never writes down a total energy and so never has the chance to choose the wrong one.

The distinction survives in three places.

Hand checks of a nonlinear result. An engineer checking a program’s deflection of a member near its proof stress by differentiating an energy will get the program’s answer from the complementary energy and a much larger one from the strain energy — and may conclude, wrongly, that the program is unconservative.

Secant-stiffness shortcuts. Replacing a nonlinear member by a linear one of secant stiffness $N/e$ reproduces the right extension at one load, and that linear member’s strain energy is then $Ne/2$ — neither of the real member’s energies. It gives the right deflection at that load and the wrong derivative at every other.

Energy-based stability and collapse methods. Rayleigh’s method and the energy criteria for buckling are written in strain energy, which is right because they are stationary conditions on displacement fields. Rewriting them in terms of forces requires the complementary energy, and an analyst who carries a force-based energy argument across the proof stress without switching is making the error drawn here.

What the figures do not show #

Unloading. Every curve here is a loading curve. The Ramberg–Osgood law describes a material that yields, and a yielded material unloads along a straight line rather than back down the curve, so neither energy is recovered on unloading. The theorems here are theorems of nonlinear elasticity, and a real alloy satisfies them only for monotonic loading.

Compression. The bars are in tension. The same law in compression is a material model for a stocky member; a slender aluminium strut buckles before its proof stress matters.

Geometric nonlinearity. The apex drops tens of millimetres on bars 2,500 mm long, and the bars’ angle changes by a few per cent. The force in each bar is computed on the undeformed geometry. That is a separate nonlinearity, and a separate reason a real truss departs from the solid line.

A fitted law. The Ramberg–Osgood exponent is a curve fit, and the error ratio depends on it directly. The exponents 5, 10 and 20 are representative of the range of structural stainless steels and aluminium alloys; the number for a particular alloy comes from its own test curve.

Where the model stops #

Elastic, in the sense of path-independent. Both energies are functions of the current state only if the material returns along its loading curve. Real metals past their proportional limit do not, and for them the correct tool is an incremental plasticity analysis, not either theorem.

Axial members only. Bending members have the same distinction, with moment and curvature in place of force and extension, and a moment–curvature relation that is nonlinear for the same reasons. The arithmetic is longer and the conclusion is identical.

One load. Everything here differentiates with respect to a single load. With several loads acting, each derivative of the complementary energy gives its own displacement, and the strain energy’s derivatives are wrong in proportion to how nonlinear the members carrying each load are.

Still open: the member that stiffens #

Every member here softens: its strain grows faster than its stress, and the strain energy is the larger of the two areas. A cable sagging under its own weight, a rubber bearing, and a membrane under pressure all do the opposite — they get stiffer as they deform — and for them the complementary energy is the larger area and the strain-energy route understates the deflection instead of overstating it. Whether the understatement has a limit, what plays the part of the exponent, and whether least work in the wrong energy then errs toward the stiffer member rather than away from it, is the same question with its sign reversed.