The sag the strain energy halves

The other area under the curve found where Castigliano’s second theorem stops being true. Differentiate a structure’s strain energy with respect to a load and the answer is the deflection at that load only while the material is linear; past that, the right energy is the complementary one, the area on the other side of the stress–strain curve. On aluminium at its proof stress the strain energy overstated a deflection four times over, and the overstatement tended to the curve’s exponent.

Every member in that essay softened: its strain grew faster than its stress, so its load–deflection curve bent over, and the strain energy was the larger of the two areas. It closed on the members that do the opposite. A cable loaded across its span, a membrane under pressure, a rubber bearing squeezed flat and anything that has to take up some play before it carries load all grow stiffer as they deform. For them the curve bends up, and the answer to the obvious question — does the same error simply reverse its sign — turns out to be yes, with a precise size, a limit, a case where it swallows the whole answer, and a consequence for redundant structures that points the opposite way from the softening one.

A cable that stiffens as it sags #

The member throughout is a steel cable 10 m long between fixed anchors, pretensioned to 60 kN, with an axial stiffness $EA$ of 64 MN — a strand of about 400 mm² of the kind that holds up a glass façade — loaded across its span at midpoint by a force $P$, as a wind load arrives through a fitting. Its stiffness against that load has two sources. The pretension alone resists a sideways deflection $\delta$ in proportion to it: the two halves, each of length $a = 5$ m, lean by $\delta/a$ and their tensions resolve to $2T_0\delta/a$. And the deflection stretches the cable, so its tension grows, by $EA$ times the strain $\approx \delta^2/2a^2$, and the growth resists in proportion to $\delta$ again. The load is therefore

$$P \approx \frac{2T_0}{a}\,\delta + \frac{EA}{a^3}\,\delta^3,$$

linear at first and cubic once the stretching dominates. The figures use the exact geometry, but this is the whole of the mechanism.

The picture is the mirror of the softening member’s. The load–deflection curve bends upward, so of the two areas that together fill the rectangle under a point on it, the one between the curve and the load axis — the complementary energy — is now the larger: at 10 kN, 1,323 J against 802 J of strain energy actually stored. The strain energy is the work the load has done, and because the cable was soft at first and stiff later, most of that work was done early, at small loads.

Two derivatives of two energies #

Crotti and Engesser’s theorem says the deflection is the derivative of the complementary energy with respect to the load, whatever the material. Castigliano’s second theorem says it is the derivative of the strain energy, and holds only on a straight line. On this cable the two can be compared directly.

The complementary route lies on the truth at every load, as it must. The strain-energy route lies below it, and not by a little: at 10 kN it reports 107 mm against a real 212 — almost exactly half — and at 20 kN, 128 mm against 294. The error is on the side that matters for a façade. A deflection is checked against a limit that protects the glass and the seals, and the strain-energy route passes cables that sag twice as far as it says.

The third curve is the calculation that ignores the stiffening altogether and treats the cable as a spring of stiffness $2T_0/a$. It overstates the sag, by a factor of two at 10 kN and nearly three at 20, which is the familiar error and the safe one. A designer who knows the cable stiffens and reaches for an energy method to include it has, with the strain energy, turned a conservative answer into an unconservative one of about the same size.

The error is one ratio, in both directions #

The size of the error has a one-line derivation that covers every single-load member, stiffening or softening. For a member carrying load $P$ at deflection $\delta$, the strain energy’s derivative with respect to $P$ is $dU/d\delta \cdot d\delta/dP = P/k_t$, where $k_t = dP/d\delta$ is the tangent stiffness; the deflection itself is $P/k_s$, where $k_s = P/\delta$ is the secant stiffness. So

$$\frac{\partial U/\partial P}{\delta} = \frac{k_s}{k_t}.$$

The strain-energy route returns the deflection multiplied by the ratio of the secant stiffness to the tangent stiffness. For a straight line the two are equal and the theorem is exact. For a member that softens the tangent is below the secant and the ratio exceeds one; for one that stiffens the tangent is above it and the ratio falls below one.

For a power law $P \propto \delta^m$ the ratio is exactly $1/m$, which answers the question of whether the understatement has a limit. A cable with no pretension carries load purely through stretching, $P \propto \delta^3$, and the strain-energy route gives one third of its deflection at every load. A pretensioned cable starts linear, with a ratio of one, and approaches a third as the stretching term takes over; the more pretension, the longer it stays near one, which is why with 120 kN of pretension the route still returns 0.56 of the truth at 20 kN of load. A flat membrane under pressure, whose load is also cubic in its deflection once its pretension is overcome, has the same limit, and so does a net of cables prestressed against each other in two curvatures of opposite sign.

What plays the part of the exponent, then, is the reciprocal of the stiffening exponent — and for the stiffening that geometry produces, which is almost always a cube, the limit is a third.

The mirror, drawn once #

On a logarithmic axis the two families sit either side of one, and the symmetry is the argument. The aluminium bars of the earlier essay hold at one while they are linear and climb toward their exponent once the knee of their curve is passed; the cable with no pretension sits at a third throughout, and the pretensioned cable leaves one and descends toward it. Each curve is its member’s $k_s/k_t$ and nothing else. The theorem is not approximately right for mildly nonlinear members; it is right exactly to the extent that the secant and tangent stiffnesses agree, and a member whose curve bends at all, either way, has a known and computable error.

The mirror also explains why the softening case gets the attention. Materials soften near their limit, which is where strength is checked, and an overstated deflection there is embarrassing rather than dangerous. Geometry stiffens in service, which is where deflection is checked, and the understatement arrives at exactly the check that governs a cable, a membrane or a net.

Where structures stiffen as they deflect #

The cable is the cleanest case, but it is not a rare one. Any member that carries load by changing its shape stiffens as the shape changes, and several of the situations in which a deflection matters most are of that kind.

An inflated roof held up by the air inside resists a local load by stretching its membrane, and its load–deflection curve is the cable’s in two dimensions: soft at first, then cubic. A floor slab that has lost a supporting column hangs from its reinforcement as a net, and the tension that develops as it sags — the force nobody put in the model — is what holds it up; the deflection at which it does so is a large one, and it is found by exactly the kind of energy argument that picks the wrong area if it is not careful. A laminated rubber bearing stiffens in compression as its layers bulge and lock against the steel plates. A glass pane clamped at its edges and pushed by wind stops bending like a plate and starts stretching like a membrane once its deflection passes its thickness, which is why the rules for glass under wind use charts rather than beam formulas.

In each of those the hand check that reaches for an energy method is reaching for it precisely because the member is nonlinear — nobody integrates an energy to find the deflection of a straight line — and in each the strain energy gives an answer on the unsafe side. The complementary energy is not harder to write: for a member whose deflection is known as a function of its load, it is the load times the deflection minus the strain energy, and the extra term is the one that carries the error.

The member with play in it #

The extreme case of stiffening is a member that carries nothing at all until it has moved a certain distance, and then carries load linearly: a hanger with a clearance in its pin, a tie with slotted holes, a bracing rod installed slightly slack. Its load–extension curve is a flat line followed by a sloping one, the stiffest kind of upward bend there is.

The strain energy of such a member is exactly the energy of its elastic stretch: $N^2/2k$, from the moment the play is taken up. The play stores nothing, so its derivative cannot contain the play, and the strain-energy route returns $N/k$ — the stretch alone — for every load. At 20 kN the hanger has extended 4.18 mm and the route reports 1.18; the slack is not understated, it is absent. The complementary energy is $N e_0 + N^2/2k$, where the first term is the rectangle of the force times the play, and its derivative is the whole extension.

That makes the slack member the limit of the ratio in the other direction: secant over tangent is $(e - e_0)/e$, which tends to zero as the play dominates. In a determinate structure the missing play is a missing deflection; in a redundant one it is a missing redistribution, because a member built to the wrong length picks up load only after the others have deflected by its misfit.

Least work, and the load handed to the wrong member #

The theorem of least work finds the redundant forces in an indeterminate structure as those that make its energy stationary — for a linear structure the same thing as the compatibility condition of the force method — and the earlier essay found that minimising the strain energy of a truss with a softening bar put too little force into that bar. The stiffening case runs the other way.

Put the cable beside a mullion — a light aluminium section, 40 N/mm at the point where the cable crosses it — so that the two resist the same load at the same point and share it by their stiffnesses. Compatibility says both deflect alike, and at 10 kN it gives the cable 4.6 kN and the mullion 5.4; minimising the complementary energy gives the same split to every figure the search returns. Minimising the strain energy gives the cable 6.2 kN and the mullion 3.8.

The forces are in equilibrium, and they are impossible: the cable at 6.2 kN would sag 164 mm and the mullion at 3.8 kN would deflect 95, and the two are connected at the same point. The reason is the one-line derivation again. Stationary strain energy sets the strain energy’s derivative of each member equal, and for the cable that derivative is a third to a half of its real deflection; to make the two equal, the cable has to be loaded until its understated sag matches the mullion’s honest one. The wrong energy hands the load to the member that stiffens — and so leaves the mullion, which is the member whose deflection limit was the reason for the calculation, a third lighter than it really is.

How much the load is shared decides the size of the error #

The size of the misallocation depends on how the load is really shared. A mullion ten times stiffer than the one above carries nearly all of the load whatever energy is minimised, and the error falls to a few per cent. A mullion much softer leaves nearly all of it to the cable, and the cable’s share cannot grow past the whole. Between the two the cable’s excess peaks, at 41 per cent with a mullion of 56 N/mm, where the cable’s true share is about a third. That is also where such systems are designed to sit, because a cable and a mullion are combined on a façade precisely so that each carries a useful share: the case in which the wrong energy errs most is the case the combination exists for. The same logic runs through every redundant system in which a member gains stiffness from its own load, as a stay does from the stiffness that comes from its shape — the error is small where one path dominates and largest where the stiffest path takes the load only partly.

The numbers at 10 kN, by hand #

The cable at 10 kN has deflected $\delta = 212.5$ mm. Each half, 5,000 mm long, is now $\sqrt{5000^2 + 212.5^2} = 5{,}004.5$ mm, stretched by 4.5 mm, a strain of $9.0 \times 10^{-4}$. Its tension is $60 + 64{,}000 \times 9.0 \times 10^{-4} = 117.8$ kN, and the load it carries is $2 \times 117.8 \times 212.5/5{,}004.5 = 10.0$ kN, which closes the loop.

The secant stiffness is $10{,}000/212.5 = 47.1$ N/mm. The tangent stiffness, from the approximate form, is $2T_0/a + 3EA\,\delta^2/a^3 = 24 + 3 \times 64 \times 10^{6} \times 212.5^2/5{,}000^3 = 24 + 69.4 = 93.4$ N/mm. Their ratio is 0.504, and the strain-energy route gives $0.504 \times 212.5 = 107$ mm — the value the figure reports.

The pretension alone would give $P\,a/2T_0 = 10{,}000 \times 5{,}000/120{,}000 = 417$ mm. The truth is between the two, nearer the stiff answer than the soft one, and only the complementary energy lands on it.

The energies, and the path each is taken along #

The deflection in each figure is found from the geometry: the midpoint drop at which the exact tension of the stretched halves, resolved, balances the load. The strain energy is the pretension times the extension plus half the axial stiffness times the extension squared, over both halves — the work done from the pretensioned state — and its derivative with respect to the deflection is the load exactly. The complementary energy is the load times the deflection minus that. Both derivatives with respect to the load are taken numerically by central differences on the energies themselves, so neither is the geometry restated. The shared load is found three ways: by bisection on equal deflections, and by golden-section minimisation of each energy over the cable’s force.

With the pretension set to nearly nothing the strain-energy route returns a third of the deflection to three figures, and at small loads the exact geometry reproduces the linear-plus-cubic form above to within its approximation.

Small strains, a straight chord and a static load #

The cable’s material is linear. Its stiffening is all geometric; a strand that also softens as its wires bed in would carry both effects, and its $k_s/k_t$ could cross one.

The load is at midpoint and the cable is straight between anchors that do not move. Real façade cables are attached to anchors that deflect, which adds a linear spring in series with the stiffening cable and moves the curve toward a straight line, and they carry several loads, which makes the deflection at one of them a function of all.

The load is static. A cable under gusting wind is stiffer on average than its static curve says, and a slack member under a reversing load impacts at each reversal; neither is in the energies here.

The energies are for one load path. For a single load the error is $k_s/k_t$ exactly. With several loads the strain-energy route’s error at one of them depends on how the others were applied, because a nonlinear member’s strain energy depends on the path and its complementary energy’s derivatives are the deflections only along the path actually taken.

What the pictures cannot show #

Why an analyst would use the strain energy at all. The argument here does not touch a program that solves for the deformed shape directly, which is what any geometrically nonlinear analysis does; it touches the hand check, the spreadsheet and the textbook energy method, which is where a cable is first sized and where the wrong area is easiest to pick up. It also cannot show a material that stiffens: the geometry of a sagging member does it here, but a rubber bearing in compression stiffens because of its constraint, and a soil under a footing because of its confinement, with curves that are not cubes and ratios that are not a third.

Still open: the net whose members stiffen by different amounts #

The shared-load figures put one stiffening member beside one linear one. A cable net is nothing but stiffening members sharing load, each with its own pretension and its own curve, and there the wrong energy’s bias does not simply hand load to “the stiffening member” — it hands it to whichever members are furthest along their curves, the ones already most stretched. Whether that concentrates the error in the members that are most loaded, and so understates the peak cable force as well as the deflection, is the question a prestressed net asks of any energy method used to check it.