The load goes to the cable furthest along

The sag the strain energy halves found what goes wrong when Castigliano’s theorem is applied to a member that stiffens. The theorem’s derivative of strain energy gives a member’s deflection only if the member is linear. For a stiffening member, it gives the deflection times the ratio of the secant stiffness to the tangent stiffness, and for a cable with no pretension that ratio tends to a third. That essay also put one stiffening member beside a linear one and found the wrong energy handing the stiffening member more load than compatibility would allow.

Its last figure raised a further question. A cable net is nothing but stiffening members sharing load, each with its own pretension and its own curve. If a net is checked by a force method, using the wrong energy, the wrong-energy bias cannot simply favour “the stiffening member”, since all of them are. It has to choose among them. The essay’s guess was that it favours the members most stretched, and that this concentrates the error where the forces are highest, understating the peak force as well as the deflection. Half of that is right. The other half has its sign fixed by something other than stretch.

A hub on four cables #

A real net has hundreds of nodes, and the point here is about sharing, so the structure is reduced to the one thing that shares and nothing else. A rigid hub is held in space by four cables running level to fixed anchors, and it carries a vertical load. Every cable’s end is at the hub, so compatibility is one number, the hub’s movement $\delta$. Each cable has axial stiffness $EA = 128$ MN, a length $L$ and a pretension $T_0$. Moved by $\delta$ its tension and its vertical force are

$$T = T_0 + \frac{EA}{L}\left(\sqrt{L^2 + \delta^2} - L\right), \qquad V = \frac{T\,\delta}{\sqrt{L^2 + \delta^2}}.$$

For a small movement the vertical force is the pretension’s $T_0\delta/L$, a linear spring set entirely by geometry. For a large one the stretch takes over and $V \approx EA\,\delta^3/2L^3$, cubic and set by the cable’s own stiffness. In between, every cable is somewhere along a curve from one regime to the other, and how far along it is at a given $\delta$ depends on its length and pretension.

The figure shows the first of two nets, four lengths with a single pretension. A short cable is steeper at the start, since $T_0/L$ is bigger, and it turns upward sooner, since its stretch goes as $\delta^2/2L^2$. So at any movement the shortest is the furthest along its curve and it also carries the most: 25.4 kN of the 50 at 0.78 m, half of the load in one cable.

Two ways of dividing a load #

There are two things one can ask of the shares $V_i$ that sum to the load $P$. Compatibility asks that each share imply the same hub movement: $\delta_i(V_i)$ all equal. A force method reaches the same answer if it minimises the complementary energy $\sum\int\delta\,dV$, whose stationarity with respect to the shares says exactly that, as the other area under the curve showed. That is the theorem of Crotti and Engesser.

Minimising the strain energy $\sum\int V\,d\delta$ is a different thing, and its derivative is a different quantity:

$$\frac{\partial U_i}{\partial V_i} = V_i\,\frac{d\delta_i}{dV_i} = \frac{V_i}{k_i},$$

force over tangent stiffness. The minimum over the shares sets $V_i/k_i$ equal for all the cables. On a linear member that ratio is the movement, and the two conditions coincide, which is why anyone can use either in an elastic truss. On a stiffening member, $V/k$ is not the movement. It is the movement multiplied by the ratio of secant to tangent stiffness: near one in the linear regime, near a third in the cubic. A cable far along its curve reports only a third of its movement. It looks as though it can take more, and the minimum gives it more.

On the first net the shortest cable carries 51 per cent of the 50 kN by compatibility and 67 by the wrong energy. It was already the most heavily loaded and it is the furthest along, and the two reasons point the same way.

The second net separates them. All four cables are 20 m long, with pretensions from 20 to 240 kN. Now the tightest cable carries most, because it has most vertical stiffness to begin with, 38 per cent of the load by compatibility. But it is still almost in its linear regime, with a secant nearly equal to its tangent, while the slackest cable is far along: at a small movement, the 20 kN cable has already started to feel its stretch. So the wrong energy takes load off the tightest and gives it to the slackest, and the order of the four cables reverses: 0.33, 0.28, 0.21, 0.18. What was the most heavily loaded member is now reported as the least.

The peak force, either way #

The peak tension is what a net’s cables and their end fittings are sized for, and so this is the figure that matters. On the first net the wrong energy overstates it, by up to 19 per cent at about 35 kN. That at least would be conservative, if an accident. On the second, it understates it by up to 25 per cent at about 75 kN. The same error, in the same theorem, moves the design force in either direction, and nothing in the method’s output shows which. The peak is off whenever the cable carrying most and the cable furthest along are not the same one.

Across the range of load, the shape in both cases is a hump. At a low load every cable is near its linear regime and all the secants match their tangents, so there is nothing to favour. At a high load every cable is deep in the cubic, where all report the same third, so again nothing separates them, and the error dies away to a percent or two. It is largest in the transition, when some of the cables have got there and others have not. A prestressed roof net is designed to spend its working life somewhere near that transition, with the pretension and the load comparable, and so its service loads are in the middle of the hump rather than in either of the regimes where this would not matter.

The deflection, and the cables that no longer meet #

The hub’s deflection behaves as the earlier essay’s single cable did, on both nets. Near the linear regime the wrong energy gets nearly the right answer, 0.85 of it at 5 kN. As the load rises, the answer falls below half the real deflection, 0.45 at 50 kN, and approaches a third. The two nets are almost on top of each other, in spite of their different cables, since this ratio is a statement about the regime rather than the distribution. A check of a net’s deflection by this route reports less than half of the real movement at the service load.

A check is available, and it is one that the wrong energy would fail at once if anyone looked. Take the shares the minimisation gives and work out the movement each one implies for its cable. By compatibility those are all one number. By the wrong energy they spread by as much as the hub’s whole movement, 0.96 of it at 10 kN on the second net. The cables have not met at the hub. A force method’s answer can always be tested by going back to the members with the forces and computing the displacements, and any force solution reached by the wrong energy will fail that test in a way visible on the back of an envelope.

This is exactly what the check that cannot see the beams meant by a check that reads the right thing: a check on a calculation that computes something the calculation did not already assume. Compatibility was never imposed by the wrong energy, so testing it is a real test. A force method for a net that works in the proper complementary energy will pass it to rounding, and that is what distinguishes the two when the program was written by somebody else.

Which net gets the force right #

Take the pretension away altogether and the error vanishes, at least as far as the forces go. With no pretension every cable is the cubic $V = EA\,\delta^3/2L^3$, for which $V/k = \delta/3$ exactly. The wrong energy then equalises a third of the movement, and that is the same thing as equalising the movement. The shares come out correct, whatever the lengths. Only the deflection is wrong, and by an exact factor of three. This is what the earlier essay’s limiting one-third was, and it is one of the rare cases where a wrong theorem gets some of the answer right by symmetry. An unprestressed net of a uniform material, loaded a good way past its sag, is therefore safe to share out by the wrong energy, as long as nobody reads a deflection from it.

Any pretension at all parts the curves. A tenth of the pretension already brings the worst error in the peak tension to 1.11 on the first net and 0.84 on the second. Eight times as much leaves it nearly unchanged from the original, at 1.18 and 0.75, and all it does is shift the hump along the load axis. Pretension does not make the problem go away; it decides at what load the problem lies. The common advice that a well-prestressed net can be treated as linear is true at loads below its transition, and a designer who has made sure of that needs no energy argument at all. Where the load approaches the pretension’s scale, the problem is present at whatever its size.

The route that needs no energy at all #

There is a way of reading the deflection that does not care whether a member is linear, and it is worth setting beside the two energies because it also tells on the wrong one. Any structure will carry the unit load showed that the unit-load theorem is a statement of virtual work, true of any material: the deflection is the sum of each member’s virtual force times its real elongation, and the virtual forces need only be in equilibrium with a unit load. So for this hub the unit load may be put wholly on any one cable, and the answer is that cable’s real movement. With the true shares, it makes no difference which is chosen. Every cable gives 0.78 m.

With the wrong energy’s shares, the choice matters, and that is the tell. At 50 kN on the first net, the shares imply 0.42 m for the longest cable and 0.87 m for the shortest, so a unit-load check gives one or the other depending on which path the checker happened to pick, and neither is the 0.35 m the energy’s own derivative reported. Three answers for one movement, from one set of shares, and a checker has only to try a second virtual path to see them.

That makes the mistake easy to find and also explains why it survives. The deflection that is a derivative is a tidy statement that holds for a linear truss and has been learned as the general one, and a tabulated cable curve fed to a spreadsheet produces forces and deflections that look entirely reasonable. A net’s stiffness comes from its shape, as the stiffness that comes from the shape put it, and so what makes it stiffen also makes the two energies differ, and it is the very reason for building a net in that form. The structure which most needs its force method to be the right one is the one for which the wrong one is most wrong.

The hand check #

Take the first net at 50 kN and check the wrong share by its defining equation. The shortest cable, $L = 12$ m and $T_0 = 120$ kN, at the true hub movement of 0.78 m: stretch $\sqrt{144.61} - 12 = 0.0253$ m, tension $120 + 128{,}000 \times 0.0253/12 = 390$ kN, vertical force $390 \times 0.78/12.03 = 25.3$ kN, matching the 25.4 in the first figure. The longest, $L = 30$ m, has stretch 0.0101 m, tension 163 kN and vertical force 4.2 kN. So their true shares are 51 and 8 per cent.

Their tangent stiffnesses at that movement are near $T/L + (EA/L)(\delta/L)^2$: for the short cable $32.5 + 45 = 77.5$ kN/m, and for the long one $5.4 + 2.9 = 8.3$ kN/m. So the ratio $V/k$ is $25.3/77.5 = 0.33$ m for the short cable and $4.2/8.3 = 0.51$ m for the long one, where both should have read 0.78. The wrong energy is looking at those two numbers and sees the short cable as much less displaced than the long one. So it moves load to the short cable until they match, about 0.35 m in the figure for the deflection. The peak rises from 391 to 459 kN.

And the no-pretension limit, in one line: $V = c\delta^3$ gives $k = 3c\delta^2$, and $V/k = \delta/3$ for any $c$.

What the hub leaves out #

One movement shared. A real net’s nodes move separately, so the shares here are a stripped-down version of what happens at each node between the two families of cables meeting there. Two curvatures of opposite sign describes the real structure, where a hogging family loses tension as a sagging family gains it. This hub has no such exchange and can’t lose a cable to slackening, and its nearest counterpart is a net’s central cable ring with its radial cables. What it does have is the simplest form of the sharing question, without any of those features getting in the way.

Level cables. All four are level at rest, so a sag or rise in the net’s profile, which makes the linear stiffness larger and the transition later, has been left out. The cable that is a spring sets out the corresponding curve for a cable with a sag, and it has the same two regimes.

A force method that really uses this energy. The point of the essay is what would happen, and modern net analysis works directly on displacement equilibrium, which can’t make this mistake, since the displacement method has compatibility built in. The risk lies with hand checks and with force-method spreadsheets: someone minimising an energy over assumed force shares, with a stiffening law tabulated from a single cable test. Software that works this way, with its cables taken from a load–extension table, is the case that deserves checking.

Perfect elasticity. Cables have a seating stretch on first loading and an effective modulus that rises with tension. Both of those make a cable stiffen by another route, and both would add to the spread between one cable and the next, particularly with cables of different diameters.

Still open: the net with a hard member in it #

Everything in this essay stiffens. Real roofs have members that do the opposite: a slender strut that buckles, a compression ring that softens, a cable that has gone slack in its other family. On a softening member, secant is greater than tangent, $V/k$ overstates the movement, and the wrong energy takes load away. So a net with a softening member in parallel has its bias working in both directions at once, and a load will be moved out of the member that is softening and towards the one that is stiffening, by a method whose error then exceeds either one alone. Whether a force check on a guyed mast or a cable-braced frame, where some members stiffen and others buckle, overstates the buckling member’s safety at the worst possible moment, is what the softening half of the curve would ask.