The redundancy only the sun can find

The cut that needs a joint first applied the method of sections to one face of a K-braced tower and found that no horizontal cut through it has a moment centre: four members are cut, three equations are available, and the section is solvable only after the joint in the middle of the horizontal has been solved first. That joint receives the two diagonals of one K and nothing else with a vertical component, so its vertical equation says that the two diagonals carry equal and opposite forces — the fourth equation the section lacked.

The tower was determinate because every K had a joint of its own. A K-braced bridge girder built symmetric about mid-span does not. Its K’s open toward the centre from both ends, and at the central vertical the two middle K’s meet at the same joint, which then receives four diagonals. Its vertical equation relates four forces instead of fixing a ratio between two. The question is what solves the middle of such a girder: symmetry, stiffness, or a detail that removes the problem.

Counting, and where the extra member is #

The girder drawn below has eight panels of three metres and is three metres deep. Every vertical except the two end posts has a joint at mid-depth, where a K’s two diagonals arrive; left of mid-span the K’s point right, right of it they point left, and at the central vertical both middle K’s arrive at once.

The count is the one that decides whether statics can answer: two equations a joint, one unknown a member and one a reaction. Twenty-five joints give fifty equations; forty-eight members and three reactions give fifty-one unknowns. The girder is indeterminate to the first degree.

Everywhere but the middle, the argument of the tower still works. Cut any panel away from the centre and four members are cut; the mid-depth joint of that panel’s K supplies the ratio between its two diagonals; the section is then solvable. At the central joint the ratio is not supplied, because the joint’s vertical equation now contains the vertical components of four diagonals and says only that they sum to nothing.

The forces it can hold with nothing on it #

An indeterminate frame has, for each degree of indeterminacy, a set of forces it can carry with nothing applied: a state of self-stress. Its members pull and push on its joints in a pattern that balances at every one, and the frame’s actual forces under any load are any one solution of the equilibrium equations plus some amount of that state. Statics cannot say how much. That is what indeterminacy means, stated as a thing that can be drawn — and it is why a determinate truss, which has no such state, has no second path when a member goes.

The state here is small and local. It lives entirely in the two middle panels: the four diagonals at the central joint in tension, pulling the central vertical’s mid-point four ways at once and balancing there; the four central chord members in compression, holding the outer ends of those diagonals apart; and the halves of the two verticals at the outer side of the middle panels, also in compression, closing the loop at the joints where the diagonals start. Nothing outside those two panels takes part.

And it is symmetric about mid-span. Each member and its mirror image carry the same share. That settles the first candidate answer at once. Symmetry is a way of removing unknowns when the unknowns that are left can only be antisymmetric — a symmetric load cannot excite an antisymmetric pattern, so such a pattern’s amount must be zero. Here the pattern is symmetric. A symmetric load can call on it exactly as freely as any other load, and a uniform load does: the two diagonals of the left middle panel carry 48.4 kN in tension and 63.4 kN in compression where a K on its own would carry equal and opposite forces. Symmetry does not supply the missing equation.

What decides it: compatibility #

The equation that does decide it is the one statics never contains: the members must fit together after they have stretched. Each member stretches by its force times its length over its axial stiffness, $N L/EA$, and in an indeterminate frame the stretches must be consistent with one set of joint positions. For one redundancy that is a single equation. Written against the self-stress state $t$, it says that the amount $x$ of the state present is whatever makes the state’s members, taken together, do no net work:

$$\sum_i t_i\,\frac{(N^0_i + x\,t_i)\,L_i}{EA_i} + \sum_i t_i\,e_i = 0 \quad\Longrightarrow\quad x = -\frac{\sum_i t_i\,\big(N^0_i L_i/EA_i + e_i\big)}{\sum_i t_i^2\,L_i/EA_i},$$

where $N^0$ is any one set of forces that balances the load and $e_i$ is any length change the member has from something other than force: a misfit, a temperature. That formula is the whole of the girder’s middle. It says what excites the state — the extensions of its own twelve members — and what does not.

A load calls on it only through its own members #

The consequence is sharper than a “small effect”. Put a single 400 kN load at each top joint in turn and measure how much of the state each calls on. For a load at five of the seven joints the answer is exactly zero, at every ratio of chord to web area. Only a load at joints three and five — the joints at the outer side of the middle panels, whose upper half-verticals are members of the state — calls on it at all, and there it is at most 9 per cent of the middle diagonals’ force, when the chords are ten times the web’s area.

The formula says why. A girder bending under a load at any joint shortens its top chord and lengthens its bottom chord, and in the middle panels the two diagonals of each K stretch by equal and opposite amounts. The state’s chord members are all compressed and its diagonals all in tension, the same top and bottom, so the stretches of the top and bottom halves cancel term by term in the sum. The state is symmetric about mid-depth as well as about mid-span, and bending is antisymmetric about mid-depth. They are orthogonal, in the exact sense that the sum above is zero, and no area makes them otherwise.

What breaks the orthogonality is a load that enters one of the state’s members directly. A load on the top joint at three goes down the upper half of the vertical there to reach the mid-depth joint, compressing that half and not the lower one. That is a top-to-bottom asymmetry inside the state, and it calls on the state in proportion. Hang the same load from the bottom joint instead and the amount is the same with the opposite sign: 15.1 kN of it either way for 400 kN at joint three.

For a designer this is a strong result, and it is the practical answer to the question the tower essay ended on. Solve the middle of the girder by sections, treating each middle K as if it had its own joint — equal and opposite diagonals — and the answer is exact for every load not applied at the two joints beside the middle panels, and within a tenth for loads that are. The redundancy the count found is, for loads, very nearly an idle one.

What finds it: the fitter and the sun #

The same formula has a second term, and there the orthogonality does not help. A length change $e_i$ in any of the state’s members that is not the pattern of bending calls on the state directly, and every such change is by definition not a load.

A middle diagonal fabricated one millimetre too long — a tolerance well inside ordinary practice for a member three metres long — must be forced into place, and the state takes up the misfit: 23.6 kN in each of the four central diagonals, with the chords and half-verticals compressed to match, and not a kilonewton anywhere else. The effect is a truss built to the wrong length, confined here to one panel’s width either side of the centre.

The sun does the same with no fitter involved. A top chord twenty degrees warmer than the rest of the girder — an ordinary summer afternoon on a bridge whose deck shades its web — lengthens the four central top-chord members; the state is compressed in them, so their lengthening does negative work against it and calls on it: 30.4 kN in each central diagonal. Both chords warmer than the web, 60.7 kN. A movement that nobody applied produces force only where it is restrained, and in this girder the only restraint is the extra member at the centre.

Against those, the uniform load of 100 kN at every joint puts 63.4 kN into the larger middle diagonal. The forces that the load calculation does not contain are of the same size as the forces it does.

The state and the millimetre, by hand #

Everything in the last figure can be done with two joints and a sum. Each middle diagonal runs three metres along the girder and a metre and a half up or down to the central joint, so it makes an angle with the chords whose cosine is 0.894 and whose sine is 0.447. Put a unit tension in all four. At the central joint the four pulls balance by symmetry. At the top joint where one of them starts, the diagonal pulls the joint along the girder by 0.894 and down by 0.447; the only other members of the state there are the top chord member toward the centre and the half-vertical below, so the chord must push back along the girder with 0.894 — a compression — and the half-vertical must push up with 0.447, another compression. The whole state is those two equations, repeated four times: diagonals at one, chords at 0.894, half-verticals at 0.447.

Its flexibility is the sum of each member’s share squared times its own flexibility, $L/EA$. The four diagonals, 3.35 m long at 2,000 mm², contribute 0.032 mm per kN; the four chord members, 3 m at 6,000 mm² but carrying only 0.8 of the state’s squared share, contribute 0.008; the four half-verticals, 1.5 m at 2,000 mm² and a fifth of the share, 0.003. Together, 0.042 mm per kN. A millimetre of misfit divided by that is 23.6 kN, which is the figure’s number. Three-quarters of the flexibility is in the diagonals, so it is the web area, not the chords’, that sets the force a misfit makes.

The sun is the same sum with a different numerator. Twenty degrees on three metres of steel is 0.72 mm; the top chord has two members in the state, each at a share of 0.894, so the work term is 1.29 mm and the force is 1.29 divided by 0.042 — 30.4 kN.

Why the stiffer girder is not the safer one #

The misfit’s force is the misfit divided by the flexibility of the state — the sum of $t_i^2 L_i/EA_i$ over its twelve members — so it is proportional to their stiffness. Make the web members heavier to carry more load and the same millimetre produces more force; make the chords heavier and the sun’s twenty degrees produce more. It is the stiffest path taking the load, with the load replaced by an imposed movement: a strength designer strengthening the middle of the girder is also stiffening the thing the misfit and the temperature act against.

That is the ordinary behaviour of a redundant member, and here it is unusually sharp because the redundancy does nothing else. In most indeterminate structures the extra member shares the load, and its self-strain forces are the price paid for that sharing. This one shares almost none of the load — the section calculation was nearly exact — so its self-strain forces are a price with nothing bought.

What a load test cannot see #

There is a practical corollary for anyone who tries to check the middle of such a girder by measurement. A load test — a known load placed on the finished girder, with gauges on the members — measures the change of force the test load produces. The gauges are zeroed on the girder as it stands, with its misfit forces already locked in and whatever temperature difference the day has given it. The test then confirms, to the precision of the gauges, exactly what the section calculation predicted, because for a load at most joints the section calculation is exact; and it is blind to the 24 or 30 kN that were in the middle diagonals before the first gauge was read. A girder that passes its load test perfectly can be carrying, in its four most important web members, forces the test was built not to see.

Designing the joint out of existence #

The third candidate answer is the simplest. Turn every K the same way, so that each middle-depth joint receives one K and never two. The last panel then has a K pointing at the end post, where there is no mid-depth joint to receive it, and takes a single diagonal instead. The count comes out at fifty unknowns against fifty equations: determinate, with no self-stress state at all.

The girder has lost its symmetry, and that is all it has lost. Each middle diagonal carries 55.9 kN under the uniform load, half the panel’s shear divided by the sine of the diagonal’s angle, exactly what a section says. A millimetre of misfit is taken up by the girder moving by a millimetre somewhere, freely; the warm top chord makes it hog a little and stresses nothing. The asymmetry has a cost of its own — two kinds of end panel to detail, and a girder whose appearance is lopsided — but it is a cost that can be seen on the drawing, which the central redundancy’s is not.

The other way out is to accept the redundancy and give it room: slotted holes at one end of one middle diagonal, closed only after the girder has taken its dead load, so that the fit-up misfit is released and only what arrives afterwards is locked in. That is the same decision as stitching a continuous deck only after its beams carry their own weight, and it answers the misfit without answering the sun.

Where the numbers come from #

The girder is solved as a pin-jointed frame by the stiffness method with its real areas — chords 6,000 mm², web members 2,000 mm², steel — which gives the member forces under any load without choosing which member is redundant. The self-stress state is the null vector of the equilibrium matrix, found directly. The amount of the state in any solution is read off the left middle panel: its two diagonals carry $s + x$ and $-s + x$ for some split $s$, so $x$ is their average, and the “by sections” forces are the solution with $x$ removed. The self-strain forces come from the compatibility equation above with the state’s twelve members’ flexibilities, and the figure’s statement that a load at five of seven joints calls on none of the state holds at every area ratio tried, to the precision of the arithmetic.

Pins, a flat girder, and a temperature that is uniform #

Pinned joints. The members are pin-ended. A welded or gusseted girder has joints that resist rotation, which add secondary bending moments everywhere and give the girder a great many more redundancies than one; the argument here is about the axial forces, which the pins govern.

A girder that stays in its plane. The girder is two-dimensional. A real bridge has two girders joined by bracing, and a top chord warmer than the bottom on one girder only twists the pair.

Temperatures that are uniform along each chord. A chord partly in shade has a temperature that varies along it, and only the change in the four central chord members matters to the state.

Still open: the joint where more than two K’s meet #

The central joint here receives two K’s and adds one redundancy. A lattice girder whose bracing is a double K — two K’s stacked in each panel, meeting at the third-points of the verticals — has two such joints in every panel where the pattern turns, and a space frame whose faces are K-braced has corners where the K’s of two faces meet on one leg. Whether each shared joint adds exactly one self-stress state, whether those states remain orthogonal to bending as this one was, and whether a load can call on them only through their own members, is a question about how the extra equations a joint lacks add up when the joints are no longer alone.