The springings that make shortening worse

The arch that gets shorter found that a funicular arch carries its load with no bending at all until its rib shortens under its own thrust, and that the tenth of a per cent of thrust the shortening costs is the whole of the arch’s moment. It did the calculation for a two-hinged arch, with one redundant, and named the case it had not done: “the fixed arch, which is indeterminate to the third degree and has three of these corrections rather than one.”

There is an intuition from the two-hinged calculation that is easy to carry across to that case, and it is the wrong way round. In a two-hinged arch the secondary moments are zero at the springings by definition, so shortening, temperature and shrinkage look like problems of the crown, well away from the springings where a fixed arch is designed. In a fixed arch they are largest at the springings.

The same rib, held at both ends #

The rib is the earlier essay’s: 60 m span, 6 m rise, a concrete section with a bending stiffness of $420{,}000\ \mathrm{kN\,m^2}$ and an axial stiffness of 20.4 million kN, carrying 60 kN/m. Its funicular thrust is $wL^2/8f = 4{,}500$ kN, and a rib that did not shorten would carry that thrust and no moment anywhere.

Two-hinged, the rib loses 0.10 per cent of its thrust, and the missing 4.6 kN acting at the crown’s 6 m lever arm leaves 29 kN·m of sagging — the earlier essay’s 27.9 by its own integration, the small difference being the straight segments the frame solve is built from. Fixed at the springings, the same rib loses 0.63 per cent of its thrust, 28.4 kN, and the moment diagram changes shape. It sags by 58 kN·m at the crown, twice the two-hinged rib’s, and hogs by 112 kN·m at each springing, where the two-hinged rib has nothing.

Where the lost thrust acts #

The reason is where a fixed arch’s redundants sit. A two-hinged arch has one, the thrust, and it acts along the springing line, because that is where the pins are. Any change in it bends the rib by the change times the height of the axis above that line: most at the crown, nothing at the springings.

A fixed arch has three redundants, and the standard way to find them is to move them to a point where they uncouple. That point is the elastic centre, the centroid of the rib with each length weighted by its flexibility, $1/EI$. Taken there, the horizontal redundant and the moment redundant are independent, and for a symmetric arch under a symmetric load the vertical one is zero. A change in thrust then bends the rib by the change times the height of the axis above the elastic centre, not above the springings.

For a parabola of constant section the elastic centre is two thirds of the way up. On this rib it is 3.96 m above the springing line, 0.66 of the 6 m rise. So a loss of thrust $\delta H$ puts $\delta H \times 2.04$ m of sagging at the crown and $\delta H \times 3.96$ m of hogging at the springings. The springing moment is twice the crown’s and of the opposite sign, and that ratio is fixed by the shape of the arch, not by its size or its stiffness.

With $\delta H = 28.4$ kN that is 58 kN·m at the crown and 112 at each springing — the two numbers the frame solve found by assembling the rib’s stiffness and solving it. The elastic-centre integrals and the frame solve are two independent calculations, and they agree to a fraction of a per cent.

Three redundants, and why two of them move #

A fixed arch has six support reactions and three equations of equilibrium, so three of the reactions cannot be found from statics. Taken at the elastic centre they are a horizontal force, a vertical force and a moment. A uniform shortening of a symmetric rib is symmetric, so it cannot produce the vertical one, which would have to push one half up and the other down; that leaves the thrust and the moment.

The moment redundant is the one the two-hinged arch does not have, and it is what the elastic centre is for. Shortening does not rotate the springings of a symmetric rib, so on its own the moment redundant would be zero — but only when the thrust is taken at the elastic centre. Taken at the springing line instead, as a two-hinged calculation would take it, the same state of the rib needs a springing moment to hold the tangents still, and that moment is exactly the thrust change times the elastic centre’s height. The two descriptions are the same physics. The elastic centre is simply the point about which the springings’ fixity costs nothing to write down.

The vertical redundant is the one that waits. Anything asymmetric — an abutment that settles more on one side, a temperature difference between the sunlit and shaded halves, a live load on half the span — engages it, and with it the antisymmetric bending that the arch that leans instead of squashing found to be the rib’s softest mode.

Six times the loss #

The other half of the answer is how much thrust is lost. The flexibility equation has the same form as the two-hinged one: the rib’s axial shortening on top, and the rib’s flexibility against the redundant underneath. The shortening on top is the same. What changes is underneath. The two-hinged rib resists a change in thrust by bending with a lever arm of $y$, the height above the springings; the fixed rib bends with a lever arm of $y - y_c$, the height above the elastic centre, which is much smaller everywhere. A rib that bends less for a given change of thrust has a smaller denominator, and the same shortening costs it more thrust.

For a flat parabola of constant section the two integrals can be done by hand. With $i$ the radius of gyration,

$$\frac{\delta H}{H_0}\bigg|_{\text{two-hinged}} \approx \frac{15}{8}\left(\frac{i}{f}\right)^2, \qquad \frac{\delta H}{H_0}\bigg|_{\text{fixed}} \approx \frac{45}{4}\left(\frac{i}{f}\right)^2$$

and the ratio of the two is exactly six.

Computed along the real parabola at every rise from 2 to 30 per cent of the span, the ratio runs between 5.3 and 7.3, close to the six of the closed forms, which treat the arch as flat and drop the axial term from the denominator. At the rib’s 10 per cent rise the fixed rib loses 0.63 per cent of its thrust and the two-hinged 0.10.

Put the two results together and the fixed rib’s moments follow in two lines. Six times the thrust loss at a third of the lever arm gives twice the crown moment. Six times the loss at two thirds of the lever arm gives four times the two-hinged crown moment at the springings. Fixing the springings is the stiffer construction, and it makes every effect of the rib’s own shortening larger. That is the second refutation, and it is the same statement as the earlier essay’s observation that a deeper rib is more sensitive: in both cases the structure has been made stiffer in bending while its axial stiffness stayed where it was, and the correction is a ratio of the two.

Where the factor of six comes from #

The six is worth taking apart, because it says what fixity does. The shortening on top of the flexibility equation is the rib’s axial flexibility, the same in both arches. Underneath is the rib’s flexibility against a change of thrust: for the two-hinged rib, the thrust bends it about the springing line, and the integral of $y^2$ along a flat parabola is $8f^2L/15$; for the fixed rib it bends it about the elastic centre, and the integral of $(y - y_c)^2$ is $4f^2L/45$. The second is a sixth of the first.

So the fixed rib is six times stiffer against a change of thrust than the two-hinged one. That is exactly what fixity is for — it is why a fixed rib’s buckling load is twice the two-hinged rib’s and why it deflects less under load. And it is exactly why the rib loses six times as much thrust to shortening: a stiffer restraint against a length change is a larger force for the same length change. The movement nobody applied is the general statement, that an imposed deformation produces a force in proportion to the stiffness resisting it; the fixed arch is the case where the stiffness in question was added deliberately and for good reasons, and the force came with it.

The weather, six times over #

Everything that changes the rib’s length enters the same equation over the same denominator. A uniform temperature rise lengthens the rib, pushes the springings apart and raises the thrust; in a fixed arch the rise acts at the elastic centre too, so it hogs the crown and sags the springings — the opposite signs to shortening — with the same factor of about six between the two ribs.

On this rib a 10 degree rise puts 50 kN·m into the fixed springings, 26 into the fixed crown and 13 into the two-hinged crown. A seasonal swing of 30 degrees either side of the temperature at which the arch was closed is therefore 150 kN·m at the springings in each direction, larger than the shortening moment itself. Shrinkage, which is a permanent shortening of a few hundred microstrain, adds to the rib shortening with the same signs and a magnitude comparable with a 30 degree fall. Creep is the subtle one. On a rib that creeps uniformly it multiplies the bending and the axial strains alike, so the ratio that sets the loss is unchanged; what it does relax is the moment from strains imposed on the rib rather than produced by its load — shrinkage and temperature — which it reduces over the years in which they act.

The earlier essay’s historical observation reads more exactly with this in hand. It noted that twentieth-century fixed concrete arches are the ones that crack near their springings, from shortening, shrinkage and temperature together. That is not a combination the two-hinged calculation can predict, because in a two-hinged arch all three put nothing at the springings. It is exactly what the fixed calculation predicts: all three act at the elastic centre, and all three land at the springings at twice their crown value.

The rib that could not have been two-hinged #

There is a second reason the earlier essay’s rib is a fixed arch, and it is more basic. Its first buckling mode is antisymmetric, with a coefficient of 29.1 as a two-hinged rib, which on this section is $29.1 \times 420{,}000/60^3 = 56.6$ kN/m. The rib carries 60. As a two-hinged arch this rib is past its elastic buckling load under its own design load, and would not stand. Fixed, its buckling load is 119 kN/m, and it carries its load at half of it.

At half its buckling load, the fixed rib is in the range where the thrust acting on the rib’s own deflection matters. The load that makes itself worse is the general statement. Here it has a particular shape. Shortening lets the crown drop, and the thrust acting across that drop adds sagging at the crown and relieves the hogging at the springings. Solved to second order, the crown moment rises from 58 to 76 kN·m and the springing moment falls from 112 to 74. The total amount of bending is similar; the second-order solve moves it from the ends towards the middle. As the load approaches the buckling load the crown moment keeps growing and the springing moment turns back towards zero.

That makes the fixed arch’s design moment a second-order quantity in exactly the way the half-loaded rib was. A first-order analysis of the shortening puts the design section at the springings; a second-order one moves some of the moment to the crown. Both sections have to be checked, and the redistribution between them depends on how close the load is to a buckling load that the two-hinged version of the same rib has already passed.

Why the fixed arch is built anyway #

None of this is an argument against fixing the springings. Fixity doubles the rib’s buckling load and removes two hinges that are expensive to build and maintain in concrete, and under a live load on half the span it costs nothing: 10 kN/m on the left half of this rib gives a peak moment of 563 kN·m fixed against 564 two-hinged. The earlier essay’s list of what is actually done — build it and then adjust it, jacking the crown gap closed so the redundants are set rather than computed — is the practical answer, and it is aimed at exactly the effects above: jacking at the crown restores the thrust that shortening, shrinkage and early creep took away, and does so after they have happened.

What the fixed calculation does change is where the effects are looked for. A designer who has taken the two-hinged intuition — secondary moments at the crown, nothing at the ends — into a fixed arch will check the crown for them and not the springings, which is where they are largest and where the live load’s own hogging moment already is. The hinge put in on purpose is the other answer, and it is the only one that removes the effects rather than managing them: a three-hinged arch is determinate, has no compatibility equation, and shortens, expands and shrinks without any moment at all.

By hand #

The whole calculation fits on a page. The radius of gyration is $\sqrt{EI/EA} = \sqrt{420{,}000/20{,}400{,}000} = 0.143$ m, so $(i/f)^2 = 5.7 \times 10^{-4}$. Two-hinged, $15/8$ of that is a loss of 0.107 per cent; fixed, $45/4$ of it is 0.64 per cent. Of the 4,500 kN thrust that is 4.8 and 28.9 kN. The two-hinged crown moment is $4.8 \times 6 = 29$ kN·m. The fixed crown and springing moments are $28.9 \times 2 = 58$ and $28.9 \times 4 = 116$ kN·m. The exact integration along the curved rib gives 28.4 kN, 58 and 112.

Where the model stops #

The section is constant. Real fixed arches are deeper at the springings than at the crown. That moves the elastic centre up, towards the crown, since the stiff springings count for less flexibility, lengthening the springings’ lever arm and shortening the crown’s; the classical choice of a section whose $I$ grows as $\sec\theta$ towards the springings is the one for which the closed forms above are exact.

The abutments are rigid. An abutment that rotates or spreads releases some of the fixity and some of the thrust, and in the limit turns the fixed arch back into a two-hinged one. The earlier essay treated abutment movement as a fourth term in the denominator; for a fixed arch it is three. A tied arch replaces the abutments by a tie, and the tie’s own extension enters the same denominator as one more length change.

The load is uniform and the rib elastic. A live load on part of the span produces bending on its own, which adds to all of the above, and a cracked concrete rib has a lower bending stiffness than the one used here, which lowers the loss and the buckling load together. And a much shallower rib meets the arch’s other instability first: below about a tenth of the span, snap-through rather than the antisymmetric buckle is the limit.

What the pictures cannot show #

That the moments from shortening, shrinkage, creep and temperature are self-stresses. They are in equilibrium with no load at all, they exist because the rib cannot change its length freely, and they vanish if it is allowed to. A crack at the springing is the arch releasing some of that restraint, and it reduces the self-stress it relieves. That is why these moments are treated more leniently than those from load in an ultimate check, and why they matter most for cracking and durability, at exactly the sections where water collects.

Still open: the abutment that moves #

Every number here assumes the springings are held perfectly. A real abutment spreads a little under the thrust and rotates a little under the springing moment, and both release some of what the fixity imposes. For a two-hinged arch an abutment spread is one more length change in the same equation; for a fixed arch the abutment’s rotational stiffness enters as well, and it can move the springing moment anywhere between the fixed value and zero. How stiff an abutment has to be before the arch behaves as fixed, and whether real foundations on real ground are that stiff, is the next question.