The centre that hangs in the air

A single-bay portal frame drawn as a column section has an elastic centre inside it, in mid-air between the beam and the feet, and the horizontal thrust a gravity load produces is that section’s bending stress about the axis through that point. The construction needed three things from the frame: the area of its centreline drawn at a width of $1/EI$, the position of that area’s centroid, and its second moment about the centroid.

A frame of three straight members makes those three quantities easy to write down. It does not make them easier to be. An area is an integral, a centroid is an integral, and a second moment is an integral, and an integral does not know whether the path it runs along is straight. Curve the frame and every step survives.

A fixed arch is the case where that matters most, because a fixed arch is three times redundant and there is no way to reduce it to a beam.

The section is the arch’s own centreline #

Take the arch’s centreline and draw it as a band of width $ds/EI$ at every point, narrow where the arch is stiff and wide where it is not. That band is a section. It has an area,

$$A = \int \frac{ds}{EI},$$

a centroid at $(\bar x, \bar y)$, and second moments $I_x$ and $I_y$ about axes through that centroid. For a symmetric arch the product of inertia is zero and the two axes are the vertical and the horizontal, which is why the construction stays simple.

The centroid is the elastic centre, and for an arch it is nowhere on the structure. On a portal frame it hung inside the rectangle; here it sits on the axis of symmetry, above the springings and below the crown, in the air under the arch. That is not a curiosity — it is the whole mechanism. Applying the three redundants at that point is what makes the three compatibility equations uncouple, so that instead of a three-by-three system there are three divisions.

Release the arch by cutting it at one springing. What is left is a curved cantilever fixed at the other, carrying no thrust whatever, and its moment at any station is the moment of the loads to one side — which for vertical loads on any shape of member is the moment a straight cantilever of the same span would have carried, because a vertical force has a horizontal lever arm whichever way the member runs. The arch’s whole difference from a beam is the thrust, and the release is the operation that removes it.

Load that released diagram $M_s$ onto the analogous column as a distributed load of intensity $M_s$ per unit of section area, and the three “stresses” are

$$M_0 = -\frac{\int M_s \,ds/EI}{A}, \qquad H = -\frac{\int M_s (y - \bar y)\,ds/EI}{I_x}, \qquad V = -\frac{\int M_s (x - \bar x)\,ds/EI}{I_y},$$

which is $P/A$ and the two bending stresses of a section under an eccentric load. The arch’s moment at a station is then

$$M = M_s + M_0 + H(y - \bar y) + V(x - \bar x).$$

$M_0$ is the redundant couple, $H$ is the horizontal thrust and $V$ the redundant vertical shear at the cut. The thrust of an arch is the gradient of a bending stress on a section that is a drawing of the arch, and for the arch above it comes out at 375.0 kN.

Two thirds of the rise, exactly #

The flat line is worth stopping on, because it is the one exact result in the whole construction and it comes from an assumption made for a different reason.

An arch built to a constant depth normal to its axis has a horizontal projection that thins toward the springings, and the classical simplification is to assume instead that $I = I_0 \sec\theta$, where $\theta$ is the slope and $I_0$ is the value at the crown — a section whose depth grows exactly as the arch steepens. That assumption was made because it makes $ds/EI = \sec\theta\,dx / (EI_0 \sec\theta) = dx/EI_0$, so every integral along the arc becomes an integral along the span and can be done by hand. With it, for a parabola $y = 4f x(L-x)/L^2$,

$$\bar y = \frac{1}{L}\int_0^L \frac{4f\,x(L-x)}{L^2}\,dx = \frac{4f}{L^3}\cdot\frac{L^3}{6} = \frac{2f}{3},$$

independent of the span, of the rise, and of $EI_0$. Two thirds of the rise, always.

That number is worth carrying because it is the one an arch designer can use without computing anything. It also marks the size of the assumption: with $EI$ genuinely constant along the arc the centre drops to 0.643 of the rise at a rise-to-span of 0.2 and 0.605 at 0.4, and the difference feeds straight into the thrust, because the thrust is a stress computed about that axis.

The diagram that is nothing #

A parabolic arch under a load uniform over its span is the funicular shape for that load, so it carries the load in pure compression and its bending moment is zero at every station. The analogy returns that: $M_s + M_0 + H(y-\bar y)$ cancels to zero identically, not approximately, and not at selected points.

That is the strongest check the method has, and it is a better one than agreeing with another calculation. A method that returns the right nonzero number might be right, or might be wrong in two compensating ways; a method that returns an identically zero function over a continuum is being asked to cancel three separate integrals against a fourth at every station, and cannot do it by accident. The construction passes.

The circle is the interesting half of the figure. Its centreline differs from the parabola’s by at most 240 mm over a 30 metre span — a difference no one would see on a drawing and a difference within the tolerance most arches are built to. It carries 80.4 kN·m at its springings, which is 3.6 per cent of the free moment. Three and a half per cent of a free moment is not a large number for an arch and it is an infinitely large number compared with zero, and the whole of it comes from 240 mm of shape.

What the thrust line does with a section that has a depth #

The moment diagram and the thrust line are the same information twice, related by one division: the thrust line sits $e = M/H$ from the centreline, so a diagram of moments becomes a picture of where the resultant passes. Drawing it that way turns the arithmetic into a question a mason would recognise.

It also decides which magnification is honest. The excursion here is 0.21 m against a rise of 6 — a pencil width at any scale that fits the arch on a page — so an exaggerated drawing is the usual answer. The exaggeration destroys the only question the picture is for. Given a section 0.75 m deep, 0.21 m of eccentricity is 168 per cent of the way to the edge of the middle third, which means one face of the section is in tension at the springings — and for a masonry arch, which has no tensile strength to spend, that is where it stops being an arch and starts being a mechanism with a hinge in it.

A point load is what an arch is bad at, and the figure says why in a form the moment diagram hides. The load breaks the symmetry, so the analogous column is now bent about both axes: $V$, which vanished by symmetry under the uniform load, is the second bending stress and it tilts the whole thrust line. The line leaves the section altogether, which for a concrete or steel arch means a large bending moment and for a masonry one means the arch has become a four-bar mechanism.

The whole of it, once, by hand #

For the parabolic arch with $I = I_0\sec\theta$, $L = 30$ m, $f = 6$ m and $w = 20$ kN/m, and with $EI_0$ taken as one because it cancels:

The analogous column has $ds/EI = dx$, so its area is simply the span, $A = 30$, and its centroid is at $\bar x = 15$ m, $\bar y = 2f/3 = 4$ m. Its two second moments are

$$I_x = \int_0^L (y - \bar y)^2 dx = \frac{4f^2 L}{45} = 96, \qquad I_y = \int_0^L (x - \bar x)^2 dx = \frac{L^3}{12} = 2{,}250 .$$

Release at the left springing, so the arch is a curved cantilever fixed at the right one and $M_s(x) = -wx^2/2$. Three integrals:

$$\int M_s\,dx = -\frac{wL^3}{6} = -90{,}000, \quad \int M_s(y-\bar y)\,dx = \frac{wfL^3}{90} = 36{,}000, \quad \int M_s(x-\bar x)\,dx = -\frac{wL^4}{24} = -675{,}000 .$$

Three divisions:

$$M_0 = \frac{90{,}000}{30} = 3{,}000\ \text{kN·m}, \qquad H = -\frac{36{,}000}{96} = -375\ \text{kN}, \qquad V = \frac{675{,}000}{2{,}250} = 300\ \text{kN}.$$

$H$ is the horizontal thrust, 375 kN, and the closed form for a parabolic arch under a load uniform over its span is $H = wL^2/8f = 20 \times 900 / 48 = 375$. The two routes share no step after the shape of the arch. $V$ is 300 kN, which is $wL/2$ — the vertical reaction, arrived at as the second bending stress on a section rather than by summing forces.

Now read the moment anywhere. At the crown,

$$M = M_s(15) + M_0 + H(f - \bar y) + V(15 - \bar x) = -2{,}250 + 3{,}000 - 750 + 0 = 0,$$

and at the left springing,

$$M = 0 + 3{,}000 + (-375)(0 - 4) + 300(0 - 15) = 3{,}000 + 1{,}500 - 4{,}500 = 0 .$$

Four numbers of order a thousand cancelling to nothing, twice, at two different stations, and at every station between them. That cancellation is the check, and it is the reason to do the arithmetic once by hand rather than only reading it off a curve: a construction that produces zero this way is producing it from the structure of the problem, and a construction that produced a small number instead would be one with a sign in it.

The third stress resultant is the one to watch. Under a symmetric load $V$ comes out at $wL/2$ and contributes nothing at the crown, so it is easy to treat as bookkeeping. Under the point load at the quarter point it is doing real work: it is the term that tilts the whole thrust line, and it is the term that vanishes on every symmetric example a method is usually demonstrated with.

The analogy against a stiffness solution #

Two implementations is the only check of a sign, and an arch is where that matters most: the thrust is the answer, the thrust is a sign as much as a magnitude, and a thrust of the wrong sign draws a perfectly plausible arch.

The residue of 0.03 per cent is not the analogy’s error. It is the check’s: the stiffness model lumps the distributed load at its own nodes, which is exact for the total load and reactions and loses the local sagging between them, of order $w(\Delta x)^2/8$. At 48 segments that is 1 kN·m, which is what the comparison reports.

One thing about that check was wrong before it was right, and it is the kind of error worth recording. The frame solution the arch was checked against takes a load transverse to a member. A vertical load on a sloping member has a component along it as well, and dropping that component quietly scales the total applied load by $\cos^2\theta$ — which on a 6 m rise came back as a thrust of 3,161 kN against the 375 the arch actually carries. Nothing warned; the model was well formed, the solution converged, and the number was eight times too large. Lumping at the nodes fixes it because a nodal load has no direction to lose.

Which free body produced the number #

The free body is the whole arch, cut at the left springing, with the three components of the cut’s action — a horizontal force, a vertical force and a couple — carried on a rigid arm to the elastic centre and applied there. The three compatibility conditions are that the cut does not open: no relative horizontal displacement, no relative vertical displacement, no relative rotation. Applying the redundants at the elastic centre is what makes those three conditions involve one redundant each instead of all three, and it is the only reason the answer is three divisions rather than a matrix.

The check is the stiffness solution of the same arch with no release at all and no elastic centre in it, and the agreement is to 0.03 per cent of the free moment.

What it replaced, and what replaced it #

The column analogy is Hardy Cross’s, published in 1930, five years before the moment distribution he is better remembered for. Both had the same purpose and the same constraint: an engineer with a slide rule, a drawing board and no matrix algebra, who has to solve a structure that is three times redundant.

The alternative in 1930 was the force method — release three redundants, compute nine flexibility coefficients by virtual work, invert a three-by-three matrix by hand, and hope. The analogy replaces that with an area, a centroid, two second moments and three divisions, and it does so by a change of variable rather than by an approximation: applying the redundants at the elastic centre is a choice of where to put them, and a choice that makes the off-diagonal flexibility terms vanish is a choice that diagonalises the matrix before it is written down.

What replaced it is a stiffness solver, which is what the check on this page is, and which does not care where the elastic centre is or whether one exists. The analogy survives for two reasons that are not nostalgia. It gives the thrust as a formula in the arch’s own geometry rather than as an output, so the effect of raising the rise or stiffening the haunches can be read rather than re-run. And it puts the answer in a language — a section, a centroid, a bending stress — that a designer already has intuitions in, which is why $2f/3$ is a number worth carrying and a stiffness matrix’s inverse is not.

What a small change of shape is worth #

The 240 mm between the circle and the parabola is worth returning to, because it is the most useful number on the page for anybody who builds arches rather than analyses them.

Over a 30 m span the two centrelines are never more than 240 mm apart, which is under one per cent of the span and comparable with the setting-out tolerance of a large concrete arch. The parabola carries the uniform load with no bending at all. The circle carries 80.4 kN·m at its springings — and 80.4 kN·m on a 0.75 m section is an eccentricity of 0.21 m, which puts the thrust outside the middle third.

So the sensitivity is not small, and it runs the way that matters: the funicular shape is a point, not a region. Any departure from it produces bending in proportion, and the constant of proportionality is the thrust rather than anything about the shape — 375 kN acting through 0.21 m. That is why a shape that carries itself is a statement about one load case and why an arch built for its dead load is bending under everything else.

It is also why a real arch is built to a shape nobody’s formula names. The funicular of the dead load alone is one curve; of dead plus half the live load, another; the shape chosen is a compromise that is funicular for none of them and is within a few hundred millimetres of all of them, which is exactly the tolerance this figure prices.

What the picture cannot show #

The arch getting shorter. Every integral here is a bending flexibility. The arch’s own axial shortening under 375 kN of thrust relieves some of that thrust, and the rib shortening is a separate calculation with its own sign. On a shallow arch it is not small.

Anything about buckling. The thrust is computed from an undeformed shape, and an arch under 375 kN of compression can lean out of that shape rather than squash along it.

The springings moving. Both feet are assumed to be where they were drawn. An arch is the structure most sensitive to that assumption, because its thrust is what an unyielding abutment is for, and a foundation that spreads by a few millimetres sheds thrust in proportion.

Whether the section is real. The middle-third reading in the thrust-line figures assumes a section of a stated depth with a linear stress distribution and no tension. For a concrete or steel arch none of that is the governing check; for a masonry one it is the only check.

The assumption that decides the number #

That $I$ varies as the secant of the slope. It is not a property of any arch anybody builds — it is a fiction chosen because it turns $ds/EI$ into $dx/EI_0$ and makes every integral elementary, and it is stated in every textbook treatment as though it were a description.

Its effect is measurable rather than rhetorical. With it the elastic centre is at exactly two thirds of the rise; without it, at a rise-to-span of 0.2, the centre is at 0.643 of the rise and the section’s $I_x$ changes with it. A designer who takes $2f/3$ from a handbook and applies it to an arch of constant depth has moved the axis the thrust is computed about by about 6 per cent of the rise, and the thrust is a stress about that axis.

The parabola-under-uniform-load check passes under both assumptions, which is worth knowing before it is trusted too far: a diagram of zero is insensitive to where the elastic centre is, because a funicular arch has no moment to distribute.

Still open: the frame that is not symmetric #

Every construction on this page has leaned on a symmetry. A symmetric arch’s analogous column has its principal axes horizontal and vertical, its product of inertia is zero, and the two bending stresses are computed about those two axes independently. That is why $H$ and $V$ came out of two separate divisions.

An unsymmetric structure has no such axes handed to it. A portal with one column taller than the other, or an arch springing from two levels, has an analogous column with $I_{xy} \ne 0$ — and a section with a product of inertia does not bend about the axis it is loaded about. The elastic centre is then not enough: the axes have to be rotated to the principal ones before the stresses can be read, which is the same construction an angle in bending needs and is the question after this one.