Structural Engineering & Statics
where the load goes
Structural form

The hinge put in on purpose

An arch with two pinned feet cannot be solved by statics. Add a third hinge at the crown — deliberately weakening it — and the whole structure falls out of one moment equation.
15 min read 6 figures Drawing as calculationOne support too many

An arch on two pinned feet has four unknown reaction components — a horizontal and a vertical at each springing — and three equations. It is one degree redundant, and no amount of care with free bodies will produce its thrust.

The remedy adopted for a century of long-span arches is to put a hinge at the crown. That makes the structure weaker in the obvious sense — it can no longer carry moment at its highest point — and it makes it solvable, insensitive to settlement, and indifferent to temperature. Three benefits from deliberately removing something.

A three-pinned arch, rise 2.6 on span 9A three-pinned arch under a uniform load. One moment equation about the crown hinge gives a horizontal thrust of 23.37, with no stiffness and no assumption about the section. The thrust line lands on the axis everywhere, so there is no bending anywhere in the arch.crown hinge — no moment here, by constructionH = 23.4H = 23.427.027.0thrust line and axis coincide — the definition of funicular
Fig. 1 A three-pinned arch under a uniform load. The crown hinge carries no moment by construction, so taking moments there on half the arch gives the horizontal thrust in one line, with no stiffness and no assumption about the section.

The hinge is an equation

Counting explains the whole arrangement. Four unknowns and three equations leave one short; a hinge supplies exactly one more.

A hinge is a point that cannot transmit moment. So the statement “the bending moment at the crown is zero” is an additional equation about the structure’s internal forces, available without knowing anything about stiffness. Four unknowns, four equations, solved.

That is a general device rather than an arch trick. Any internal release — a hinge in a member, a slotted connection that cannot carry axial force, a bearing that cannot carry moment — adds an equation and reduces the degree of redundancy by one. The counting rule has to be told about them, and a count that ignores releases will call a determinate structure redundant.

The arithmetic is worth doing in the other direction as well. A fixed-base arch has six reaction components and three equations: three times redundant. A two-pinned arch has four: once redundant. A three-pinned arch has four unknowns and four equations: determinate. Each hinge added takes the structure one step down that ladder, and the last step is the one that changes what body of theory the problem belongs to.

One equation, and the thrust

With the crown hinge in place the calculation is short enough to write out.

Take the whole arch and sum vertically: the two vertical reactions carry the total load, and for a symmetric arrangement they are half each. Then take half the arch as a free body, cut at the crown, and sum moments about the crown hinge. On that half act the vertical reaction, the horizontal thrust, and the portion of the load the half carries. The crown carries no moment, so

H×f=RL2−(moments of the loads on that half about the crown),H \times f = R\frac{L}{2} - \left(\text{moments of the loads on that half about the crown}\right),

with ff the rise. For a uniform load ww this reduces to

H=wL28f,H = \frac{wL^2}{8f},

which is the same expression as a cable’s thrust with the sag in place of the rise, and for the same reason: the right-hand side is the equivalent beam’s mid-span moment, and the thrust is that moment divided by the rise.

The reciprocal in it is the whole economics of arch design. Halving the rise doubles the thrust. A shallow arch is elegant, occupies little headroom and pushes twice as hard on its abutments; a deep one is the reverse. The choice is nearly always settled by what the ground will take rather than by anything structural.

The thrust line, and where the shape stops being right

Once HH is known, the bending moment anywhere in the arch follows immediately, and it has a geometric reading.

The thrust line is the locus of the compressive resultant, and for a three-pinned arch it is exactly the equivalent beam’s moment diagram divided by HH:

ythrust(x)=Mbeam(x)H.y_{\text{thrust}}(x) = \frac{M_{\text{beam}}(x)}{H}.

Where the thrust line coincides with the arch’s own axis, the compression passes straight down the middle of the section and there is no bending. Where it does not, the moment in the arch at that station is the thrust times the gap:

March(x)=H×(ythrust(x)−yaxis(x)).M_{\text{arch}}(x) = H \times \left(y_{\text{thrust}}(x) - y_{\text{axis}}(x)\right).

For a uniform load and a parabolic arch the two coincide everywhere, and the arch carries pure compression from springing to springing. That is not a coincidence — the parabola is the funicular shape for a uniform load, so an arch drawn as a parabola has been drawn as its own thrust line.

A three-pinned arch, rise 2.6 on span 9A three-pinned arch under a uniform load with a point load added. One moment equation about the crown hinge gives a horizontal thrust of 42.60, with no stiffness and no assumption about the section. The thrust line has left the axis by up to 0.74 near x = 2.5, which is a bending moment of 31.6 in a shape chosen to have none.40crown hinge — no moment here, by constructionH = 42.6H = 42.655.938.1thrust line off the axis by 0.74 — a moment of 31.6
Fig. 2 The same arch with a point load added. The thrust line has left the axis, and the gap between the two — multiplied by the thrust — is a bending moment in a shape chosen to have none. A single lorry does this to an arch bridge.

The figure makes the central limitation of arch action visible in one picture. The shape was chosen for one load case. Any other load case moves the thrust line, and the arch pays in bending for the difference. The deeper the arch and the heavier it is relative to the variable load, the smaller the movement — which is why a masonry arch bridge is buried under a deep fill, and why the fill is structural.

What the hinge buys

The obvious cost of a hinge is capacity, and the benefits are all of a kind that never appears in a strength calculation.

Insensitivity to settlement. If one springing of a two-pinned arch sinks a few millimetres, the arch is forced to change shape while its ends are held, and enormous forces develop — the structure is redundant, so it fights the movement rather than accommodating it. A three-pinned arch simply rotates a little about its three hinges and carries on, with no change in its internal forces whatsoever. For a bridge on anything other than rock this is decisive.

Insensitivity to temperature. The same argument. A restrained arch that warms tries to lengthen against its abutments and develops thrust it was not designed for; the three-pinned version lifts its crown slightly and develops none. Thermal forces in indeterminate arches are large enough that they routinely govern.

Erectability. Two halves can be built out from the springings and joined at the crown, with the hinge as the closure detail. Nothing has to be jacked into position or fitted to a tolerance, because the last connection is one that cannot transmit moment and therefore cannot lock anything in.

Analysability. Historically the largest benefit and now the smallest. Before moment distribution and computers, a determinate arch was a calculation and a redundant one was a research project.

Counting unknowns against equationsThree frames differing by one member. Two equilibrium equations per joint, one unknown per member and one per restraint: fewer unknowns than equations is a mechanism, equal is solvable by statics, more needs stiffness.m 4 + r 3 − 2j 8 = -1a mechanismm 5 + r 3 − 2j 8 = 0statically determinatem 6 + r 3 − 2j 8 = +1one member too manystatics can answer only the middle case
Fig. 3 The counting rule that decides which case a structure belongs to. A hinge moves an arch one step left along this row, from a structure needing stiffness to one statics can finish.

What it costs

The hinge is not free, and the costs are the reason two-pinned and fixed arches continue to be built.

A larger crown moment nearby. The moment is zero at the hinge and not zero either side of it, and removing the restraint at the crown pushes moment outward into the haunches. A fixed arch distributes its bending more evenly and is therefore lighter in material for the same span.

More deflection. Three hinges make a more flexible structure than two, and a fixed arch is stiffer than either. For a bridge carrying a railway, where deflection under a moving load matters a great deal, the flexibility is a real objection.

A detail that has to work. A structural hinge is a physical component — a pin, a rocker bearing, a narrow throat of concrete — that must rotate freely for a century and carry the full thrust while doing so. It is a maintenance item in a structure that otherwise has none, and a hinge that has corroded solid has quietly turned the arch back into a redundant one, with all the temperature and settlement sensitivity that implies and none of the analysis that would have been done had it been designed that way.

Robustness. Determinate means no alternative path. A three-pinned arch that loses a hinge is a mechanism, and the count that made it solvable is the same count that leaves it no spare.

The trade is the same one that runs through the whole subject, and this is an unusually clean instance of it: determinacy buys analysability and insensitivity, redundancy buys efficiency and survival, and the choice depends on which risks are real for the structure at hand. Nineteenth-century bridge engineers chose determinacy overwhelmingly, on ground they did not trust, with analysis they could not afford to make harder.

The same answer, drawn

There is a graphical route to the thrust that needs no algebra at all, and it is worth having because it shows why the crown hinge is doing what it does.

Take half the arch as a free body. Exactly three forces act on it: the load carried by that half, with a known line of action through its resultant; the force transmitted through the crown hinge, which can be in any direction but has no moment; and the reaction at the springing. Three forces on a body in equilibrium must be concurrent — their lines of action meet at a point.

Two of the three are known enough to place. The load’s line is vertical through the resultant of that half. For a symmetric arch under symmetric load the crown force is horizontal, so its line is the horizontal through the crown hinge. They cross at one point, and the springing reaction has to aim at it.

Three forces must meet at a pointA body held by two supports and one load. The reaction at the roller is vertical and the load's direction is given, so their lines of action fix a meeting point — and the pin reaction has to point at it.the loadroller: vertical onlypin: any directionall three lines meet here
Fig. 4 Three forces on a body, with two lines of action known and the third obtained by intersecting them. Half a three-pinned arch is exactly this problem: the load, the crown force, and the springing reaction that must point at where the first two meet.

Drawing the force triangle then gives both components to scale in one construction. The whole analysis of a three-pinned arch — thrust, vertical reaction, and the inclination of the resultant at each foot — is two lines and a triangle, and it was done that way for eighty years.

The construction also makes visible what the hinge contributed. Without it, the crown transmits a moment as well as a force, the crown action is no longer a single force through a known point, the three-force theorem has nothing to work with, and the drawing cannot begin. The hinge is precisely the device that turns half an arch into a three-force body.

What the abutment is asked for

The springing reaction is not vertical, and its inclination is the whole difficulty of arch construction.

The resultant at each foot has a vertical component equal to half the load and a horizontal component equal to the thrust, so it leans inward at an angle set by the ratio of the two. For an arch with a rise of a quarter of its span under uniform load, the thrust is half the total load and the reaction leans at about forty-five degrees from vertical.

That angle has to be within what the foundation can supply, and the three ways of failing to supply it are all distinct. The ground can be unable to resist the horizontal component and the abutment slides. It can resist the force but not the overturning moment the inclined resultant produces about the toe of the abutment, and the abutment rotates. Or the pressure under one edge of the base can exceed what the soil takes, and the abutment settles unevenly — which, on an arch, immediately flattens the geometry and raises the thrust.

A closed force polygonThe forces on a joint, laid tip to tail. Equilibrium is the statement that the polygon closes, and the gap when it does not is the out-of-balance force, to scale.thrust 55half the load 60reaction 81.4starts and ends here
Fig. 5 The three forces at one springing, laid tip to tail. The reaction that closes the triangle is the one the abutment has to deliver — inclined, large, and permanent — and its direction is fixed by the arch’s proportions rather than by anything the foundation designer chooses.

The historical answer to all three is mass. A masonry abutment is enormous because its own weight steepens the resultant: adding vertical force to an inclined one turns it toward the vertical and brings it back inside the base. That is why the piers of an arch bridge are so much larger than the loads suggest, and why the outermost arch of a viaduct — which has no neighbour to balance its thrust — sits on a pier several times the size of the internal ones.

The mechanism it fails by

An arch does not usually crush. It hinges, and the counting rule says how many hinges it takes.

A three-pinned arch has three. One more — anywhere — makes four, and four hinges in a plane arch is a mechanism: the two halves and the two segments either side of the new hinge form a four-bar linkage, and the structure moves. So a three-pinned arch is one hinge from collapse by construction, and a two-pinned arch is two, and a fixed arch is three.

The new hinge forms wherever the thrust line reaches the edge of the section. In masonry that is where the joint opens, since masonry cannot carry tension and a resultant outside the middle third would require it. In steel or concrete it is where the section yields. Either way the failure sequence is visible in advance if the thrust line is drawn: it walks toward one face under increasing load, touches it, forms a hinge, and the structure then has one fewer degree of redundancy and a thrust line free to move somewhere new.

A three-pinned arch, rise 1.5 on span 9A three-pinned arch under a uniform load with a point load added. One moment equation about the crown hinge gives a horizontal thrust of 70.50, with no stiffness and no assumption about the section. The thrust line has left the axis by up to 0.28 near x = 3.0, which is a bending moment of 19.8 in a shape chosen to have none.30crown hinge — no moment here, by constructionH = 70.5H = 70.547.037.0thrust line off the axis by 0.28 — a moment of 19.8
Fig. 6 A shallower arch under an asymmetric load. The thrust is far higher because the rise is smaller, and the thrust line’s excursion from the axis is correspondingly more expensive — the same gap, multiplied by a larger number.

That sequence is why a masonry arch gives warning. Cracks appear where hinges are forming, and a crack in an old arch is usually a record of a successful adjustment rather than a symptom of failure. The dangerous condition is not the presence of cracks; it is having enough of them, in the wrong places, to complete a mechanism.

Where the model stops

Small deflections. All of the above is written on the undeformed shape. A shallow arch under load flattens, which raises the thrust, which flattens it further — a load making itself worse, and for very shallow arches it culminates in snap-through, an instability with no counterpart in beams.

Rigid abutments. The thrust is assumed to be resisted without movement. Abutments spread, and an arch whose springings have moved apart is a flatter arch with a higher thrust than the one that was designed.

Axial force only. The essay treats the arch as carrying compression along its axis. A real section carries shear too, and near the springings of a steep arch the shear on a vertical plane is substantial.

Buckling. An arch is a compression member curved in its own plane and it can buckle both in that plane and out of it. Nothing in an equilibrium calculation contains that, and for a slender steel arch it is often the governing check.

One plane. Everything here is two-dimensional. A real arch bridge is a pair of arches with a deck between them, and the out-of-plane behaviour — where the thrust of one arch is resisted partly through the deck — is a different problem.

The figures share a distortion worth naming. The thrust line is drawn as a line, which suggests the compression travels along a wire inside the arch. It does not: it is the centroid of a distribution of compressive stress over the whole section, and a thrust line touching the edge of the arch does not mean a force has arrived at the surface — it means the stress distribution has become triangular with its zero at the far face. The line is a summary of a distribution, exactly as a shear force is, and the same caution applies.

The ladder from here

Later rungs on this anchor: two-pinned and fixed arches, and the compatibility that resolves them. Arch thrust under asymmetric load. Temperature and rib shortening. Snap-through and in-plane buckling. Tied arches. Masonry arch assessment and the geometrical factor of safety. The line of thrust in a buttressed wall. Arch bridges in cast iron, wrought iron, steel and concrete, and how the material changed the shape. And the transition to the shell, where the same argument becomes two-dimensional and much harder.

The three-pinned arch was the standard long-span form in Europe from the 1860s. The Garabit viaduct of 1884 and the Maria Pia bridge of 1877, both by Eiffel’s office, are three-pinned crescents chosen for exactly the reasons above: unreliable ground, no confidence in an indeterminate analysis, and two halves that had to be built outward and met in the air.

Where this goes next

cable: pure tensionreflected herearch: pure compression The shape that carries itself, and the arch that is its reflection Hang a chain and it takes the one shape that carries its load in pure tension. Turn the shape upside down and it carries the same load in pure compression. That is what an arch is. m 4 + r 3 − 2j 8 = -1a mechanismm 5 + r 3 − 2j 8 = 0statically determinatem 6 + r 3 − 2j 8 = +1one member too manystatics can answer only the middle case Counting the unknowns, and finding out whether statics can answer Two equations per joint, one unknown per member, one per restraint. Subtract, and the sign of the answer says whether the structure is a mechanism, solvable, or beyond what equilibrium alone can settle. simply supportedstatics alonesag 32.0propped at one endneeds stiffnesssag 18.0hog 32.0built in at both endsneeds stiffnesssag 10.7hog 21.3the load never changes; only what is holding the endsthe built-in case peaks at two-thirds of the simple span's moment One support too many, and what it costs to know Add a redundant restraint and the load has two routes to the ground. Equilibrium cannot say how it splits, and the answer turns out to depend on stiffness — which is a different kind of question.