Transforming a beam into an arch - more about funiculars

One of the themes of this site is to emphasise the relationships between different kinds of structures, and to demonstrate the continuity between them.

The example here concerns a Warren truss with nine panels, with a support at each end. The diagrams below show the forces in the top chord and the bottom chord for various angles of inclination of the supports. Red represents compression, with blue for tension. The horizontal white line represents the zero of force.

The parameter on each diagram is the angle between the supports and the vertical. The purple line, dotted in the first example, is the funicular, which is easy to calculate for a beam of constant mass per unit horizontal distance, it being a parabola. In some of the diagrams the funicular goes off the top of the diagram. A beam is simply an arch whose funicular is infinitely high.

Note that when the funicular just touches the top chord (angle = 18.6 degrees), there is still some tension in the bottom chord, and the funicular must be well inside the truss (eg angle = 14.4 degrees) to eliminate all tension. In the case of angle = 12.4 degrees the funicular just reaches the edge of the middle third at the top, but of course it is outside the middle third near the ends. Finally, with angle = 9.3 degrees, the funicular passes through the centre of the truss.

We could imagine the same thing with a concrete beam, creating a crude form of pre-stressed beam.

We could invert the whole thing and call it a suspension bridge with a stiff deck acting as the cable too. We couldn't in this case use concrete, but we could insert wires along the funicular to take the tension. This would be a form of pre-stressing. An even simpler case has the concrete copying the shape of the cable: this is called a stressed ribbon.

We can summarize all this in one diagram, shown below, in which the vertical axis represents the height of the funicular curve, and the horizontal axis represents the angle of the forces at the ends of the structure. The two extremes represent simple beams with vertical supports, piers at the right, and hangers at the left. Hangers correspond to the case of the suspended girders of cantilever bridges such as the Quebec bridge. The regions near the ends of the curve represent beams with inclined supports. The central region of the graph represents arches and suspension spans, albeit rather unusual ones in the examples given earlier.

The method described earlier isn't a good way of making an arch, because the shallow angle implies an immense outward thrust, but it does illustrate the arch-beam relationship. In reality, we would separate the functions of support and stiffness and make a deck-stiffened arch or a normal suspension bridge. Categories such as arch, beam and cantilever are of course very useful, but many real structures, especially those with fully three dimensional complexity, cannot be simply categorized. Many natural forms in the animal and vegetable world are of this type: they evolved for efficiency without knowledge of categories and names. All structures are three dimensional, of course, but in many bridges, the structure is mainly two-dimensional, with a lateral depth in which not much happens except, perhaps, bracing against wind and water, and, in curved bridges, against the thrust caused by the curvature of the path of the traffic on the road or rails.

In addition to the graph shown above, we could also have drawn a curve showing the strength of the horizontal component of thrust at the abutments, which would be outward for an arch, zero for a beam, and inward for a suspension bridge. Here is the graph, showing how the outward thrust would go to infinity if the height of the funicular would reach zero, meaning of course that you cannot stretch a cable exactly horizontal or make a perfectly flat arch.

There is an important point buried in this rather silly example, namely that fact that the sharing of forces can be controlled. Looking at the Bayonne bridge, the Hell Gate bridge, and the Sydney Harbour bridge, we see that the top chords have no abutments, and therefore have little force in them at their extremities. But by suitable dimensioning of the parts of the bridges, the designers could have arranged that around the centre of these bridges, the forces could be shared equally between the top and bottom chords. Furthermore, engineers can employ jacking during construction. Temperature changes, of course, can change the distribution of forces to some extent. This usually annoying result is used deliberately in bimetallic strips. 

The movement of some of the force from bottom chord (at the abutments) to top chord (at the centre) takes place through the ties and struts between them), even though the funicular is not aligned with any of these ties and struts. A similar thought can be applied to load paths. Can you actually draw a line which represents the load path in any structure?

Real arches

Let us now actually look at a practical arch, as opposed to a squeezed Warren truss. From the diagram below, of a two-pin arch, we detect a problem that was glossed over in the previous discussion. There we imagined a truss which was supported by inclined struts, and we were able calculate all the forces. But imagine that the truss had simply rested on piers, and that we had applied the required horizontal forces using jacks. Knowing these forces, we could still have calculated the forces. But if someone had fixed the ends in concrete, we would not subsequently be able to calculate anything, because the temperature might have changed since the fixture. With fixed ends, the structure is statically indeterminate. With jacks, it is not, because we are setting forces rather than positions.

The arch below is statically indeterminate for the same reason. We can escape from this by making a gap in the centre of the top chord, so that all the thrust is in the bottom chord. The arch is now three-pinned.

In order to start calculating the forces, consider one half of the arch.

Definitions -

   S = Span

   R = Rise from abutment to crown of bottom chord

   D = Depth of truss

   W = Weight of arch, equally distributed per horizontal unit of distance

   Tb = Thrust at crown of of bottom chord

   Tt = Thrust at crown of top chord

   Ta = Horizontal component of thrust at abutment

We can see that Ta = Tb + Tt, for horizontal equilibrium.

Currently Tb =0 because we made a gap in the top chord.

Taking moments about the abutment pin,

   R x Tt = 0.25 x S x 0.5 x W, so

   Tt = 0.125 W S / R = Ta.

In the example as drawn, the rise R is one fifth of the span S, and so

   Tt = 0.625 W = Ta

The angle of the total abutment thrust from the horizontal is given by

   tan (Aa) = 0.5 W / Ta = 0.5 W / 0.625 W = 0.8, giving A = 38.7 degrees.

The funicular is coincident with the bottom chord, so the other members come into play only to ensure rigidity against buckling, and to counteract live loads.

Now we imagine placing a jack in the top chord and slowly increasing the force between the two halves, pushing them very slightly apart. If we are pushing the halves of the arch apart, we might expect an increase in the abutment thrust, but no, this is not the case, as we shall see.

Taking moments, as before, about the abutment pin,

   R x Tt + (R + D) x Tb = 0.25 x S x 0.5 x W, so

   R x Tt + (R + D) x Tb = 0.125 W.

Now W doesn't change, but as we increase Tb, the centre of force of Tt and Tb rises, until, when Tb = Tt, it is on the centre line of the arch. That means that the mean line is now at R + 0.5 D above the abutment, and the total thrust must have decreased to balance the constant moment 0.125 W. Therefore, as we increase the jacking force, the thrust at the abutment decreases and becomes steeper.

In the special case Tb = Tt, we have

   R x Tt + (R + D) x Tb = 0.125 S W = Tt (R + R + D), so

   Tt = 0.125 S W / (2 R + D) = Tb

As before, Ta = Tt + Tb = 0.25 W / (2 R + D), or

   Ta = 0.125 W / (R + 0.5 D), as compared with the previous Ta = 0.125 W / R.

In the example as drawn, D = S / 20, so R + 0.5 D = 0.2 S + 0.025 S = 0.225 S.

Then we have

   Tt = 0.125 S W / 0.225 S = 0.555 W.

The angle of the total abutment thrust from the horizontal is given by

   tan (Aa) = 0.5 W / Ta = 0.5 W / 0.555 W = 0.9, giving A = 42 degrees.

Where does the funicular go now? Does it pass halfway between the chords at the crown?

We have not mentioned the fact that real materials are elastic, and will slightly change their dimensions when stressed. In an indeterminate arch, will this tend to equalize the forces in the two chords? If there were no give, would we require absolutely zero inaccuracy in design and construction for indeterminate structures to obtain the desired forces?

Since the arch is pinned at the abutments, why not taper it like the Garabit viaduct? And in a three-pin arch, why not taper it to the centre as well? Here are pictures of three-pinned arches.

The next two diagrams show graphically the behaviour of the funicular and the forces at the centre of the span. The distance along the x axis represents the angle of the abutment thrust from the horizontal, in degrees, from 0 to 90. The value 90 is the case of a simple beam, resting freely on two piers. The upper graph shows the height of the apex of the funicular above the abutments. The purple lines show the angles at which the funicular matches the chords at the crown of the arch.

In the lower graph we see the behaviour of the forces in the two chords at the centre of the span, showing how deviation from the funicular creates greater forces. 

In both graphs, the white lines at left, and the horizontal white lines, are the axes.

Note how each force becomes zero at the crown when the funicular passes through the other chord, and the equality of the forces when the funicular bisects the truss. A beam is represented by the values at angle = 90. Red represents compression, and blue represents tension. When one force is zero, the corresponding chord could do without its central segment, and the arch could become a three pinned one.

From the formulas derived earlier, we can demonstrate that arches can be made much longer than beams, using the arch depicted above.

For the arch, Ta = Tb = 0.125 W / (R + 0.5 D) with central funicular.

For the beam, Ta = -Tb = 0.25 W / D.

Thus Ta(beam) / Ta (arch) = 2 (R + 0.5 D) / D.

In the example we have used, Ta(beam) / Ta(arch) = 2 (0.2 + 0.5 x 0.05) / 0.05, so

Ta(beam) / Ta(arch) = 9. The forces at the centre of the beam are nine times the forces at the centre of the arch, for the same weight of material. That is why arches can be made so much longer than beams. One might expect to be able to make longer beams could be made by making them deeper, but this route leads to greater weight, and possibly to the tied arch.

Jen28.jpg (32831 bytes)  3PinGlos1.jpg (79053 bytes)  3PinM6.jpg (20340 bytes)

Definition of the funicular


How is the funicular defined? Consider a two dimensional object with weight, supported at two points which are not on the same vertical line. At each end the funicular is tangential to the supporting force, and between the ends, the second derivative of the funicular is proportional to the mass per horizontal distance. This leads to a parabola when the mass per horizontal distance is uniform, and to a cosh curve (catenary) when the mass distribution is uniform along the curve.

If the supporting forces are vertical, then the curve is entirely at infinity, and has no practical use.

Furthermore, if we split the structure between the two supports into two parts, we can deduce something about the relative strengths of the forces in those parts. The ratio of the forces is inversely related to the ratio of the distances of the chords from the funicular. Thus, if the funicular bisects the structure, the two two chords share the load equally, while if one chord lies along the funicular, the other chord takes no load. If the funicular is at infinity, the ratio of the distances is minus one, and the chords take equal forces but of opposite sign, which is typical of beams.

While this must not be taken too literally with complicated structures, we can see that an arch which deviates far enough from the funicular will experience large beam-like forces, and will need to be able to resist them. For concrete this means reinforcement or pre-stressing. The extreme of this is the bent or portal frame.

In real life, we cannot always defer, even slightly, to the funicular. Rooms in many cultures, for example, are expected to have horizontal floors and ceilings, and vertical walls - all highly non-funicular.

Having related the beam, arch and cable, we might think about a cantilever. The truss we used above was supported at each end, and has weight through out its length. It could be used as a part of cantilever bridge by supporting it at the centre, and connecting one end to the ground. The free end would be used to support a suspended span. Exactly like the beam, the bending moment at the centre is opposed by opposite forces in the two chords, but they are interchanged as compared with a beam. In the cantilever, the tension is in the top and the compression in the bottom.

Perhaps we could summarize the three structures as follows.

Trussed Arch

More or less funicular

Top chord in compression

Bottom chord in compression

Distributed downward self load

Moving downward live load

Needs upward vertical reaction component at ends.

Needs inward thrust component at ends.


Trussed Beam


Top chord in compression

Bottom chord in tension

Distributed downward self load

Moving downward live load

Needs upward vertical reaction at ends.

No horizontal force at ends


Trussed Cantilever


Top chord in tension

Bottom chord in compression

Distributed downward self load

Moving downward live load

Downward vertical force at free end from suspended span

Needs vertical reaction at fixed end - polarity depends on design and on live load.

No horizontal force at ends

Trussed Suspension System

You can work this out from the case of the arch. The Tower Bridge is perhaps the only well known example. The suspension structures seem to resemble inverted three pin arches.