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
prestressed 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 prestressing. 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 archbeam relationship.
In reality, we would separate the functions of support and stiffness and
make a deckstiffened 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
twodimensional, 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 twopin 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 threepinned.
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 threepin arch, why not taper it to the centre as well?
Here are pictures of threepinned 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.
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 beamlike forces, and will need to be able to resist
them. For concrete this means reinforcement or prestressing. 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 nonfunicular. 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 Nonfunicular 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 Nonfunicular 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. 