Cables and arches

Beams carry load by bending, which is inefficient because most of the cross-section works at low stress. Cables and arches do better by changing shape. A cable hangs in whatever shape puts it in pure tension under its loads, and an arch built to the same shape turned upside down is in pure compression. These forms let suspension bridges span more than a kilometre and masonry arches stand for two thousand years. For the operations centre, Priya's team is comparing a cable-supported option and a steel arch option for the pedestrian bridge across the drainage channel, and the comparison needs the methods of this lesson.

Cables: assumptions

A cable is assumed perfectly flexible: it cannot resist bending, shear or compression, only tension along its length. It is also assumed inextensible for finding its shape, although its elastic stretch matters for deflections and for very long spans. Because it has no bending stiffness, a cable under point loads forms straight segments between the loads, and under a distributed load it forms a smooth curve. At every point, the tension acts along the tangent to the cable.

A consequence of these assumptions is that the horizontal component of cable tension, H, is the same everywhere along a cable carrying only vertical loads. Horizontal equilibrium of any piece of the cable requires it. The vertical component varies, and the tension is largest where the cable is steepest, normally at the higher support.

The general cable theorem

For a cable with supports at the same level carrying vertical loads, there is a neat link with beams. Imagine a simply supported beam of the same span carrying the same loads, and let \(M(x)\) be its bending moment. Take moments about a point on the cable at distance x from the left support, for the part of the cable to the left. The vertical reaction is the same as the beam's, and the horizontal tension H acts with a lever arm equal to the sag \(y(x)\) below the chord. Therefore

The cable's sag at every point is proportional to the moment in the equivalent beam. The cable takes the shape of the moment diagram, hung upside down. If you know the sag at one point, H follows at once, and then the sag everywhere else. This is the general cable theorem, and it also holds when the supports are at different levels if the sag is measured from the chord joining them.

Cables under uniform load

A suspension bridge cable carries a deck whose weight is roughly uniform per unit horizontal length. The equivalent beam's moment is the parabola \(M = wx(L - x)/2\), so the cable is a parabola too. With maximum sag h at midspan, the cable theorem gives

Here the maximum tension occurs at the supports, where the vertical component is half the total load. The equation for H shows the key design trade-off: halving the sag doubles the horizontal tension, and with it the force on the anchorages. Shallow cables are elegant but expensive to anchor, which is why suspension bridges typically use sag-to-span ratios of about 1:8 to 1:12.

A cable hanging under its own weight, which is uniform along its length rather than along the horizontal, takes the shape of a catenary. For sag ratios below about 1:10 the difference from a parabola is small, and the parabolic formulas are used for design. The cable's length for a shallow parabola is approximately \(L[1 + \tfrac{8}{3}(h/L)^2]\), which you need when ordering the cable and checking its stretch.

Cables have one more weakness: they change shape when the load pattern changes. A cable shaped for uniform load will sag noticeably under a point load. Suspension bridges control this with stiffening girders or trusses at deck level, which share moving loads and distribute them to many hangers.

Arches

Turn the cable upside down and make it rigid, and you have an arch. An arch shaped to match the funicular (the inverted cable shape) for its load carries that load in pure compression, with no bending at all. A parabolic arch is funicular for a load uniform along the span; a catenary arch for its own weight; and a polygonal arch for point loads. Because the arch pushes outward at its supports, it needs abutments or a tie to resist the horizontal thrust. A tied arch, with a tension tie between its feet, delivers only vertical reactions to its supports, which is convenient on poor ground.

Arches are classified by their hinges. A three-hinged arch, with hinges at both supports and the crown, is statically determinate and insensitive to settlement and temperature. A two-hinged arch, with hinges only at the supports, is indeterminate to the first degree. A fixed arch is indeterminate to the third degree but stiffer. Most steel and concrete highway arches are two-hinged or fixed, while three-hinged forms appear in timber and steel buildings.

Three-hinged arches under non-funicular load

Real arches carry moving loads, so they are never funicular for every load case. Under a load that does not match their shape, arches carry bending as well as compression. For a three-hinged arch, find the reactions as for the three-hinged frame in lesson 3, then compute the moment at any section from the forces on one side. The moment at a point on the arch at height \(y\) above the springing line is the equivalent simple beam's moment minus the thrust times the height:

This is the cable theorem in another form. Where the arch follows the moment diagram, the two terms cancel and the arch is in pure compression. Where they differ, the difference is the bending the arch must carry.

The thrust line

Engineers since Robert Hooke in the seventeenth century have used the idea of a thrust line: the path of the resultant compressive force through an arch. For a masonry arch, which cannot carry tension, the arch is safe as long as some thrust line for the loads can be drawn within the masonry, ideally within its middle third so that no joint opens. For a steel or concrete arch, the distance between the thrust line and the arch axis is the eccentricity of the compression, and the moment is the thrust times that eccentricity, the same result as the formula above. Choosing an arch shape close to the thrust line for the dominant load keeps bending small and the arch slender.

Computing a cable's shape

Sofia's program uses the cable theorem to find the shape of a cable under any set of vertical point loads once one sag is specified, and the tension in every segment. Changing the input to a single heavy point load shows how much a cable changes shape under non-uniform loads.

import math
L = 12.0
loads = [(4.0, 20.0), (8.0, 30.0)]          # (x from A, load down) in m, kN
x_spec, y_spec = 8.0, 2.0                    # specified sag at one point

RB = sum(P * x for x, P in loads) / L; RA = sum(P for _, P in loads) - RB
def M(x):                                    # equivalent simple beam moment
    return RA * x - sum(P * (x - xp) for xp, P in loads if xp < x)
H = M(x_spec) / y_spec
pts = [0.0] + [x for x, _ in loads] + [L]
print(f"H = {H:.2f} kN")
V = RA
for a, b in zip(pts[:-1], pts[1:]):
    ya, yb = M(a) / H, M(b) / H
    T = math.hypot(H, V)
    print(f"segment {a:4.1f}-{b:4.1f} m: sag {ya:4.2f} to {yb:4.2f} m, tension {T:6.2f} kN")
    V -= dict(loads).get(b, 0.0)

H = 53.33 kN
segment  0.0- 4.0 m: sag 0.00 to 1.75 m, tension  58.21 kN
segment  4.0- 8.0 m: sag 1.75 to 2.00 m, tension  53.44 kN
segment  8.0-12.0 m: sag 2.00 to 0.00 m, tension  59.63 kN

The output reproduces the hand calculation and shows the tension in each segment. The segment next to B is the steepest and carries the most tension, as predicted.

Choosing between the options

Owen summarizes the comparison for Priya. The cable option is light and slender, but it needs two heavy anchorages for the horizontal pull, and its deck needs stiffening so that a group of people walking to one side does not change the cable shape noticeably. Pedestrian bridges are also sensitive to vibration, and a light, flexible cable structure can respond to the rhythm of footsteps; that check belongs to a later dynamics course but it influences the choice now. The arch option needs abutments for its thrust, or a tie, but carries the deck from below, keeps the structure stiff under moving crowds, and fits the shallow channel banks. Both options carry their dominant load mainly by axial force, which is why either one is lighter than a plain girder of the same span. The team will carry both forward to cost estimates, and the decision will likely turn on the foundations rather than the superstructure, as it so often does.