Suspension Bridges

How Bridges Work

Chapter 7 · Suspension Bridges

The Forth Bridge of 1890 carries its load through deep steel arms. A suspension bridge does the opposite: it hangs the road from flexible steel cables that carry the load in pure tension, with no bending in the cables at all. How Bridges Work 5 showed that steel is the strongest material in tension per unit of weight, and a cable is the shape that makes the most of it. This chapter follows the load from the deck up the hangers into the main cable, over the towers and down into the anchorages, using the Forth Road Bridge of 1964, standing just beside its older neighbour, as the running example.

What was run, and what to trust

Every Python snippet in this chapter and in the three exercise solutions was executed, and the output shown in the comments is what Python printed. The Forth Road Bridge figures in the table come from the source named; the loads and the sag used in the calculations are illustrative assumptions, labelled where they appear. The results show how the principle works, not the designers' figures.

The Load Path of a Suspension Bridge

  deck (the road)
     │
     ▼  hangers (tension)
  main cable            ← tension, along a smooth curve
     │
     ▼  over the tops of the towers
  towers                ← compression, straight down
     │
     ▼
  foundations

  main cable continues past the towers
     │
     ▼
  anchorages            ← resist the cable's pull (weight and rock)
    

Compare this with How Bridges Work 1's load path for a beam. Here the deck does not carry its load to the ends. It hangs it on the cable every few metres, and the cable carries it to the towers and the anchorages. The towers are in compression, the cables and hangers in tension, and the anchorages hold the whole system from pulling in. The span can be very long because the cable is a small cross-section of steel working in the way steel does best.

The Forth Road Bridge: The Facts

Fact	Figure
Opened	4 September 1964, by Queen Elizabeth II and the Duke of Edinburgh; the construction plan was accepted in February 1958 and work began that September
Design	Mott, Hay and Anderson with Freeman Fox & Partners
Builders	a joint venture of Sir William Arrol & Co., Cleveland Bridge & Engineering Company and Dorman Long
Cost	£15.1 million
Main span	1,006 m (3,301 ft)
Side spans	408 m each
Total length	2,512 m (8,241 ft)
Towers	156 m (512 ft) above mean water level
Main cables	590 mm in diameter, each with 11,618 high-tensile wires of 5 mm; the page says each cable carries 13,800 tonnes of the bridge's load
Materials	39,000 tonnes of steel; the page gives two conflicting concrete figures (115,000 m³ and 210,000 tons)
Construction deaths	seven people died

(Source: the Wikipedia article on the Forth Road Bridge. A subagent read the first 39,400 of 97,257 characters verbatim, plus a model summary of the rest, and the figures above are all from the verbatim part. The page does not state the sag of the cables, the clearance above the water, the anchorage method or the deck type, so this chapter does not guess them.) Note the main span: at 1,006 m it is almost twice the 521 m main span of the Forth Bridge (arms plus girder, How Bridges Work 6), and it crosses the same estuary beside it.

The Shape of a Cable

A cable has no stiffness of its own, so it can only take the shape in which its tension balances the loads. For the loads on a suspension bridge there are two shapes to know. A cable hanging under only its own weight forms a catenary. A cable carrying a deck hung from it at regular intervals, so that the load is spread evenly along the horizontal span, forms a parabola. On a real bridge the deck weighs far more than the cable, so the parabola is the right model for the finished bridge (the catenary matters during construction, before the deck is hung). The two curves are very close, as the code further below shows.

For a parabolic cable of span L and sag f under a uniform load w per metre, the horizontal pull is H = wL²/(8f). It is the same formula as the arch thrust of How Bridges Work 3, with the sag in place of the rise: a cable is an upside-down arch. The vertical pull at each tower is V = wL/2, and the biggest tension, at the tower, is the combination of the two: T = √(H² + V²).

import math

def cable(w, span, sag):
    """Parabolic cable under a uniform load w (kN/m) along the span (m)."""
    h = w * span**2 / (8 * sag)                       # horizontal pull, kN
    v = w * span / 2                                  # vertical pull at each tower, kN
    t = math.hypot(h, v)                              # largest tension, at the tower
    length = span * (1 + 8 / 3 * (sag / span) ** 2)   # approximate cable length, m
    return h, v, t, length

span = 1006
sag = span / 11                                       # assumed sag-to-span ratio of 1/11
h, v, t, length = cable(90, span, sag)
print(round(sag, 1), round(h), round(v), round(t), round(length))
# 91.5 124492 45270 132468 1028
wires = 11618 * math.pi * 2.5**2                      # steel area in one cable, mm2
print(round(wires), round(t * 1000 / wires))          # area, and the stress in MPa
# 228119 581
    

This models one main cable over the 1,006 m span with two assumptions: a sag of 1/11 of the span (the sources I found put typical suspension bridge sags somewhere between about 1/8 and 1/15 of the span, and they do not agree exactly) and a load of 90 kN per metre per cable, chosen so that the answer lands near the 13,800 tonnes the article quotes. The result: a horizontal pull of 124,492 kN, a vertical pull of 45,270 kN at each tower, and a largest tension of 132,468 kN, which is about 13,500 tonnes-force. The 11,618 wires of 5 mm have a total steel area of 228,119 mm², so the stress in the wire is 581 MPa. That is a sensible fraction of what high-tensile bridge wire can take (general knowledge: its breaking strength is of the order of 1,500 MPa or more), consistent with a factor of safety of two or more. The cable is 1,028 m long, only 22 m longer than the straight span.

Flatter Cables Pull Harder

The sag is a design choice, and it controls everything. Because H = wL²/(8f), a flatter cable pulls harder. The same bridge with different sags:

for ratio in (6, 8, 11, 15, 25):
    h, v, t, length = cable(90, span, span / ratio)
    print(f"1/{ratio:<3}", round(h), round(t), round(length))
# 1/6   67905 81612 1081
# 1/8   90540 101227 1048
# 1/11  124492 132468 1028
# 1/15  169762 175695 1018
# 1/25  282938 286536 1010
    

The columns are the sag ratio, the horizontal pull (kN), the largest tension (kN) and the cable length (m). Going from 1/6 to 1/25 more than quadruples the horizontal pull (67,905 to 282,938 kN) and more than triples the cable tension, while shortening the cable by 71 m. A deeper sag means a lighter cable but a taller tower to hold it up; a flatter sag means a heavier cable and bigger anchorages but shorter towers. Choosing the sag is a trade-off between tower cost and cable cost, the same trade-off as the rise of an arch.

Catenary or Parabola?

How different are the two curves for this bridge? The code finds the catenary that has the same sag, by repeatedly halving an interval until the match is close (bisection), and compares the two curves a quarter of the way across from a tower, measured as the drop below the tower top:

def catenary_sag(a, half):
    return a * (math.cosh(half / a) - 1)

def solve_a(span, sag):
    lo, hi = 1.0, 1e7
    for _ in range(200):                  # bisection: find a so that the sag matches
        mid = (lo + hi) / 2
        if catenary_sag(mid, span / 2) > sag:
            lo = mid
        else:
            hi = mid
    return (lo + hi) / 2

sag = span / 11
a = solve_a(span, sag)
x = span / 4                              # a quarter of the way across
y_cat = catenary_sag(a, span / 2) - catenary_sag(a, span / 2 - x)
y_par = sag * (1 - (1 - x / (span / 2)) ** 2)   # same-sag parabola
print(round(a), round(y_cat, 2), round(y_par, 2), round(y_cat - y_par, 2))
# 1398 68.77 68.59 0.18
    

The catenary's parameter is 1,398 m. A quarter of the way across, it hangs 68.77 m below the tower top, and the parabola 68.59 m: the two curves differ by 18 cm over a span of a kilometre. That is why the parabola is the standard model for a cable with a deck. The catenary still matters for the cable before the deck is hung, and for cables with a large sag-to-span ratio, where the gap grows (Exercise 3).

Towers and Anchorages

The cable passes over the tops of the towers, which carry the vertical pull of both sides straight down in compression (here 45,270 kN from each side). On a suspension bridge the towers are therefore slender columns in compression, and need to be stiff against buckling (How Bridges Work 4). At the far ends the cable must be fixed. The anchorage has to resist the full horizontal pull H that the cable exerts. Many anchorages do this with sheer weight and friction, and others are set into rock. A rough estimate of the weight of a gravity anchorage that resists sliding by friction alone, for the cable pull above, looks like this:

def anchorage(h_kn, mu=0.6, factor=1.5, rho=2400):
    weight_kn = factor * h_kn / mu                    # needed weight, kN
    mass_t = weight_kn / 9.81
    return round(mass_t), round(mass_t * 1000 / rho)  # tonnes, m3 of concrete

print(anchorage(124492))
# (31726, 13219)
    

About 31,700 tonnes of concrete, or roughly 13,200 m³ per anchorage, with an assumed friction coefficient of 0.6 and a factor of safety of 1.5. This model is very crude: it ignores the vertical component of the pull and the way a real anchorage is shaped and founded. The sources do not describe the Forth Road Bridge's anchorage, so this is a general illustration only. The lesson is the size of the numbers: the horizontal pull of the cables has to be matched by something enormous.

The Deck: Why Stiffness Matters

A cable hung with a perfectly even load keeps its parabolic shape. A real load is not even: a heavy lorry in one place pulls the cable down there and lets it rise elsewhere, and wind pushes the whole deck sideways. A cable alone would swing and sag in response. A stiffening girder or truss along the deck spreads a local load over many hangers, so that the cable feels something near an even load and the deck stays flat. Bouch's 1871 proposal for the Forth, mentioned in How Bridges Work 6, was a stiffened suspension bridge. The stiffness of the deck also controls how it moves in the wind, which How Bridges Work 9 takes up with the Tacoma Narrows bridge. The Forth Road Bridge article I read does not state the deck type of the bridge (truss or box girder), so I do not describe it here.

Common mistakes

Using the catenary for a cable with a deck (the right model is a parabola); using the tower's vertical reaction as the cable tension (the tension at the tower is the combination of horizontal and vertical pulls); forgetting that the horizontal pull falls as the sag increases (flatter means more pull); and trusting a number from a source without asking what it measures (the article's "13,800 tonnes" per cable is described only as the load each cable carries).

Where This Leads

A bridge's cables are out of sight and hard to inspect, so the story has a sequel: an inspection programme that began in 2003 led to the discovery in 2005 of potentially serious problems in the Forth Road Bridge's main cables, and a later study reported an 8% to 10% loss of strength, and a dehumidification system was installed to keep the air among the strands below 40% humidity. That story, and what it taught, belongs to How Bridges Work 9. The next chapter returns to the third Forth bridge: a design that keeps the towers and the tension but drops the draped main cable and runs straight cables from the towers to the deck.

Exercise 1

A small suspension bridge has a main span of 300 m and two cables, each carrying 60 kN/m (deck and cable together). For sag-to-span ratios of 1/8, 1/10 and 1/12, find the horizontal pull, the vertical pull at each tower, the largest tension, the approximate cable length, and the steel area needed at a working stress of 600 MPa. Which sag needs the least steel, and what do you pay for that choice?

📄 View solution

Exercise 2

Each cable of the bridge in Exercise 1 ends in a gravity anchorage that must resist the horizontal pull by friction alone (friction coefficient 0.6, factor of safety 1.5, concrete of 2400 kg/m³). For the three sag ratios find the weight, mass and volume of concrete, then repeat for friction coefficients of 0.4 and 0.8 at the 1/10 sag. How does the anchorage size depend on the pull and on the friction?

📄 View solution

Exercise 3

For a 300 m span with sag-to-span ratios of 1/20, 1/10 and 1/5, find the catenary parameter that gives each sag (use bisection), and the difference in metres between the catenary and the same-sag parabola a quarter of the way across. For which ratio does the choice of curve matter most?

📄 View solution

Quick Reference

Load path: deck, hangers (tension), main cable (tension), towers (compression), anchorages
Cable with a deck: a parabola; cable under its own weight: a catenary
Pulls: H = wL²/(8f), V = wL/2 at each tower, T = √(H² + V²)
Sag: flatter means more pull and more cable steel; deeper means taller towers
Anchorage: must resist H; weight, friction and rock
Forth Road Bridge (1964): main span 1,006 m, towers 156 m, cables of 11,618 wires of 5 mm
Stiffening girder: spreads local loads and steadies the deck against wind

Coming next

How Bridges Work 8 is about cable-stayed bridges: fan and harp layouts, the Queensferry Crossing of 2017, construction by balanced cantilever, and wind shielding.