Arches, Cables & the Line of Thrust

Two structures that carry load the same way, one in pure tension and one in pure compression. Hang a cable and choose what it carries, or build a stone arch and look for a line of thrust that stays inside it. The table under the drawing works each figure out.

Worked example — Arches, Cables & the Line of Thrust

Deck load	128.00 MN
Cable weight	52.86 MN
Horizontal tension, parabolic estimate	200.50 MN
Horizontal tension H	201.44 MN
Vertical reaction per support	90.43 MN
Tension at the supports	220.80 MN
Cable length	1,321.48 m
Deck share of the load	70.8%

These are the values the calculator shows when the page opens. Change any input above to recompute them.

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Lesson

The theory — Arches, Cables & the Line of Thrust

The line of thrust is the path the compressive force takes through an arch: at each joint between stones, the point where the resultant of everything above it crosses that joint. An arch of blocks that cannot take tension stands if some line of thrust lies inside the masonry everywhere, which is Jacques Heyman's safe theorem, and each point where the line touches a face is a potential hinge. A cable is the same thing in tension. Its shape is the funicular of its load: a parabola for a load spread evenly along the span, a catenary for a load spread evenly along the cable itself.

What each symbol means

H: the horizontal thrust, in kN per metre of width for the arch and kN for the cable. The same at every section, because nothing pushes sideways
t: the arch's thickness. The page finds the smallest one that still holds a line of thrust
R: the radius of a circular centre line. For a semicircle, t ÷ R is the number to compare with Milanković's 0.1075
w_d: a load per metre of span, like a deck. On its own it hangs a cable as a parabola
w_c: a load per metre of cable, like its own weight. On its own it hangs a cable as a catenary

Assumes: The stones cannot take tension across a joint, are strong enough in compression never to be crushed, and cannot slide on one another. These are Heyman's three assumptions, and the second insight tests the middle one. The page also treats each stone as rigid and lets the arch carry only its own weight. The cable has no bending stiffness, and its supports are level.

The deck, not the cable, decides which curve the cable hangs in 🖖

Press A Golden Gate-sized cable. It hangs 164.7 mm from a parabola and 407.7 mm from a catenary, on a sag of 143 m. Press Cable, no deck and the same cable lies on the catenary. A chain carrying only itself has its load spread along its own length, which is heavier per metre of span where it is steep, and that makes a catenary. A deck hung from it is spread evenly along the span, and that makes a parabola. Galileo drew parabolas with a hanging chain, which only approximates one, and the hyperbolic functions page tells that story; the suspension bridge is the case where the parabola is the better answer. The two curves are only 572.4 mm apart at this sag, 0.40% of it, which is why the upper panel shows one line and the lower one has to plot the gaps in millimetres. Make the cable sag half its span and they are 5% apart.

A thin arch falls with its stone hardly stressed 🖖

Press Semicircle, t/R 0.12. A line of thrust fits inside the stone, and each springing carries the 94.25 kN per metre of width of its half-arch straight down onto 0.6 m of stone. That is an average of 0.16 MPa, a tiny fraction of what any building stone can take in compression. Press Semicircle, t/R 0.09 and no line of thrust fits at all: the arch opens hinges at the crown, at the haunches and at the springings and folds, without a single stone being crushed. An arch stands or falls on its geometry. At its default 40 voussoirs this page puts the semicircle's limit at t/R = 0.1075, which is the figure Milutin Milanković derived in his papers on the line of thrust, published between 1904 and 1910, years before the orbital theory of the ice ages that carries his name. Crushing strength does not enter it: stone ten times stronger falls at the same thickness.

Turn a hanging chain upside down and the arch needs almost no thickness 🖖

Press Thin catenary arch: 0.1 m of stone across a 10 m span, and it stands. Then Thin parabolic arch, same span, same rise, same thickness, and nothing fits. The catenary works because a chain hanging under its own weight is in pure tension along its whole length, and turning it over turns every force into compression along the same line. The thrust line of an inverted catenary runs down the middle of its own stone, so it needs no thickness to contain it. Robert Hooke saw this in 1675 and hid it in an anagram: as hangs the flexible line, so but inverted will stand the rigid arch. Gaudí designed the church of the Colònia Güell, of which only the crypt was built, with strings and little bags of lead shot hung from a board, and turned the photographs upside down. The comparison table lists all four shapes. At a rise of half the span the pointed arch is the semicircle, the parabola needs 0.1444 m and the semicircle 0.5373 m.

Problem solved in full

The tension in a suspension bridge's main cable 5 steps
A main cable spans 1,280 m between towers with a sag of 143 m. It carries 100 kN of deck per metre of span and weighs 40 kN per metre of its own length. Estimate the tension at the towers without solving the curve. Then press A Golden Gate-sized cable and see how close the estimate is.

Open this problem in the tool
1. Treat the whole load as though it were spread evenly along the span. That counts the cable as 1,280 m long, which it is not quite:
  
  $w \approx w_{d} + w_{c} = 100 + 40 = 140 kN/m, w L = 140 \cdot 1280 = 179.20 MN$
2. A load spread evenly along the span hangs a cable as a parabola, and for a parabola the horizontal tension follows from one moment of the half-cable about the tower:
  
  $H = \frac{w L ^{2}}{8 f} = \frac{140 \cdot 128 0 ^{2}}{8 \cdot 143} = 200.50 MN$
3. Each tower carries half the total load straight up:
  
  $V = \frac{w L}{2} = 89.60 MN$
4. At the tower the cable pulls along its own direction, and that tension combines the two components:
  
  $T = H^{2} + V^{2} = 200.5 0^{2} + 89.6 0^{2} = 219.61 MN$
5. The page solves the curve exactly. The cable is 1,321.48 m long, so its own weight is 52.86 MN rather than 51.20, and the true figures are a little higher:
  
  $H = 201.44, V = 90.43, T = 220.80 MN$
Answer

About 219.6 MN at each tower by the parabola, against 220.80 MN from the exact curve, 0.54% low. The parabola is off because the cable is 3.2% longer than the span, and the cable's weight is 29% of the load. For a first sizing of the cable that is as good as the exact answer, and it needed one moment and one square root.

$\frac{220 . 80 - 219 . 61}{220 . 80} = 0 . 54%$

Learning path

Where the load goes

References (5)

The paper that put the analysis of masonry on the footing this page uses: blocks that cannot take tension, cannot slide and are never crushed, and an arch that is safe if any line of thrust fits inside it: J. Heyman, "The stone skeleton." International Journal of Solids and Structures 2(2), 1966, pp. 249-279. DOI 10.1016/0020-7683(66)90018-7
Insight 2's figure t/R = 0.1075 is Milankovitch's, for radial joints, and this paper is the modern treatment of it. Its abstract gives 0.1075 for radial cuts and 0.1095 for vertical ones, which is why this page states which joints it uses: N. Makris and H. Alexakis, "The effect of stereotomy on the shape of the thrust-line and the minimum thickness of semicircular masonry arches." Archive of Applied Mechanics 83(10), 2013, pp. 1511-1533. DOI 10.1007/s00419-013-0763-4
That the astronomer of the ice ages is the same Milankovitch, and that his thrust-line papers date from 1904 to 1910: F. Foce, "Milankovitch's Theorie der Druckkurven: Good mechanics for masonry architecture." Nexus Network Journal, 2007, pp. 185-209. DOI 10.1007/978-3-7643-8699-3_3
The catenary itself, cosh and Galileo's mistake, are derived on another tool of this site rather than here: The hanging chain y = a cosh(x/a) and the 1638 story: see /math/hyperbolic on this site.
The Golden Gate-sized preset uses the bridge's span and sag and ESTIMATED loads — see the note: No source. The 1,280 m main span and a sag of about 143 m are the bridge's published dimensions, recalled; the 40 kN/m cable weight is derived from a recalled wire count and wire diameter; the 100 kN/m deck load is an estimate.

By the 42Math team · Reviewed against published sources. Corrections welcome.