Cable sag and pretension for a hanging cable

A cable with no bending stiffness carries tension along its own curve and nothing else. No horizontal load acts on it, so the horizontal component of the tension, H, is the same at every point, and the vertical component at a point is the load carried between the lowest point and there. H is the pretension in the usual sense, and with the loads it fixes the whole shape. Whichever value you give (the sag at mid-span or below the lower support, H itself, the tension at the higher or the lower support, or the unstretched length), the calculator turns it into the H that produces it, and every other result follows from H.

When the load is uniform along the horizontal, as for a deck or a pipe hung from the cable, the curve is a parabola. For a level span L and a mid-span sag d, H = qL²/(8d), with q the load per unit of horizontal length. Each support carries a vertical reaction qL/2, and the largest tension, at the supports, is T_max = √(H² + (qL/2)²). The length of the cable is S = L(1 + 8/3 (d/L)² − 32/5 (d/L)⁴ + …), which is why a shallow cable is barely longer than its span.

On an inclined chord, with one support a height h above the other and L still the horizontal distance, the sag d is the vertical offset of the cable below the chord at mid-span, and H = qL²/(8d) holds unchanged. What changes is the split of the load: the higher support carries qL/2 + Hh/L and the lower one qL/2 − Hh/L. The tension at a support is √(H² + V²), with V its vertical reaction. When h is greater than 4d the lower reaction is negative, and the cable then leaves that support upward.

A cable loaded by its own weight w per unit of its LENGTH hangs as a catenary, y = a(cosh(x/a) − 1), measured from the lowest point, with a = H/w. For a level span the sag is d = a(cosh(L/2a) − 1), the length is S = 2a sinh(L/2a), and the tension at a point a height y above the lowest is T = wa cosh(x/a) = H + wy, so that T_max = H + wd at the supports. There is no closed form for a from d, and the calculator solves for it numerically; on an inclined chord the same curve is cut by the two supports, and its lowest point may lie outside the span.

With the cable’s weight and a load along the span together, neither curve is the exact one. The calculator solves the equilibrium of an element of cable, H z″ = q + w√(1 + z′²), and gives that exact shape, with the stretch of the cable in it when a modulus and an area are given. Next to it comes the parabola that results from spreading the cable’s weight over the horizontal, q + w per unit, so the two can be compared.

The two curves agree closely for a shallow cable. For a level cable under its own weight, the parabola’s H, largest tension or mid-span sag departs from the exact value by 1 % at a sag of about 0.085 of the span, and by more beyond. The page then notes it, and the exact column is the one to read.

A cable that is elastic stretches under tension: an element of unstretched length ds becomes ds(1 + T/EA), with E the modulus and A the metallic area. The stretched length is added up along the cable, element by element, and not taken from an average tension.

A change of state follows one cable from state 1 to state 2: another load, another temperature, or both. What links the two states is the unstressed length, which changes only with temperature, L₀₂ = L₀₁(1 + αΔT), α being the coefficient of thermal expansion. For a level parabola, taking the loaded length as L + q²L³/(24H²) and the strain as H/EA, equal unstressed lengths give a cubic in the new horizontal tension: (K/EA)H₂³ + (K − 1)H₂² − q₂²L²/24 = 0, where K = (1 + αΔT)(1 + q₁²L²/(24H₁²))/(1 + H₁/EA). A heavier load or a lower temperature raises H; a lighter load or a higher one lowers it. The calculator solves the exact lengths of both states for the catenary and the combined cable, without those approximations, and shows the parabola’s result beside them.

A cable that carries separate vertical loads is straight between them: a funicular polygon. The bending moment M(x) that the same loads cause in a simply supported beam of the span L equals H times the vertical offset s(x) of the cable below the chord, H s(x) = M(x). One sag therefore fixes H, H = M(x₀)/s₀ at the position x₀ where it is given, and x₀ need not carry a load. The sag at each load is then M/H, the slope of each segment follows from the sags on either side, and its tension is T = H√(1 + slope²). The largest tension is in the steepest segment, which is next to a support when every load is downward. H, or the largest tension, can be given instead of a sag.

Two cases whose answers are known in closed form, to check the page against.

One load P at mid-span of a level span L, with the cable a sag d below the chord under it: H = PL/(4d), each segment has a vertical component P/2, and the tension in both is T = √(H² + (P/2)²).

A level cable of weight w per unit of length, with a = H/w chosen so that L/(2a) = 1: the sag is d = a(cosh 1 − 1), the length S = 2a sinh 1, and the tension at the supports T_max = wa cosh 1. That is one case where a, and not the sag, is the starting value.