The Catenary

The statics of a hanging chain

The curve described by a uniform, flexible chain hanging under the influence of gravity is called the catenary. Let us choose the coordinates to describe the curve as shown in the Figure, with the origin at the lowest point of the chain, x to the right, and y upwards. Let one end of the chain be supported at x = L/2 and y = h, and let w be the weight per unit length of the chain. The tension T in the chain is in the direction of the derivative dy/dx. Its horizontal component is H, and its vertical component is V. If we cut the chain at some point as shown, and consider the equilibrium of the portion of the chain between the right-hand end and this point, we see that H is constant at all points of the chain, since the weight of the chain acts vertically downward. Therefore, the vertical component of the tension in the chain is Hy', where y' stands for dy/dx.

Consider the equilibrium of the short length of chain subtending a distance dx on the x-axis. The horizontal forces are equal, as we have already seen, and the sum of the vertical forces is V + dV - V - w ds = 0. Since V = Hy', dV = Hy" dx, and we obtain a differential equation that is first order in y'. This equation is integrated as shown in the Figure to find y'(x), and again to find y(x), using the conditions that y = y' = 0 at x = 0. The catenary is a hyperbolic cosine curve, and its slope varies as the hyperbolic sine. The arc length, measured from the origin, is s = (H/w) sinh (wx/H), which shows that the slope is proportional to arc length. The total length of the chain is S = (2H/w) sinh (wL/2H), and h = (H/w)[cosh (wL/2H) - 1]. We can use these two equations to find that H = (w/8h)(S² - 4h²), in terms of the total length of the chain and the sag. Then, the span L = (2H/w) sinh^-1 (Sw/2H). The vertical force at a support is V = wS/2, so the tension there is given by T² = H² + (wS/2)². We have assumed that the chain is supported at equal heights on the two sides, but this is not necessary, and the equations can be modified to suit any case.

If we expand the hyperbolic cosine in powers of x, the lowest term yields y = wx²/2H. Therefore, to this approximation, we have h = wL²/8H, a familiar result for the parabolic curve of a weightless chain supporting a uniform load, easily obtained by taking moments about the right-hand support of the chain. If we assume the weight is uniformly distributed in x, the differential equation is w dx = Hy", which, when integrated twice, gives just the result above.

If a vertical downward force F acts at a point of the chain, the slope is discontinuous at that point, jumping by an amount F/H. This will add a term Fδ(x - x₀) for a force F at x = x₀ to the differential equation. The shape of the whole chain changes, illustrating a well-known problem with suspension bridges. If the deck is not sufficiently rigid, the bridge can oscillate, one portion sinking as another rises. This can happen in opposite phase on the two sides of the bridge, resulting in a torsional oscillation.

If the chain is imagined to become rigid, and is turned over, tensions will become compressions, and we will have an arch. The curve shows how the stress is propagated to the supports, provided the structure remains rigid. In this case, a thin structure tends to buckle, a problem not encountered with the chain. The results must be used with care for arches, since the loading condtions are usually quite different. The usual circular arch has heavy haunches, which helps to guide the force along the voussoirs. A segmental arch is helped by pierced spandrels, since this reduces the force on a flatter portion of the arch and distributes the load more uniformly.

Composed by J. B. Calvert
Created 10 July 2000
Last revised