Article 33
The Elastic Catenary.

An elastic string of uniform section and density in its natural state is suspended from two points. Find its equation of equilibrium.

Let the element dσ stretch into ds; then, by Hooke’s law, ds = dσ(1 + λT), where λ is the elastic constant of the string; hence the weight of the stretched element ds = gρωdσ = gρωds (1 + λT). Accordingly, as before,

d(Tsinφ) = gρωds (1 + λT),  and Tcosφ = H = gρωc,  hence cd(tanφ) = ds (1 + μsecφ),  in which μ stands for λH, the extension at the lowest point; therefore ds = c(sec2φ + μsec3φ)dφ, s c = tanφ + 1 2μ(secφtanφ + gd1φ), [prob. 20, p. 37

which is the intrinsic equation of the curve, and reduces to that of the common catenary when μ = 0. The coordinates x, y may be expressed in terms of the single parameter φ by putting

dx = dscosφ = c(secφ + μsec2φ)dφ, dy = dssinφ = c(sec2φ + μsec3φ)sinφdφ.  Whence x c = gd1φ + μtanφ,y c = secφ + 1 2μtan2φ.

These equations are more convenient than the result of eliminating φ, which is somewhat complicated.