33 The Elastic Catenary.

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 $d σ$ stretch into $d s$ ; then, by Hooke’s law, $d s = d σ (1 + λ T)$ , where $λ$ is the elastic constant of the string; hence the weight of the stretched element $d s = g ρ ω d σ = \frac{g ρ ω d s}{(1 + λ T)}$ . Accordingly, as before,

\begin{array}{l} d (T sin φ) & = \frac{g ρ ω d s}{(1 + λ T)}, \\ and \\ T cos φ & = H = g ρ ω c, \\ hence \\ c d (tan φ) & = \frac{d s}{(1 + μ sec φ)}, \\ in which μ stands for λ H, the extension at the lowest point; therefore \\ d s & = c ({sec}^{2} φ + μ {sec}^{3} φ) d φ, \\ \frac{s}{c} & = tan φ + \frac{1}{2} μ (sec φ tan φ + {gd}^{- 1} φ), & [prob. 20, p. 37 \end{array}

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

\begin{array}{l} d x & = d s cos φ = c (sec φ + μ {sec}^{2} φ) d φ, \\ d y & = d s sin φ = c ({sec}^{2} φ + μ {sec}^{3} φ) sin φ d φ . \\ Whence \\ \frac{x}{c} & = {gd}^{- 1} φ + μ tan φ, \frac{y}{c} = sec φ + \frac{1}{2} μ {tan}^{2} φ . \end{array}

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

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Article 33The Elastic Catenary.

Article 33
The Elastic Catenary.