2. Benchmark solution

2.1. Calculation method used for the reference solution

Analytical solution:

For an extendable (elastic) cable, subject to its own weight, the displacement is equal to:

\(x(s)=a\mathit{Argsh}\left(\frac{s}{a}\right)+\frac{\rho g}{E}as\)

\(z(x)=a\sqrt{1+\frac{{s}^{2}}{{a}^{2}}}+\frac{\rho g}{E}\frac{{s}^{2}}{2}-a\sqrt{1+\frac{{l}_{0}^{2}}{{a}^{2}}}-\frac{\rho g}{E}\frac{{l}_{0}^{2}}{2}\)

\(a\) solution of equation \(L=a\mathrm{Argsh}(\frac{{l}_{0}}{a})+\frac{\rho g}{E}a{l}_{0}=f(a)\)

With \(s\) curvilinear abscissa, \(s\in [-{l}_{o},{l}_{o}]\). Here we are interested in the arrow in the center (point \(C\)):

\(z(C)=a-a\sqrt{1+\frac{{l}_{0}^{2}}{{a}^{2}}}-\frac{\rho g}{E}\frac{{l}_{0}^{2}}{2}\)

\(a\) solution of equation \(L=a\mathrm{Argsh}(\frac{{l}_{0}}{a})+\frac{\rho g}{E}a{l}_{0}=f(a)\)

The only difficulty in calculating this solution is solving equation \(L=f(a)\). This resolution was done numerically (Fortran program using the zero search routine from Aster ZEROFO).

Note:

In the case of thermal expansion, the solution is the same as before, considering that the initial length \(2{l}_{0}\) is equal to its initial length \(\mathrm{2L}\) increased by the linear expansion: \({l}_{0}=L(1+\alpha T)\)

2.2. Benchmark results

Move to \(Z\) to point \(C\)

2.3. Uncertainty about the solution

Semianalytic solution: the numerical resolution of equation \(L=f(a)\) gives a value to within \({10}^{-3}\).

2.4. Bibliographical references

[1] C. CONEIM « On the approximation of the equations of the statics of aerial cables in the presence of electromagnetic force fields ». Thesis and note HI/3640-02 (February 1981)