2. Benchmark solution

2.1. Formal solution

The movements in the cable and in the concrete are continuous and homogeneous. So the horizontal displacement on the interval \(\mathrm{[}0;L\mathrm{]}\) is equal to \(u(x)\mathrm{=}\frac{{u}_{L}}{L}x\)

The normal stresses in the concrete plate and the steel cable are written respectively (elasticity hypothesis):

\(\mathrm{\{}\begin{array}{c}{\sigma }_{b}\mathrm{=}{E}_{b}\frac{\mathrm{\partial }u}{\mathrm{\partial }x}\mathrm{=}{E}_{b}\frac{{u}_{L}}{L}\\ {\sigma }_{a}\mathrm{=}{E}_{a}\frac{\mathrm{\partial }u}{\mathrm{\partial }x}+{\sigma }_{0}\mathrm{=}{E}_{a}\frac{{u}_{L}}{L}+{\sigma }_{0}\end{array}\)

where \({\sigma }_{0}\mathrm{=}\frac{{F}_{0}}{{S}_{a}}\) is the initial prestress in the cable

and \({u}_{L}\) is the horizontal displacement at abscissa \(L\).

The balance of the plate and cable assembly is written as: \({\sigma }_{b}{S}_{b}+{\sigma }_{a}{S}_{a}\mathrm{=}0\mathrm{\Rightarrow }{u}_{L}\mathrm{=}\frac{\mathrm{-}L{F}_{0}}{{E}_{b}eH+{E}_{a}{S}_{a}}\)

The normal effort in the cable is worth: \({N}_{a}\mathrm{=}{E}_{a}{S}_{a}\frac{{u}_{L}}{F}+{F}_{0}\mathrm{=}{F}_{0}\frac{{E}_{b}eH}{{E}_{b}eH+{E}_{a}{S}_{a}}\)

The total effort on the vertical section of the concrete plate is equal to:

\({N}_{b}\mathrm{=}{E}_{b}eH\frac{{u}_{L}}{L}\mathrm{=}\mathrm{-}{F}_{0}\frac{{E}_{b}eH}{{E}_{b}eH+{E}_{a}{S}_{a}}\)

The linear density of normal force on the concrete plate is deduced.

\({N}_{\mathit{xx}}\mathrm{=}\mathrm{-}{F}_{0}\frac{{E}_{b}e}{{E}_{b}eH+{E}_{a}{S}_{a}}\)

2.2. Numeric reference values

The numerical reference values are:

\({u}_{L}\mathrm{=}–\mathrm{1,11013974}{.10}^{\mathrm{-}5}m\)

\({N}_{a}\mathrm{=}\mathrm{1,99825153}{.10}^{5}N\)

\({N}_{b}\mathrm{=}–\mathrm{1,99825153}{.10}^{5}N\)

\({N}_{\mathit{xx}}\mathrm{=}–\mathrm{9,99125765}{.10}^{4}N\mathrm{/}m\)