2. Benchmark solution#

2.1. Calculation method#

In the case of a sliding cable without friction (modeling C), with embedded concrete and cable sheath (only the sliding of the cable is then authorized), the following analytical solution is shown.

The imposed slip is noted \(\stackrel{̃}{g}\), the length of the cable \(L\), its section \(A\)

The deformation of the cable is uniform (no friction) and is then worth \(\frac{\tilde{g}}{L}\), the stress \(E\frac{\tilde{g}}{L}\) and the tension \(T\mathrm{=}AE\frac{\tilde{g}}{L}\).

The reaction on the whole concrete is \(R=\mathrm{2T}\).

On a cable section, the reaction is \(S\mathrm{=}T\Delta \alpha\) with \(\Delta \alpha\) the angular deviation.

If the cable is discretized into \(n\) elements of the same size, we have \(S\mathrm{=}T\frac{\pi }{n}\).

For tension calculations BPEL, analytical references are obtained using the formulas given in [R7.01.02].

2.2. Reference quantities and results#

2.2.1. Voltage and reactions for C modeling#

Taking \(\tilde{g}\mathrm{=}0.1\), \(L\mathrm{=}r\pi\), \(A=2.5e-3\), \(E=185000\mathit{MPa}\), we find:

  • voltage \(T=2.9443664{10}^{6}N\)

  • the reaction on all \(R=5.888734{10}^{6}N\) concrete and on a \(S=462500N\) cable section with \(n\mathrm{=}20\).

2.2.2. Voltage BPEL for E, F, and G models#

The Gauss points tested are the Gauss points \(1\) of the \(105\) and \(96\) meshes and for the E models for the E and E models. Any point \(P\) belonging to the cable can be defined by the value of the angle \(a\mathrm{=}\stackrel{ˆ}{\overrightarrow{({\mathit{OA}}_{1},\mathit{OP})}}\) or \(b\mathrm{=}\stackrel{ˆ}{\overrightarrow{({\mathit{OA}}_{2},\mathit{OP})}}\) where \(59\) \(50\) \(O\) is the center of the semicircle, \({A}_{1}\) is ANCR1, and \({A}_{2}\) is ANCR2.

The first node in stitch \(96\) is defined by \(a=\mathrm{67,5}\) or \(b=\mathrm{112,5}\). The first Gauss point in cell \(96\) is then defined by \({a}_{96}=\mathrm{67,5}+\frac{\mathrm{4,5}}{\sqrt{(3)}}\) or \({b}_{96}=\mathrm{112,5}-\frac{\mathrm{4,5}}{\sqrt{(3)}}\).

The first node in stitch \(105\) is defined by \(a=\mathrm{148,5}\) or \(b=\mathrm{31,5}\). The first Gauss point in cell \(105\) is then defined by \({a}_{96}=\mathrm{148,5}+\frac{\mathrm{4,5}}{\sqrt{(3)}}\) or \({b}_{96}=\mathrm{31,5}-\frac{\mathrm{4,5}}{\sqrt{(3)}}\).

It is necessary to calculate the cumulative angular deviation \(\alpha\) of these points which will then be used to calculate the tensions. The cumulative angular deviation used in voltage formula BPEL should be calculated using the closest active anchor. In the E model, two calculations are made: one with ANCR1 passive and ANCR2 active and the other with both anchors active. Mesh \(105\) is closer to ANCR2 which is still active, so we have in both cases: \({\alpha }_{105}\mathrm{=}{b}_{105}\mathrm{\ast }\pi \mathrm{/}180\).

We notice that in our case the cumulative angular deviation is equal to the angles \(a\) or \(b\), we are just passing in radians.

The \(96\) mesh is closer to ANCR1 so we have two different values:

\({\alpha }_{96}\mathrm{=}{b}_{96}\mathrm{\ast }\pi \mathrm{/}180\) for the PASSIF/ACTIF case

\({\alpha }_{96}\mathrm{=}{a}_{96}\mathrm{\ast }\pi \mathrm{/}180\) for the ACTIF/ACTIF case

The curvilinear axes of these points are also necessary to calculate the tension; they are also given in relation to the nearest active anchor. We then have: \({s}_{105}\mathrm{=}{\alpha }_{105}\mathrm{\ast }r\) and \({s}_{96}\mathrm{=}{\alpha }_{96}\mathrm{\ast }r\).

For the case without anchor recoil, we use the formula \({F}_{c}\mathrm{=}{F}_{0}\mathrm{exp}(\mathrm{-}f\alpha \mathrm{-}\phi s)\), with \({F}_{0}={10}^{6}N\), \(\phi \mathrm{=}\mathrm{0,01}\); \(f=\mathrm{0,03}\), we get:

  • for the case PASSIF/ACTIF: \({F}_{105}\mathrm{=}960448.709086365N\) and \({F}_{96}\mathrm{=}857741.905702382N\)

  • for the case ACTIF/ACTIF: \({F}_{105}=960448.709086365N\); \({F}_{96}=906761.8988894981N\)

The deformation field EPSI_ELGA is tested in modeling E in the case PASSIF/ACTIF, on the same Gauss points as the tension N. The calculation is elastic, therefore:

\({\mathit{EPXX}}_{105}\mathrm{=}{F}_{105}\mathrm{/}E\mathrm{/}A\)

\({\mathit{EPXX}}_{96}\mathrm{=}{F}_{96}\mathrm{/}E\mathrm{/}A\)

For the case with anchor recoil, with the notations of [R7.01.02], we show that the distance \(d\) on which the anchor recoil acts is equal to:

\(d=\frac{-1}{f}\mathrm{ln}(1-\sqrt{(f\ast {E}_{a}\ast {S}_{a}\ast \Delta /{F}_{0})})\)

With a setback from anchor \(\Delta =5.{10}^{4}\), here we get \(d=3.9222\). Keeping the notations defined above, it means that a point is concerned if \(a<\mathrm{44,94}\) or \(b<\mathrm{44,94}\). The mesh 105 is therefore affected by the modification due to the anchoring recoil, but not the mesh 96.

By then using \({F}_{c}\mathrm{=}\frac{{(F(d))}^{2}}{{F}_{0}}\mathrm{exp}(\mathrm{-}f\alpha \mathrm{-}\phi s)\) for the 105 mesh and \({F}_{c}\mathrm{=}{F}_{0}\mathrm{exp}(\mathrm{-}f\alpha \mathrm{-}\phi s)\) for the 96 mesh, we obtain:

\({F}_{105}=918367.3641803192\); \({F}_{96}=906761.8988894981\)

2.2.3. Reference for H modeling#

Given the imposed load, in DEFORMATION = “PETIT” there would be no constraints in the cable. On the other hand, in DEFORMATION = “PETIT_REAC”, constraints must appear. The reference solution in this case is obtained by a calculation with BARRE elements.

2.3. Uncertainty about the solution#

Néant