Reference problem ===================== Geometry --------- The concrete beam is straight and rectangular in cross section. Its dimensions are: :math:`L\mathrm{\times }h\mathrm{\times }p\mathrm{=}10m\mathrm{\times }\mathrm{0,4}m\mathrm{\times }\mathrm{0,2}m` (:math:`y\mathrm{=}h\mathrm{/}2`). The cable crosses the beam parallel to the middle fiber of the beam, halfway up. Its eccentricity with respect to the middle plane is :math:`e\mathrm{=}\mathrm{0,05}m` (:math:`z\mathrm{=}e`). The cross-sectional area of the cable is :math:`{S}_{a}\mathrm{=}\mathrm{1,5}{.10}^{\mathrm{-}4}{m}^{2}`. .. image:: images/Object_1.svg :width: 593 :height: 430 .. _RefImage_Object_1.svg: Material properties ------------------------ .. csv-table:: "Concrete material constituting the beam:", "Young's Module :math:`{E}_{b}\mathrm{=}{3.10}^{10}\mathit{Pa}`" "Steel material constituting the cable:", "Young's Module :math:`{E}_{a}\mathrm{=}\mathrm{2,1}{.10}^{11}\mathit{Pa}`" The Poisson's ratio is taken to be equal to 0 for both materials. We therefore cancel the Poisson effects in the :math:`y` and :math:`z` directions. Travel only has components in plan :math:`(x,z)`. Since voltage losses in the cable are neglected, the various parameters used to estimate them are set to 0. Boundary conditions and loads ------------------------------------- Point :math:`A` located at the bottom of the left edge of the beam, with coordinates :math:`(\mathrm{0 };–h\mathrm{/}\mathrm{2 };0)`, is blocked in translation in all three directions and in rotation around the axis :math:`y`. The blocking of the :math:`\mathit{DRY}` degree of freedom of rotation implies a zero slope in the deformation of the mean fiber in :math:`x\mathrm{=}0`. The left end of the cable, with coordinates :math:`(\mathrm{0 };\mathrm{0 };e)`, is blocked in translation in all three directions. A normal force of traction :math:`({F}_{\mathrm{0 }};\mathrm{0 };0)` or :math:`{F}_{0}\mathrm{=}{2.10}^{5}N` is applied to the right end of the cable, with coordinates :math:`(L;\mathrm{0 };e)`.