2. Reference solution#

The reference calculations are carried out using a simple elastic calculation in RdM.

2.1. Elastic center#

In simple bending, for elastic behavior, the neutral axis passes through the elastic center (barycenter of the sections weighted by the modules of the materials):

\(C\) such as \({\mathrm{\int }}_{S}E\overrightarrow{\mathit{CM}}\mathit{dS}\mathrm{=}\overrightarrow{0}\)

First, we determine the position of the centers of gravity of the concrete alone

_images/1000018E0000018D0000017216A7F0FABA3D7859.svg

And steel alone

_images/1000018E0000018D0000017233AD886678DD09D8.svg

in relation to point \(O\).

\(\begin{array}{c}{y}_{{G}_{b}}\mathrm{=}\frac{0.125\mathrm{\times }0.3\mathrm{\times }0.05\mathrm{-}0.125\mathrm{\times }0.2\mathrm{\times }0.05}{0.2\mathrm{\times }0.05+0.1\mathrm{\times }0.2+0.3\mathrm{\times }0.05}\mathrm{=}1.38888{10}^{\mathrm{-}2}m\\ {y}_{{G}_{a}}\mathrm{=}\frac{0.125\mathrm{\times }3\mathrm{-}0.125\mathrm{\times }4}{3+4}\mathrm{=}\mathrm{-}1.78571{10}^{\mathrm{-}2}m\\ {z}_{{G}_{a}}\mathrm{=}{z}_{{G}_{b}}\mathrm{=}0m\end{array}\)

We can then determine the position with respect to \(O\) of the elastic center \(C\).

\(\overrightarrow{\mathit{OC}}\mathrm{=}\frac{{E}_{a}{S}_{a}\overrightarrow{{\mathit{OG}}_{a}}+{E}_{b}{S}_{b}\overrightarrow{{\mathit{OG}}_{b}}}{{E}_{a}{S}_{a}+{E}_{b}{S}_{b}}\)

Concrete section \({S}_{b}\) is \(\mathrm{0,045}{m}^{2}\) and steel section \({S}_{a}\) is \({7.10}^{\mathrm{-}4}{m}^{2}\). The Young’s modulus of concrete is \({2.10}^{10}\mathit{MPa}\) and that of steel is \({21.10}^{10}\mathit{MPa}\). So we have

\(\begin{array}{c}{y}_{c}\mathrm{=}\frac{2\mathrm{\times }0.045\mathrm{\times }1.38888\mathrm{-}21\mathrm{\times }7{10}^{\mathrm{-}4}\mathrm{\times }1.78571}{2\mathrm{\times }0.045+21\mathrm{\times }7{10}^{\mathrm{-}4}}\mathrm{=}0.94317{10}^{\mathrm{-}2}m\\ {z}_{c}\mathrm{=}0m\end{array}\)

2.2. Quadratic moments#

The quadratic moments of the rectangular sections of concrete are calculated by the following formula:

\(\frac{b{h}^{3}}{12}+b\mathrm{\times }h\mathrm{\times }{d}^{2}\)

Where, b represents the width, h the height, and d the distance of the center of gravity of the section from the axis for which the moment is calculated.

We then obtain the quadratic moment of the concrete section with respect to the \(z\) axis passing through the elastic center:

\(\begin{array}{cc}{I}_{\mathit{béton}}& \mathrm{=}\frac{0.3\mathrm{\times }{0.05}^{3}}{12}+(0.3\mathrm{\times }0.05)\mathrm{\cdot }{(0.125\mathrm{-}0.94317\mathrm{\cdot }{10}^{\mathrm{-}2})}^{2}+\frac{0.1\mathrm{\times }{0.2}^{3}}{12}+(0.1\mathrm{\times }0.2)\mathrm{\cdot }{(0.94317\mathrm{\cdot }{10}^{\mathrm{-}2})}^{2}\\ & +\frac{0.2\mathrm{\times }{0.05}^{3}}{12}+(0.2\mathrm{\times }0.05)\mathrm{\cdot }{(0.125+0.94317\mathrm{\cdot }{10}^{\mathrm{-}2})}^{2}\mathrm{=}0.4547\mathrm{\cdot }{10}^{\mathrm{-}3}{m}^{4}\end{array}\)

The inertias of steels are calculated by the following formula:

\(\frac{\pi {\phi }^{4}}{64}+S\mathrm{\times }{d}^{2}\mathrm{\approx }S\mathrm{\times }{d}^{2}\)

Where, \(\phi\) represents the diameter of the steel, \(S\) the steel section, and d is the distance of the center of gravity of the section from the axis for which the moment is calculated. Since the diameter of steels is small, the first term is overlooked.

We then obtain the quadratic moment of the steel sections with respect to the \(z\) axis passing through the elastic center:

\(3{10}^{\mathrm{-}4}\mathrm{\times }{(0.125\mathrm{-}0.94317{10}^{\mathrm{-}2})}^{2}+4{10}^{\mathrm{-}4}\mathrm{\times }{(0.125+0.94317{10}^{\mathrm{-}2})}^{2}\mathrm{=}0.1124{10}^{\mathrm{-}4}{m}^{4}\)

For the full section of the beam, the quadratic moment weighted by the Young’s moduli of the materials is:

\(\mathit{EI}\mathrm{=}2{10}^{10}\mathrm{\times }0.4547{10}^{\mathrm{-}3}+21{10}^{10}\mathrm{\times }0.1124{10}^{\mathrm{-}4}\mathrm{=}11.4544{10}^{6}\mathit{Pa}\mathrm{.}{m}^{4}\)

2.3. Charging case 1#

For load case 1 (load concentrated in the middle of the beam), the deflection is calculated by the following RDM formula:

\(f\mathrm{=}\frac{F\mathrm{\times }{l}^{3}}{48\mathit{EI}}\)

Which gives the arrow:

\(f\mathrm{=}\frac{10000\mathrm{\times }{5}^{3}}{48\mathrm{\times }11.4544{10}^{6}}\mathrm{=}2.2735{10}^{\mathrm{-}3}m\)

The following generalized efforts can also be calculated:

  • the shear force at the start of the beam (left part) is worth \(\frac{F}{2}\mathrm{=}5000N\),

  • the bending moment in the middle of the beam is equal to: \(\frac{F\mathrm{\times }l}{4}\mathrm{=}1.25{10}^{4}\mathit{N.m}\).

2.4. Charging case 2#

For load case 2 (self-weight of the beam), the deflection is calculated by the following material strength formula:

\(f\mathrm{=}\frac{5\mathrm{\times }p\mathrm{\times }{l}^{4}}{384\mathit{EI}}\)

where \(p\) is the linear load due to the weight of the materials:

\(p\mathrm{=}g({\rho }_{a}{S}_{a}+{\rho }_{b}{S}_{b})\mathrm{=}9.8\mathrm{\times }(2800\mathrm{\times }7{10}^{\mathrm{-}4}+2400\mathrm{\times }0.045)\mathrm{=}1111.9{\mathit{N.m}}^{\mathrm{-}1}\)

Which gives the arrow:

\(f\mathrm{=}\frac{5\mathrm{\times }1111.9\mathrm{\times }{5}^{4}}{384\mathrm{\times }11.4544{10}^{6}}\mathrm{=}7.9{10}^{\mathrm{-}4}m\)

2.5. Charging case 3#

For load case 3 (homogeneous temperature rise), the beam being isostatic and the expansion coefficients of concrete and steel being identical, the solution is simple:

Generalized constraints and efforts are zero.

The elongation of the beam is: \(\Delta l\mathrm{=}\alpha \mathrm{\times }l\mathrm{\times }\Delta T\)

This gives with the values in our case:

\(\Delta l\mathrm{=}{10}^{\mathrm{-}5}\mathrm{\times }5\mathrm{\times }100\mathrm{=}5.{10}^{\mathrm{-}3}m\)