SLL 08: axial force in eccentric beam

Description

Material		Concrete C25/30
Geometry	beam	$b=200mm$ & $h=500mm$
Geometry	one way plate	$t=200mm$
Loads	Self-weight beam & plate	resulting in a total line load $q=23,6kN/m$

Independent reference results

Open handcalculations

The horizontal deformations of the beam at the bottom are zero:

$\Delta x_{at.the.bottom}=\Delta x_{.M_q}+\Delta x_{.M_H}+\Delta x_{.N_H}=0$

With $\Delta x_{.M_q}$ the horizontal displacement at the bottom due to the bending moment in the beam:

$\Delta x_{.M_q}=\int_{0}^{L} \varepsilon _{x.M_q}dx=\int_{0}^{L} \frac{\sigma _{x.M_q}}{E}dx=\int_{0}^{L/2} \frac{M_{p(x)} \frac{h}{2}}{E}dx=\int_{0}^{L/2} \frac{\left ( \frac{qL}{2} x - \frac{qx^2}{2} \right )h}{E}dx$

$=\int_{0}^{L/2} \frac{\left ( \frac{qL}{2} x - \frac{qx^2}{2} \right )h}{E}dx=\frac{h}{EI}\left [ \frac{qL^3}{16}-\frac{qL^3}{48} \right ]=\frac{qL^3h}{24EI}$

Because the horizontal reaction force is at the bottom of the beam, it causes a bending moment round the neutral axis. Also this bending moment will cause horizontal displacement:

$\Delta x_{.M_H}=\int_{0}^{L} \frac{\left ( -H \frac{h}{2} \right )\frac{h}{2}}{EI}dx=\frac{-Hh^2L}{4EI}$