SLL 08: axial force in eccentric beam

Description

Material Concrete C25/30
Geometry beam b=200mm & h=500mm
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:

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

With \Delta x_{.N_H} the prevented horizontal displacement at the bottom due to the axial force in the beam:

    \[\Delta x_{.N_H} =\int_{0}^{L}\frac{-H}{EA}dx=\frac{-HL}{EA}\]

This results in:

    \[\frac{qL^3h}{24EI}-\frac{-Hh^2L}{4EI}-\frac{-HL}{EA}=0\]

Of which H can be deducted:

    \[H=\frac{qL^3h}{24EI\left ( \frac{h^2L}{4EI}+ \frac{L}{EA}\right )}=147,6kN\]

Diamonds results and comparison

Axial force N in Diamonds

Which result Independent reference Diamonds Difference
Axial force N plate model) 146,2kN 147,6kN -1.0%
Axial force N beam model) 147,6kN 147,6kN 0.0%

References

  • Minne, P. (2024, 26 maart). Structurele analyse van betonen platen en wanden [Presentatieslides 127-134; Pdf].
  • Tested in Diamonds 2024r01.

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