asymmetrsiche buiging – BuildSoft Support https://support.buildsoft.eu BuildSoft Support Mon, 17 Mar 2025 10:27:30 +0000 en-US hourly 1 https://wordpress.org/?v=6.2.14 SLL 09: unsymmetric bending https://support.buildsoft.eu/knowledge-base/sll-09-unsymmetric-bending/ Mon, 17 Mar 2025 10:27:29 +0000 https://support.buildsoft.eu/?post_type=ht_kb&p=14960 Description

Geometry L shaped cross-section: B=75mm
H=150mm
t=9mm
\alpha=15^\circ
Span: 2m
Material: S235
Load Design forces: q=10 \text{kN}

Independent reference results

Open handcalculations
As loads always follow the path of greatest stiffness, they will follow the principle axes of the cross-section, not the path of the local axes.

  • When a cross-section is symmetrical, the principle and local axes coincide. A vertical load, will lead to only vertical deformation.
  • But in an unsymmetric cross-section, the principle and local axes don’t coincide. A vertical load will load to both vertical and horizontal deformation. The angle \alpha is the angle between the local and principle axes.

The L-shaped cross-section is an example of the latter case. We’ll need the split up the loads according to the principle axes.

    \begin{align*} q_V &=q \cdot cos(15^\circ)=9.659 \text{kN/m}\\ q_U &=q \cdot sin(15^\circ)=2.588 \text{kN/m}\\ \end{align*}

Now we can calculate the deformation parallel to the principle axis V \delta_V and axis U \delta_U. Both deformations cannot be consulted in Diamonds. Their vector sum \delta_{xyz} can be consulted in Diamonds.

    \begin{align*} \delta_V &=\frac{5}{384}\frac{q_V \cdot L^4}{E \cdot I_U}\\ &=\frac{5}{384}\frac{9.659 \text{kN/m} \cdot \left( 2m \right)^4}{210000 \text{MPa} \cdot 4877504mm^4}\\ &= 1.965mm\\ \delta_U &=\frac{5}{384}\frac{q_U \cdot L^4}{E \cdot I_V}\\ &=\frac{5}{384}\frac{2.588 \text{kN/m} \cdot \left( 2m \right)^4}{210000 \text{MPa} \cdot 512796mm^4}\\ &= 5.007mm\\ \delta_{xyz}&=\sqrt{\delta_U^2 + \delta_V^2}\\ &=\sqrt{\left(1.965mm \right)^2 + \left(5.007mm \right)^2}\\ &=5.379mm \end{align*}

The deformation parallel to the global axis Y and global axis X can be deducted from \delta_V and \delta_U.

    \begin{align*} \delta_Y &= \delta_U \cdot sin(15^\circ) + \delta_V \cdot sin(75^\circ) = 3.194mm \\ \delta_X &= - \delta_U \cdot cos(15^\circ) + \delta_V \cdot cos(75^\circ) = -4.328mm \\ \end{align*}

Diamonds results and comparison

Deformations \delta_{xyz}, \delta_{Y}, \delta_{X} calculated by Diamonds

Results Independent reference Diamonds Difference
\delta_{xyz} 5.389mm 5.376mm -0.05%
\delta_Y 3.194mm 3.192mm -0.05%
\delta_X -4.328mm -4.326mm -0.05%

References

  • HEARN, E. J. (z.d.). Mechanics of Materials 1: An introduction to the mechanics of elastic and plastic deformation of solids and structural materials.
  • VAN IMPE, R., Berekening van Bouwkundige Constructies I , Chapter 4.1
  • Tested in Diamonds 2026.
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