SLL 09: unsymmetric bending

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 2025.

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