L profiel – BuildSoft Support https://support.buildsoft.eu BuildSoft Support Fri, 03 Oct 2025 08:50:07 +0000 en-US hourly 1 https://wordpress.org/?v=6.2.14 SLL 10: elastic stresses in an L-section https://support.buildsoft.eu/knowledge-base/sll-10-stresses-in-l-section/ Thu, 10 Jul 2025 08:20:47 +0000 https://support.buildsoft.eu/?post_type=ht_kb&p=15390 Description

Geometry L shaped cross-section: B=100mm
H=200mm
t=10mm
r_1=r_2=0mm
Load Design forces: M_{y'}=125 \text{kNm}
About the stresses in Diamonds
Before we get started, we’d like to point out a few (often forgotten) properties about the stresses in Diamonds:

  • The stresses in Diamonds are always elastic stresses, calculated with W_{el}.
    Even for concrete cross-sections! Even for steel cross-section of section classes 1,2 and 4!
  • The stresses are calculated based in axial force N and bending moment M.
    • Where the bounding box of the cross-section intersects with the principal axes, the stresses in those 4 points are either due to N+M_U or due to N+M_V.
      Those 4 points are shown in red for different cross-section shapes:

      The stresses in those 4 points, correspond to the 4 buttons allowing to show stresses in the global results . Note that some of those points aren’t located on the cross-section!
    • When looking at the stresses, you can also request a detailed view of them by selecting the beam and then clicking on the button .
      In the dialog that pops up, you can hoover the mouse over the cross-section. The stresses are shown at the bottom of the dialog. These are stresses due to N+M_U+M_V. Stresses due to shear or torsion, or Von Mises stresses cannot be consulted in Diamonds.
    • When the cross-section is double symmetric, the local axes will coincide with the principal axes, U and V can then be replaced by z’ and y’.
The stresses are determined in points A until H.

  • y’ and z’ are the local axes of the cross-section
  • U and V are the principal axes of the cross-section
  • COG is the centre of gravity of the cross-section
  • The distances [mm] (shown in black in the image below) will be calculated further down this article.

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. Since for this L-section, the principle (U and V) and local axes (y’ and z’) don’t coincide, we’ll transform all coordinates to the principle axes.

Points A until D have the following coordinates in relation the center of gravity:

    \begin{align*} A'&=\begin{bmatrix} -COG_y \\ h-COG_z \end{bmatrix}=\begin{bmatrix} -20.52 \\ 129.48 \end{bmatrix}mm \\ B'&=\begin{bmatrix} b-COG_y \\ -COG_z \end{bmatrix}=\begin{bmatrix} 79.48 \\ -70.52 \end{bmatrix}mm \\ C'&=\begin{bmatrix} -COG_y \\ -COG_z \end{bmatrix}=\begin{bmatrix} -20.52 \\ -70.52 \end{bmatrix}mm \\ D'&=\begin{bmatrix} b-COG_y \\ t-COG_z \end{bmatrix}=\begin{bmatrix} 79.48 \\ -60.52 \end{bmatrix}mm \\ \end{align*}

We transform these coordinates to the principal coordinate system:

    \begin{align*} A&=\begin{bmatrix} cos\left( \alpha \right) & -sin\left( \alpha \right) \\ sin\left( -\alpha \right) & cos (-\alpha) \end{bmatrix} \cdot \begin{bmatrix} A' \end{bmatrix}=\begin{bmatrix} 14.01 \\ 130.35 \end{bmatrix}mm\\ B&=\begin{bmatrix} cos\left( \alpha \right) & -sin\left( \alpha \right) \\ sin\left( -\alpha \right) & cos (-\alpha) \end{bmatrix} \cdot \begin{bmatrix} B' \end{bmatrix}=\begin{bmatrix} 58.31 \\ -88.83 \end{bmatrix}mm\\ C&=\begin{bmatrix} cos\left( \alpha \right) & -sin\left( \alpha \right) \\ sin\left( -\alpha \right) & cos (-\alpha) \end{bmatrix} \cdot \begin{bmatrix} C' \end{bmatrix}=\begin{bmatrix} -38.22 \\ -62.71 \end{bmatrix}mm\\ D&=\begin{bmatrix} cos\left( \alpha \right) & -sin\left( \alpha \right) \\ sin\left( -\alpha \right) & cos (-\alpha) \end{bmatrix} \cdot \begin{bmatrix} D' \end{bmatrix}=\begin{bmatrix} 60.92 \\ -79.17 \end{bmatrix}mm \end{align*}

The coordinates of points E until H can easily be defined based on the coordinates of points A until D in the principal coordinate system:

    \begin{align*} E&=\begin{bmatrix} 0 \\ A_1 \end{bmatrix}=\begin{bmatrix} 0 \\ 130.35 \end{bmatrix}mm \\ F&=\begin{bmatrix} C_0 \\ 0 \end{bmatrix}=\begin{bmatrix} -38.22 \\ 0 \end{bmatrix}mm \\ G&=\begin{bmatrix} D_0 \\ 0 \end{bmatrix}=\begin{bmatrix} 60.92 \\ 0 \end{bmatrix}mm \\ H&=\begin{bmatrix} 0 \\ B_1 \end{bmatrix}=\begin{bmatrix} 0 \\ -88.83 \end{bmatrix}mm \\ \end{align*}

Also the applied bending moment should be transformed to the principal coordinate system:

    \begin{align*} M_V&= M_{y'} \cdot cos \left( 90^\circ - \alpha \right) = 32.64 \text{kNm}\\ M_U&= M_{y'} \cdot cos \left( \alpha \right) = 120.66 \text{kNm}\\ \end{align*}

This results in the following stresses:

  • In point A: M_V gives compression and M_U gives compression as well.
    Compression has a negative sign in Diamonds, tension a positive.

        \begin{align*} \sigma_A&=\frac{- M_V \cdot \left| A_0 \right|}{I_V}+\frac{-M_U \cdot \left| A_1 \right|}{I_U} \\ &= \frac{-32.64 \text{kNm} \cdot \left| -14.01mm \right|}{1378256 mm^4}+\frac{-120.66 \text{kNm} \cdot \left| 130.35 mm\right|}{13073525 mm^4}\\ &=-1534.87 \text{MPa} \end{align*}

    The result is compression.

  • In point B: M_V gives compression and M_U gives tension.

        \begin{align*} \sigma_B&=\frac{- M_V \cdot \left| B_0 \right|}{I_V}+\frac{M_U \cdot \left| B_1 \right|}{I_U} \\ &= \frac{-32.64 \text{kNm} \cdot \left| 58.31mm \right|}{1378256 mm^4}+\frac{-120.66 \text{kNm} \cdot \left| -88.83 mm\right|}{13073525 mm^4}\\ &=-561.22 \text{MPa} \end{align*}

    The result is compression.

  • In point C: M_V gives tension and M_U gives tension as well.

        \begin{align*} \sigma_C&=\frac{M_V \cdot \left| C_0 \right|}{I_V}+\frac{M_U \cdot \left| C_1 \right|}{I_U} \\ &= \frac{32.64 \text{kNm} \cdot \left| -38.22mm \right|}{1378256 mm^4}+\frac{120.66 \text{kNm} \cdot \left| -62.71 mm\right|}{13073525 mm^4}\\ &=1484.07\text{MPa} \end{align*}

    The result is tension.

  • In point D: M_V gives compression and M_U gives tension.

        \begin{align*} \sigma_D&=\frac{- M_V \cdot \left| D_0 \right|}{I_V}+\frac{M_U \cdot \left| D_1 \right|}{I_U} \\ &= \frac{-32.64 \text{kNm} \cdot \left| 90.92mm \right|}{1378256 mm^4}+\frac{120.66 \text{kNm} \cdot \left| -79.17 mm\right|}{13073525 mm^4}\\ &=-712.17\text{MPa} \end{align*}

    The result is compression.

  • In point E: M_U gives compression and M_V does nothing because E is on the neutral line

        \begin{align*} \sigma_E&=\frac{M_U \cdot \left| E_1 \right|}{I_U} \\ &= \frac{-120.66 \text{kNm} \cdot \left| 130.35 mm\right|}{13073525 mm^4}\\ &=-1203.04\text{MPa} \end{align*}

    The result is compression.

  • In point F: M_V gives tension and M_U does nothing because F is on the neutral line

        \begin{align*} \sigma_F&=\frac{M_V \cdot \left| F_0 \right|}{I_V}\\ &= \frac{32.64 \text{kNm} \cdot \left| -38.22 \right|}{1378256 mm^4}\\ &=905.27\text{MPa} \end{align*}

    The result is tension.

  • In point G: M_V gives compression and M_U does nothing because G is on the neutral line

        \begin{align*} \sigma_G&=\frac{-M_V \cdot \left| G_0 \right|}{I_V}\\ &= \frac{-32.64 \text{kNm} \cdot \left| 60.92 \right|}{1378256 mm^4}\\ &=-1442.91\text{MPa} \end{align*}

    The result is tension.

  • In point H: M_U gives tension and M_V does nothing because H is on the neutral line

        \begin{align*} \sigma_H&=\frac{M_U \cdot \left| H_1 \right|}{I_U} \\ &= \frac{120.66 \text{kNm} \cdot \left| -88.83 mm\right|}{13073525 mm^4}\\ &=819.93\text{MPa} \end{align*}

    The result is tension.

Diamonds results and comparison

The tensile stresses \sigma_t due to N+M_v calculated by Diamonds, which corresponds to the stress in point F of the cross-section

Elastic stresses due to N+M_U+M_V in point A calculated by Diamonds. The location of point A can be entered by clicking on x and y.
This is a detailed view of the stresses in the corss-section.

Results Independent reference Diamonds Difference
\sigma_A -1534.87 N/mm² -1534.87 N/mm² 0%
\sigma_B -561.22 N/mm² -561.22 N/mm² 0%
\sigma_C 1484.07 N/mm² 1484.07 N/mm² 0%
\sigma_D -712.17 N/mm² -712.17 N/mm² 0%
\sigma_E -1203.04 N/mm² -1203.04 N/mm² 0%
\sigma_F 905.27 N/mm² 905.27 N/mm² 0%
\sigma_G -1442.91 N/mm² -1442.91 N/mm² 0%
\sigma_H 819.83 N/mm² 819.83 N/mm² 0%

References

  • HIBBELER, R.C. Sterkteleer, Tweede editie.
  • Tested in Diamonds 2026.
]]> 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.
]]> Which connections can be calculated with PowerConnect? https://support.buildsoft.eu/knowledge-base/which-connections-can-be-calculated-with-powerconnect/ Fri, 22 Jun 2018 08:53:29 +0000 http://buildsoftsupport.com/?post_type=ht_kb&p=2850 PowerConnect uses the component method described in Eurocode 3 – part 8 to calculate the connections. BUT the formulas in this method
  • are only valid for certain geometries
    e.g. the connections must be 2D or in tubular connections the angle between the connecting bars is limited
  • are only given for certain types of connections
    e.g. formulas are given for commonly used moment connections. That does not imply all possible moment connections…
  • limit the different cross-section shapes that can be combined in one connection
    e.g. formulas are given for a beam-column connection with H + I cross-sections, but there are no formulas available for that same connection in H- + L- shaped cross-sections.
Therefore it is not possible
  • to calculate whatever connection you have in mind.
    The list below gives an overview of the possible connections:

  • to import/ use whatever section you have in mind.
    How to retrieve which sections I can use /import:

    • Select the desired connection in PowerConnect
    • Double click for example the beam.
    • Click on  to change the section.
      The list you see contains all possible shapes for which this connection can be calculated.
      So if you see only H, and I -shapes in the list, the connection can only be calculated when the beam is an H or I-shape.
      Therefor, if you would add a new H or I-shape to the library, it will be possible to use it in this connection.
      If you would add a T-shape, it will not be visible in the list above and can thus can not be used to calculated this connection.

If your connection cannot be calculated with PowerConnect, you can obt to calculated the connection with IDEA Statica using FEM. BuildSoft has no software supporting the FEM calculation of connections.

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