spanningen – 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.
]]> Stresses versus steel verification https://support.buildsoft.eu/knowledge-base/stresses-versus-steel-verification/ Tue, 12 Feb 2019 16:47:24 +0000 http://buildsoftsupport.com/?post_type=ht_kb&p=3955 The image below shows the tensile stress and the steel verification for 3 different beams:

  • For Beam 1: based on the stresses, you’d expect a steel verification of 100%. But the steel verification is lower.
    This is because Diamonds shows the elastic stresses (calculated with Wel), while this section is of class 1 and thus plastic properties (Wpl > Wel) will be used in the steel verification.
  • For Beam 2: based on the stresses, you’d expect a steel verification of 100%. And indeed, the steel verification is 100%.
    This is because Diamonds shows the elastic stresses and this section is of class 3 (Wel) will be used in the steel verification.
  • For Beam 3: based on the stresses, you’d expect a steel verification of 100%, but the steel verification is higher.
    This is because Diamonds shows the elastic stresses (calculated with Wel), while this section is of class 4 and thus effective properties (Weff < Wel) will be used in the steel verification.

So, it is important to do a steel verification because:

  • the stresses shown in the global window in Diamonds, are stresses caused by axial force and bending around the z’ OR y’-axis.
    It neglects shear, torsion, the risk of instabilities and combinated forces (like N + V + M or N + M + M).
  • Diamonds shows elastic stresses.

Note: it is not possible to show the stresses according to the Von mises criterium in Diamonds.

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