SLL 10: elastic stresses in an L-section

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 and bending moment .
- 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%