DS EC 06: Buckling y’ and lateral torsional buckling (Method 1) of an I-section

Description

Geometry	Cross-section:	IPE 500
Geometry	Material:	S235
Load	ULS FC:	$M_y=200 \text{kNm}$ $M_z=12.5 \text{kNm}$ $N=500 \text{kN}$
Standard		EN 1993-1-1 [- -]
Design parameters		Section class 1 for pure bending L_y=L_z=3.75m L_LT=3.75m

This article verifies the interaction of buckling and lateral torsional buckling due to axial force and bending around the strong axis. Methode 1 in Eurocode (EN 1993-1-1 Annex A) is used.

Independent reference results

Open handcalculations for individual buckling and LTB

To perform the check “Uniform member in bending and axial compression” according to EN 1993-1-1 §6.3.3, we need a few factors related to buckling resistance. We’ll determine those here without much context. The context is already can be found here.

Buckling around the strong axis y’

(6.50) $\begin{align*} N_{cr.y}&=\frac{\pi^2\cdot E\cdot I_y}{l_{cr.y}^2}=71042.7\text{kN}\\ \overline{\lambda_y}&=\sqrt{\frac{A \cdot f_y}{N_{cr.y}}} =0.195\\ \chi_{y}&=1 \end{align*}$

Because the relative slenderness $\overline{\lambda_y}$ is smaller than 0.2, $\chi_{y}$ equals 1.

Buckling around the weak axis z’

(6.50) $\begin{align*} N_{cr.z}&=\frac{\pi^2\cdot E\cdot I_z}{l_{cr.z}^2}=3156.6\text{kN}\\ \overline{\lambda_z}&=\sqrt{\frac{A \cdot f_y}{N_{cr.z}}}=0.93 \\ \phi_z&=0.5\left( 1+\alpha_z\left(\overline{\lambda_z}-0.2 \right) + \overline{\lambda_z}^2 \right) =1.05\\ \chi_z&=min\left( \frac{1}{\phi_z+\sqrt{\phi_z^2-\overline{\lambda_z}^2}};1 \right)=0.64 \end{align*}$

Torsional buckling (buckling around the x’ axis)

$\begin{align*} N_{cr.T}&=\frac{A}{I_y+I_z} \left( G \cdot I_t + \frac{\pi^2 \cdot E \cdot I_w}{L_{LT}^2} \right)=5880.5\text{kN} \end{align*}$

Lateral torsional buckling

We only included the intermediar results. If you want to learn more about the resistance again laterial torsional buckling, follow this example.

$\begin{align*} M_1&=M_5=100\text{kNm}\\ M_2&=M_4=-124.1\text{kNm}\\ M_3&= M_{max}=-198.8\text{kNm}\\ k&=k_z=1\\ A_1 &=\frac{M_{max}^{2}+9k \cdot M_2^2+16 \cdot M_3^2 + 9k \cdot M_4^2}{\left( 1+9k+16+9k \right)\cdot M_{max}^{2}} =0.69\\ A_2 &=\frac{M_{max} +4 \cdot M_1 + 8 \cdot M_2 + 12 \cdot M_3 + 8 \cdot M_4 + 4 \cdot M_5}{37\cdot M_{max}}=0.51\\ C_1 &=\frac{\sqrt{\sqrt{k} \cdot A_1 + \left(0.5 \cdot \left( 1-\sqrt{k} \right) \cdot A_2 \right)^2 }+0.5 \cdot \left( 1 - \sqrt{k} \right)\cdot A_2}{A_1}=1.21\\ M_{cr}&=C_1 \frac{\pi^2 E I_z}{(k_z L)^2} \sqrt{\left( \frac{k_z}{k_w}\right)^2 \frac{I_w}{I_z}+\frac{(k_z L)^2 G I_t}{\pi^2 E I_z}}=1085 \text{kNm}\\ \overline{\lambda_{LT}} &= max\left( \sqrt{\frac{W_y \cdot f_y}{M_{cr}}},0.2 \right)=0.689\\ \varPhi_{LT} &=0.5 (1+ \alpha_{LT}\cdot (\bar{\lambda_{LT}}-0.2)+\bar{\lambda_{LT}}^2)=0.821\\ \chi_{LT} &=min\left( \frac{1}{\varPhi_{LT}+\sqrt{\varPhi_{LT}^2-\bar{\lambda_{LT}}^2}},1 \right)=0.790\\ \end{align*}$

Open handcalculations for interaction

Moment factor C_mi.0

The general formula for combined end moment and transverse loads is used to determine $C_{my.0}$ :

$\begin{align*} \delta_y&=2.59mm \\ C_{my.0}&=1+\left( \frac{\pi^2\cdot E\cdot I_y \cdot \delta_y }{L^2 \cdot M_{y.Ed}} -1 \right)\frac{N_{Ed}}{N_{cr.y}}\\ &=1+\left( \frac{\pi^2\cdot 210\text{GPa}\cdot 482017810mm^4 \cdot 2.59mm }{\left( 3.75m \right)^2 \cdot 200 \text{kNm}} -1 \right)\frac{500\text{kN}}{71042.7\text{kN}} \\ &= 0.999 \end{align*}$

The formula for linearly distributed bending moment is used to determine $C_{mz.0}$ :

$\begin{align*} \Psi_z&=\frac{M_{z.Ed.right}}{M_{z.Ed.left}}=\frac{0}{25 \text{kNm}}=0 \\ C_{mz.0}&=0.79+0.21 \cdot \Psi_z + 0.36 \left( \Psi_z -0.33 \right) \frac{N_{Ed}}{N_{cr.z}} \\ &=0.79+0.21 \cdot 0 + 0.36 \left( 0 -0.33 \right) \frac{500 \text{kN}}{3156.6 \text{kN}}\\ &=0.771 \end{align*}$

Equivalent moment factors C_mi

$\begin{align*} M_{cr.0} &= \sqrt{\frac{\pi^2\cdot E \cdot I_z}{L_{LT}^2} \left( G \cdot I_t + \frac{\pi^2 E \cdot I_w}{L_{LT}^2} \right)} \\ &= \sqrt{\frac{\pi^2 \cdot 210 \text{GPa} \cdot 21417007 mm^4}{\left( 3.75m \right)^2} \left( 80769 \text{MPa} \cdot 892870 mm^4 + \frac{\pi^2 \cdot 210 \text{GPa} \cdot 1.2494 \cdot 10^{12} mm^6}{\left( 3.75m \right)^2} \right)} \\ &=899.4 \text{kNm}\\ \overline{\lambda_0}&=\sqrt{\frac{W_{pl.y} \cdot f_y}{M_{cr.0}}} \\ &=\sqrt{\frac{2194261mm^3 \cdot 235 \text{MPa}}{899.4 \text{kNm}}} \\ &= 0.757\\ \overline{\lambda_{0.lim}}&=0.2 \sqrt{C_1}\sqrt[4]{\left( 1-\frac{N_{Ed}}{N_{cr.z}} \right) \cdot \left( 1-\frac{N_{Ed}}{N_{cr.T}} \right)} \\ &=0.2 \sqrt{1.201}\sqrt[4]{\left( 1-\frac{500 \text{kN}}{3156.6\text{kN}} \right) \cdot \left( 1-\frac{500 \text{kN}}{5880.5\text{kN}} \right)}\\ &=0.206 \end{align*}$

If $\overline{\lambda_0}$ < $\overline{\lambda_{0.lim}}$ or if $I_t \ge I_y$ the cross-section is “torsion rigid”. Otherwise the cross-section is “torsion weak”. Different formula apply for “torsion rigid” and “torsion weak”. Since the conditions are not met for this cross-section, it is “torsion-weak”.

$\begin{align*} \varepsilon_y&=\frac{M_{y.Ed}}{N_{Ed}} \cdot \frac{A}{W_{y.el}}=\frac{200 \text{kNm}}{500 \text {kN}} \cdot \frac{11552.8mm^2}{1928071 mm^3}\\ &=2.4\\ a_{lt}&=max\left( 1-\frac{I_t}{I_y};0 \right)=max\left( 1-\frac{892870 mm^4}{482017810mm^4};0 \right)\\ &=0.998\\ C_{my}&=C_{my.0} + \left( 1- C_{my.0} \right) \cdot \frac{\sqrt{\varepsilon_y} \cdot a_{lt}}{1 + \sqrt{\varepsilon_y} \cdot a_{lt}} \\ &=0.999 + \left( 1- 0.999 \right) \cdot \frac{\sqrt{2.4} \cdot 0.998}{1 + \sqrt{2.4} \cdot 0.998}\\ &=1\\ C_{mz}&=C_{mz.0}\\ C_{mLT}&=max\left( C_{my}^2 \cdot \frac{a_{lt}}{\sqrt{\left( 1-\frac{N_{Ed}}{N_{cr.z}} \right) \cdot\left( 1-\frac{N_{Ed}}{N_{cr.T}} \right)}},1 \right) \\ &=max\left( 1^2 \cdot \frac{0.998}{\sqrt{\left( 1-\frac{500 \text{kN}}{3156.6\text{kN}} \right) \cdot\left( 1-\frac{500 \text{kN}}{5880.4 \text{kN}} \right)}},1 \right) \\ &=1.137 \end{align*}$

Plasticity factors C_ij

$\begin{align*} b_{LT}&= 0.5 \cdot a_{lt} \cdot \overline{\lambda_0}^2 \cdot \frac{M_{y.Ed}}{\chi_{LT} \cdot W_{y.pl} \cdot f_{yd}} \cdot \frac{M_{z.Ed}}{W_{z.pl} \cdot f_{yd}}\\ &= 0.5 \cdot 0.998 \cdot 0.757^2 \cdot \frac{200 \text{kNm}}{0.79 \cdot 2194261 mm^3 \cdot 235 \text{MPa}} \cdot \frac{12.5\text{kNm}}{335887mm^3 \cdot 235 \text{MPa}}\\ &= 0.022\\ c_{LT}&= 10 \cdot a_{lt} \cdot \frac{\overline{\lambda_0}^2}{5+\overline{\lambda_z}^4} \cdot \frac{M_{y.Ed}}{C_{my} \cdot \chi_{LT} \cdot W_{y.pl} \cdot f_{yd}} \\ &= 10 \cdot 0.998 \cdot \frac{0.757^2}{5+0.93^4} \cdot \frac{200 \text{kNm}}{1 \cdot 0.79 \cdot 2194261mm^3 \cdot 235 \text{MPa}} \\ &= 0.49\\ w_y&= min\left( \frac{W_{y.pl}}{W_{y.el}};1.5 \right)= min\left( \frac{2194261 mm^3}{1928071 mm^3};1.5 \right) \\ &= 1.138\\ w_z&= min\left( \frac{W_{z.pl}}{W_{z.el}};1.5 \right)= min\left( \frac{335887 mm^3}{214170 mm^3};1.5 \right) \\ &= 1.5 \\ C_{yy}&=max\left(1+ \left( w_y-1 \right) \cdot \left( \left( 2-\frac{1.6 \cdot C_{my}^2 \cdot \overline{\lambda_{max}}}{w_y} - \frac{1.6 \cdot C_{my}^2 \cdot \overline{\lambda_{max}}^2}{w_y} \right) \cdot n_{pl}- b_{LT} \right) ; \frac{W_{y.pl}}{W_{y.el}}\right) \\ &= 0.984 \\ C_{yz}&=max\left(1+ \left( w_z-1 \right) \cdot \left( \left( 2-\frac{14 \cdot C_{mz}^2 \cdot \overline{\lambda_{max}}^2}{w_z^5} \right) \cdot n_{pl}- c_{LT} \right) ; 0.6 \cdot \sqrt{\frac{w_z}{w_y}} \cdot \frac{W_{y.el}}{W_{y.pl}}\right) \\ &= 0.852\\ \end{align*}$

Interaction factors k_ij

$\begin{align*} \mu_y&= \frac{1-\frac{N_{Ed}}{N_{cr.y}}}{1- \chi_y \cdot \frac{N_{Ed}}{N_{cr.y}}} = \frac{1-\frac{N_{Ed}}{N_{cr.y}}}{1- 1 \cdot\frac{N_{Ed}}{N_{cr.y}}} =1 \\ k_{yy}&= C_{my} \cdot C_{mLT} \cdot \frac{\mu_y}{1-\frac{N_{Ed}}{N_{cr.y}}} \cdot \frac{1}{C_{yy}} \\ &= 1 \cdot 1.137 \cdot \frac{1}{1-\frac{500 \text{kN}}{71042\text{kN}}} \cdot \frac{1}{0.984} \\ &= 1.163\\ k_{yz}&= C_{mz} \cdot \frac{\mu_y}{1-\frac{N_{Ed}}{N_{cr.z}}} \cdot \frac{1}{C_{yz}} \cdot 0.6 \cdot \sqrt{\frac{w_z}{w_y}} \\ &=C_{mz} \cdot \frac{1}{1-\frac{500 \text{kN}}{3156.6 \text{kN}}} \cdot \frac{1}{0.852} \cdot 0.6 \cdot \sqrt{\frac{1.5}{1.138}} \\ &= 0.740\\ \end{align*}$

Unity check

$\begin{align*} UC&=\frac{N_{Ed}}{\chi_y \cdot N_{Rd}} + k_{yy} \cdot \frac{M_{y.Ed}}{\chi_{LT} \cdot M_{y.Rd}}+ k_{yz} \cdot \frac{M_{z.Ed}}{M_{z.Rd}}\\ &=\frac{500 \text{kN}}{1 \cdot 2714.9\text{kN}} + 1.163 \cdot \frac{200 \text{kNm}}{0.79 \cdot 515.7 \text{kNm}}+ 0.740 \cdot \frac{12.5 \text{kNm}}{78.9 \text{kNm}}\\ &=87.2\% \end{align*}$

Diamonds results and comparison

Intermediate results for lateral torsional buckling calculated by Diamonds (EN 1993-1-1 Method 1)

Results	Independent reference	Diamonds	Difference
$C_{my}$	1.000	1.000	0%
$C_{mz}$	0.771	0.771	0%
$C_{mLT}$	1.137	1.137	0%
$C_{yy}$	0.984	0.984	0%
$C_{yz}$	0.852	0.854	0.24%
$k_{yy}$	1.164	1.164	0%
$k_{yz}$	0.739	0.739	0%
Unity check	87.2%	86.9%	-0,35%

References

EN 1993-1-1: 2005+AC:2009
NBN EN 1993-1-1 ANB:2010 Annex D
Boissonnade, N., Greiner, R., Jaspart, J., & Lindner, J. (2006). Rules for member stability in EN 1993-1-1 background documentation and design guidelines
This example is inspired by “Worked example 5”, but the results are not exactly the same as in the book. The book does the verification with the maximum force internal forces, even though these internal forces do not occur in the same cross-section. While Diamonds does, resulting in M_{z.Ed} being lower. The book also uses $\Chi_{LT.mod}$ while Diamonds uses $\Chi_{LT}$ . By consequence, Diamonds is more conservative.

Tested in Diamonds 2025.

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