DS EC 07: Buckling y’ and lateral torsional buckling (Method 2) 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 2 in Eurocode (EN 1993-1-1 Annex B) 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 factors C_mi

For the bending moment $M_{y'}$ (EN 1993-1-1 Table B.3):

$\begin{align*} M_s&=-198.8\text{kNm} \\ M_h&=100.0\text{kNm} \\ \alpha_{h} &=\frac{M_h}{M_s}=\frac{100 \text{kNm}}{-198.8 \text{kNm}}=-0.5\\ \psi_{y}&=1\\ C_{my}&=0.95 + 0.95 \cdot \alpha_{h}\\ &=0.95 + 0.95 \cdot -0.5\\ &= 0.925 \end{align*}$

For the bending moment $M_{z'}$ (EN 1993-1-1 Table B.3):

$\begin{align*} \psi_{z}&=0\\ C_{mz}&=max(0.6+0.4 \cdot \psi_{z},0.4)=max(0.6+0.4 \cdot 0,0.4)\\ &= 0.6 \end{align*}$

Elastic critical moment M_cr.0

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

If $\overline{\lambda_0}$ < $\overline{\lambda_{0.lim}}$ 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”.

Interation factors k_ij

$\begin{align*} k_{yy}&= C_{my} \cdot \left( 1+\left( \overline{\lambda_{y}}-0.2 \right) \frac{N_{Ed}}{\chi_y \cdot N_{Rd}}\right) \\ &= 0.925 \cdot \left( 1+\left( 0.196-0.2 \right) \frac{500 \text{kN}}{1 \cdot 2726 \text{kN}}\right) \\ &= 0.924\\ k_{yz}&= 0.6 \cdot C_{mz} \cdot \left( 1+\left(2 \cdot \overline{\lambda_{z}}-0.2 \right) \frac{N_{Ed}}{\chi_z \cdot N_{Rd}}\right)\\ &= 0.6 \cdot 0.6 \cdot \left( 1+\left(2 \cdot 0.929-0.2 \right) \frac{500 \text{kN}}{\chi_z \cdot 2726 \text{kN}}\right)\\ &= 0.489\\ \end{align*}$

Unity check

$\begin{align*} UC&=\frac{N_{Ed}}{\chi_y \cdot N_{Rd}} + k_{yy} \cdot \frac{M_{y.Ed}}{\chi_{LT} \cdot M_{y.Rd}}+ k_{yz} \cdot \frac{M_{z.Ed}}{M_{z.Rd}}\\ &=\frac{500 \text{kN}}{1 \cdot 2714.9\text{kN}} + 0.924 \cdot \frac{200 \text{kNm}}{0.79 \cdot 515.7 \text{kNm}}+ 0.489 \cdot \frac{12.5 \text{kNm}}{78.9 \text{kNm}}\\ &=71.5\% \end{align*}$