DS EC 05: Lateral torsional buckling of an H-section

Description

Geometry	Cross-section:	HEA 280
Geometry	Material:	S355
Load	ULS FC:	300kNm
Standard		EN 1993-1-1 [- -]
Design parameters		Section class 3 L_LT=4m

Independent reference results

Open handcalculations

Moment factor C₁

Values for the moment factor C₁ is given in for example the Belgian annex NBN EN 1993-1-1 ANB:2010 Annex D. These are tables, valid for certain “standard” moment distribution. But the loads on a 3D building aren’t always “standard”. Nor is the moment distribution calculated in Diamonds. Which makes it difficult to unambiguously find a value for C₁ based on those tables. Therefor the method suggested by Serna, Lopez, Puente and Young is implemented in Diamonds. For deviating projects, the user can impose C₁ via .

$\begin{align*} M_1&=M_5=0\text{kNm}\\ M_2&=M_4=225\text{kNm}\\ M_3&= M_{max}=300\text{kNm}\\ k&=k_z\\ A_1 &=\frac{M_{max}^{2}+9k \cdot M_2^2+16 \cdot M_3^2 + 9k \cdot M_4^2}{\left( 1+9k+16+9k \right)\cdot M_{max}^{2}} \\ &=\frac{(300 \text{kNm})^{2}+9\cdot 1 \cdot (225 \text{kNm})^2+16 \cdot (300 \text{kNm})^2 + 9 \cdot 1 \cdot (225 \text{kNm})^2}{\left( 1+9 \cdot 1+16+9 \cdot 1 \right)\cdot (300 \text{kNm})^{2}}\\ &=0.78\\ A_2 &=\frac{M_{max} +4 \cdot M_1 + 8 \cdot M_2 + 12 \cdot M_3 + 8 \cdot M_4 + 4 \cdot M_5}{37\cdot M_{max}}\\ &=\frac{300 \text{kNm} +4 \cdot 0 \text{kNm} + 8 \cdot 225\text{kNm} + 12 \cdot 300 \text{kNm} + 8 \cdot 225\text{kNm} + 4 \cdot 0 \text{kNm}}{37\cdot 300 \text{kNm}}\\ &=0.6\\ C_1 &=\frac{\sqrt{\sqrt{k} \cdot A_1 + \left(0.5 \cdot \left( 1-\sqrt{k} \right) \cdot A_2 \right)^2 }+0.5 \cdot \left( 1 - \sqrt{k} \right)\cdot A_2}{A_1}\\ &=\frac{\sqrt{\sqrt{1} \cdot 0.78 + \left(0.5 \cdot \left( 1-\sqrt{1} \right) \cdot 0.68 \right)^2 }+0.5 \cdot \left( 1 - \sqrt{1} \right)\cdot 0.68}{0.78}\\ &=1.136 \end{align*}$

Elastic critical moment M_cr

The formula for the elastic critical moment is given in the Belgian annex NBN EN 1993-1-1 ANB:2010 Annex D. Diamonds assumes that the shear center coincides with the center of gravity of the cross-section (= the profile is doubly symmetrical) and that the loads act in the shear center of the cross-section. For deviating projects, the user can impose M_cr via .

$\begin{align*} k_w&=k_z=1\\ M_{cr} &=C_1 \frac{\pi^2 E I_z}{(k_z L)^2} \sqrt{\left( \frac{k_z}{k_w}\right)^2 \frac{I_w}{I_z}+\frac{(k_z L)^2 G I_t}{\pi^2 E I_z}}\\ &=1.136 \frac{\pi^2 \cdot 210000 \text{MPa} \cdot 4762cm^4}{(1 \cdot 4m)^2} \sqrt{\left( \frac{1}{1}\right)^2 \frac{785366cm^6}{4762 cm^4}+\frac{(1 \cdot 4m)^2 \cdot 80769 \text{MPa} \cdot 62 cm^4}{\pi^2 \cdot 210000 \text{MPa} \cdot 4762 cm^4}}\\ &=1099.6 \text{kNm} \end{align*}$

Resistance against lateral torsional buckling M_b.Rd

(6.56) $\begin{align*} \bar{\lambda_{LT}} &= max\left( \sqrt{\frac{W_y \cdot f_y}{M_{cr}}},0.2 \right)\\ &=max\left( \sqrt{\frac{1012.92cm^3 \cdot 355\text{MPa}}{1099.6 \text{kNm}}},0.2 \right)\\ &=0.572\\ \alpha_{LT}&=0.21 \text{EN 1993-1-1 Table 6.3 and 6.4} \\ \varPhi_{LT} &=0.5 (1+ \alpha_{LT}\cdot (\bar{\lambda_{LT}}-0.2)+\bar{\lambda_{LT}}^2)\\ &=0.5 (1+ 0.21 \cdot (0.572-0.2)+0.572^2)\\ &=0.702\\ \chi_{LT} &=min\left( \frac{1}{\varPhi_{LT}+\sqrt{\varPhi_{LT}^2-\bar{\lambda_{LT}}^2}},1 \right) \\ &= min\left( \frac{1}{0.702+\sqrt{0.702^2-0.572^2}},1 \right)\\ &=0.900\\ M_{b.Rd}&=\varPhi_{LT} \cdot W_y \cdot f_{yd} \tag{6.55}\\ &=0.900 \cdot 1013 cm^3 \cdot 355 \text{MPa} \\ &=323.7\text{kNm}\\ \end{align*}$

Unity check

(6.54) $\begin{align*} M_{b.Rd}&=\frac{M_y}{M_{b.Rd}} = \frac{300 \text{kNm}}{323.7 \text{kNm}}\\ &= 92.7% \end{align*}$

Diamonds results and comparison

Stability verification of the beam according to EN 1993-1-1

Intermediate results in Diamonds for lateral torsional buckling

Results	Independent reference	Diamonds	Difference
$C_1$	1.136	1.136	0,00%
$M_{cr}$	1099.6kNm	1099.6kNm	0,00%
$\bar{\lambda_{LT}}$	0.572	0.572	0,00%
$\varPhi_{LT}$	0.702	0.702	0,00%
$\chi_{LT}$	0.900	0.900	0,00%
$M_{b.Rd}$	323.7kNm	323.8kNm	0,00%
Unity check $\frac{M_y}{M_{b.Rd}}$	92.7%	92.7%	0,00%