8.1 Stability

We will perform some stability analyses for flexural buckling and lateral-torsional buckling for a column with double-bending, considering the interaction conditions.

Table 8.0 System and design loads (γ times)

Design values of static loads Nd = 300 kN qz,d = 5 kN/m Fy,d = 7.5 kN

The design is performed for all x-locations (see Chapter 4.5) of the equivalent member. The governing location is x = 2.00 m. RFEM or RSTAB determines the following internal forces:

$N_{cr,z}=\frac{21000·889.00·{\pi }^2}{400.{00}^2}=1151.60 \mathrm{kN}$

${\overline{\lambda }}_z=\sqrt{\frac{A·f_y}{N_{cr,z}}}=\sqrt{\frac{54.30·23.5}{1151.60}}=1.053>0.2$

→ Design for flexural buckling must be performed.

cross-sectional geometry: h/b = 1.00 ≤ 1.2; structural steel S 235; t ≤ 100 mm

• [1] Table 6.2, row 3, column 4: buckling curve c
• ⇒ α_z = 0.49   ([1] Table 6.1)

$\Phi =0.5·[1+0.49·(1.053-0.2)+1.{053}^2]=1.263$

${\chi }_z=\frac{1}{1.263+\sqrt{1.{263}^2-1.{053}^2}}=0.510$

$\frac{N_{Ed}}{{\chi }_z·A·f_y/{\gamma }_{M1}}=\frac{300}{0.510·54.30·23.5/1.0}=0.461$

$N_{cr,y}=\frac{21000·2490.00·{\pi }^2}{400.{00}^2}=3225.51 \mathrm{kN}$

${\overline{\lambda }}_y=\sqrt{\frac{A·f_y}{N_{cr,y}}}=\sqrt{\frac{54.30·23.5}{3225.51}}=0.629>0.2$

→ Design for flexural buckling must be performed.

cross-sectional geometry: h/b = 1.00 ≤ 1.2; structural steel S 235; t ≤ 100 mm

• [1] Table 6.2, row 3, column 4: buckling curve b
• ⇒ α_y = 0.34   ([1] Table 6.1)

$\Phi =0.5·[1+0.34·(0.629-0.2)+0.{629}^2]=0.771$

${\chi }_y=\frac{1}{0.771+\sqrt{0.{771}^2-0.{629}^2}}=0.822$

$\frac{N_{Ed}}{{\chi }_y·A·f_y/{\gamma }_{M1}}=\frac{300}{0.822·54.30·23.5/1.0}=0.286$

In this example, the elastic critical moment for lateral-torsional buckling is determined according to the Austrian National Annex with assumption of hinged supports free to warp.

The load application point is assumed to be in the shear center (you can adjust the application point for transverse loads in the Details dialog box, cf. Chapter 3.1.2).

$M_{\mathrm{cr}}=C_1·\frac{{\pi }^2·E·I_z}{{\mathcal{l}}^2}·\sqrt{\frac{I_{\omega }}{I_z}+\frac{{\mathcal{l}}^2·G·I_t}{{\pi }^2·E·I_z}}$

$M_{\mathrm{cr}}=1.13·\frac{{\pi }^2·21000·889}{{400}^2}·\sqrt{\frac{47940}{889}+\frac{{400}^2·8100·31.40}{{\pi }^2·21000·889}}=215.71 \mathrm{kNm}$

The program also shows M_cr,0 which is determined on the basis of a constant moment distribution.

For the results by x-location, the program also shows the M_cr,x values. Those are the elastic critical moments at the x-locations relative to the elastic critical moment at the location of the maximum moment. Using M_cr,x, the program then calculates the relative slenderness ƛ_LT.

Calculation according to [1] clause 6.3.2.2 for location with the maximum moment at x = 2.00 m:

HEB-160, cross-section class 1: W_y = W_pl,y = 354.00 cm³

${\overline{\lambda }}_{\mathrm{LT}}=\sqrt{\frac{W_y·f_y}{M_{\mathrm{cr}}}}=\sqrt{\frac{354·23.5}{215.71}}=0.621$

Calculation according to [1] clause 6.3.2.3

HEB-160: h/b = 1.0 < 2.0 ⇒ buckling curve b according to [1] Table 6.5

• Auxiliary factor:

${\Phi }_{\mathrm{LT}}=0.5·\left[1+{\alpha }_{\mathrm{LT}}·\left({\overline{\lambda }}_{\mathrm{LT}}-{\overline{\lambda }}_{\mathrm{LT},0}\right)+\beta ·{\overline{\lambda }}_{\mathrm{LT}}^2\right]$

${\Phi }_{\mathrm{LT}}=0.5·\left[1+0.34·\left(0.621-0.40\right)+0.75·0.{621}^2\right]=0.682$

Limiting slenderness:

${\overline{\lambda }}_{\mathrm{LT},0}=0.40$

Parameter (minimum value):

$\beta =0.75$

Imperfection factor according to [1] Table 6.3:

${\alpha }_{\mathrm{LT}}=0.34$

${\chi }_{\mathrm{LT}}=\frac{1}{{\Phi }_{\mathrm{LT}}+\sqrt{{\Phi }_{\mathrm{LT}}^2-\beta ·{\overline{\lambda }}_{\mathrm{LT}}^2}}=\frac{1}{0.682+\sqrt{0.{682}^2-0.75·0.{621}^2}}=0.908$

According to [1] clause 6.3.2.3, the reduction factor may be modified as follows:

${\chi }_{\mathrm{LT}\text{,mod}}=\frac{{\chi }_{LT}}{f} \text{mit } f=1-0.5·\left(1-k_c\right)·\left[1-2.0·\left({\overline{\lambda }}_{\mathrm{LT}}-0.8^2\right)\right]$

${\chi }_{\mathrm{LT}\text{,mod}}=\frac{0.908}{0.972}=0.934$

Correction factor k_c according to [1] Table 6.6 for a parabolic moment diagram:

$k_c=0.94$

$f=1-0.5·(1-0.94)·[1-2.0·(0.621-0.8{)}^2]=0.972$

Determination according to [6] Annex B, Table B.2, for structural components susceptible to torsional deformations

The equivalent moment factor C_mLT is obtained according to Table B.3 for ψ = 0 as:

$C_{my}=C_{mLT}=0.95+0.05·{\alpha }_h=0.95 \text{mit } {\alpha }_h=\frac{M_h}{M_s}=\frac{0}{10}=0$

$k_{yy}=C_{my}·(1+({\overline{\lambda }}_y-0.2)·\frac{N_{Ed}}{{\chi }_y·N_{Rk}/{\gamma }_{M1}})\le C_{my}·(1+0.8·\frac{N_{Ed}}{{\chi }_y·N_{Rk}/{\gamma }_{M1}})$

$k_{yy}=0.95·(1+(0.629-0.2)·0.286)\le 0.95·(1+0.8·0.286)=\underline{1.067}\le 1.167$

$k_{yz}=0.60·k_{zz}=0.60·1.481=\underline{0.888}$

Determination according to [1] Annex B, Table B.2, for structural components susceptible to torsional deformations

The equivalent moment factor C_mLT is obtained according to Table B.3 for ψ = 0 as:

$C_{mz}=0.90+0.01·{\alpha }_h=0.90 \text{mit} {\alpha }_h=\frac{M_h}{M_s}=\frac{0}{10}=0$

$k_{zy}=(1-\frac{0.1·{\overline{\lambda }}_z}{C_{mLT}-0.25}·\frac{N_{Ed}}{{\chi }_z·N_{Rk}/{\gamma }_{M1}})\ge (1-\frac{0.1}{C_{mLT}-0.25}·\frac{N_{Ed}}{{\chi }_z·N_{Rk}/{\gamma }_{M1}})$

$k_{zy}=(1-\frac{0.1·1.053}{0.95-0.25}·0.461)\ge (1-\frac{0.1}{0.95-0.25}·0.461)=\underline{0.892}\le 0.934$

$k_{zy}=\underline{0.934}$

$k_{zz}=C_{mz}·(1+(2·{\overline{\lambda }}_z-0.6)·\frac{N_{Ed}}{{\chi }_z·N_{Rk}/{\gamma }_{M1}})\le C_{mz}·(1+1.4·\frac{N_{Ed}}{{\chi }_z·N_{Rk}/{\gamma }_{M1}})$

$k_{zz}=0.90·(1+(2·1.053-0.6)·0.461)\le 0.90·(1+1.4·0.461)=\underline{1.525}\ge 1.481$

$k_{zz}=\underline{1.481 }$

According to [1] Eq. (6.61) the following requirement must be fulfilled:

$\frac{N_{Ed}}{{\chi }_y·N_{Rk}/{\gamma }_{M1}}+k_{yy}·\frac{M_{y,Ed}}{{\chi }_{LT}·M_{y,Rk}/{\gamma }_{M1}}+k_{yz}·\frac{M_{z,Ed}}{M_{z,Rk}/{\gamma }_{M1}}\le 1$

where

$M_{y,Rk}=W_{pl,y}·f_y=354·23.5=8319 \mathrm{kNcm}=83.19 \mathrm{kNm}$

$M_{z,Rk}=W_{pl,z}·f_y=169.96·23.5=3994.1 \mathrm{kNcm}=39.94 \mathrm{kNm}$

$\frac{300}{0.822·1276.05/1.0}+1.067·\frac{10.0}{0.908·83.19/1.0}+0.888·\frac{7.50}{39.94/1.0}=0.594\le 1$

According to EN1993-1-1 Eq. (6.62) the following requirement must be fulfilled:

$\frac{N_{Ed}}{{\chi }_z·N_{Rk}/{\gamma }_{M1}}+k_{zy}·\frac{M_{y,Ed}}{{\chi }_{LT}·M_{y,Rk}/{\gamma }_{M1}}+k_{zz}·\frac{M_{z,Ed}}{M_{z,Rk}/{\gamma }_{M1}}\le 1$

$\frac{300}{0.510·1276.05/1.0}+0.934·\frac{10.0}{0.908·83.19/1.0}+1.481·\frac{7.50}{39.94/1.0}=0.863\le 1$