Design of Steel Column (H-Section) to Eurocode 3 – Worked Example

This article presents a worked example on the design of steel column (H-section) subjected to combined Biaxial Bending and axial compression to Eurocode 3

The column to be designed has been analysed using Etabs software as part of a frame in the article, ”Analysis of Steel Frames Allowing for Imperfections”, the resulting bending moment and axial force diagram are shown below.

Internal forces

From the analysis results, the maximum of the internal forces acting on each column shall be considered to be acting on the column to be designed.

Maximum Axial Force = -256.7KN (Middle lower column)

Maximum Moment (major axis): -20.8KNm (Top rightmost column)

The frame is a planar frame, and its columns are not subjected to moment about minor axis, for the sake of this worked example we shall assume a moment of 10KNm so that the column is biaxially loaded.

Moment (minor axis): 10KNm

Column Properties  

Column section: 305 x 305 x 283 UKC S275

The properties of the steel section listed below are excerpted from SCI P363

h = 365.3mm, b = 322.2mm, t_w = 26.8mm, t_f = 44.1mm, r = 15.2mm, d = 246.7mm, i_y = 14.8cm, i_z = 8.27cm, W_ply = 5110cm³, W_plz = 2340cm³, A = 360cm², E = 210 x 10³, I_y =78900cm⁴, I_z = 24600cm⁴, I_w = 6.35dm⁶, I_t = 2030cm³

c_f/t_f = 3.0

c_w/t_w = 9.21

Cross-section Classification

We shall classify the cross-section of the column (you should click here to study a detailed article on cross-section classification of steel cross-sections).

Influence of material strength

The UK National Annex recommends that EN 10025-2:2004 replaces Table 3 of EN 1993-1-1:2005 to determine the yield strength (f_y) and ultimate strength (f_u) of steel sections

Hence, for thickness of 15.44mm for grade S275 | f_y = 255MPa, f_u = 430MPa (Table 7, EN 10025-2)

$$
\varepsilon \,\,=\,\,\sqrt{\frac{235}{255}}\,\,=\,\,0.960
$$

Flanges (Outstand Element)

Since the neutral axis of the section is assumed to be within the web, the flange is in pure compression.

(Using Table 5.2 (Sheet 2 of 3) of EN 1993-1-1-2005)

c_f/t_f = 3.29

from the table, limiting value for class 1 is 9 ℇ = 9 x 0.960 = 8.64

since 3 < 8.64, the flanges are class 1

Web (Internal Element)

(Using Table 5.2 (Sheet 1 of 3) of EN 1993-1-1-2005)

c_w/t_w = 9.21

from the table, limiting value for class 1 is:

397ℇ/(13∝ – 1)     (f0r ∝ > 0.5)

36ℇ/∝                    (for ∝ ≤ 0.5)

$
\alpha \,\,=\,\,\frac{1}{c}\left( \frac{h}{2}\,\,+\,\,\frac{1}{2}\frac{N_{Ed}}{t_wf_y}\,\,-\,\,\left( t_{f\,\,}+\,\,r \right) \right) \,\,\leqslant \,\,1
$

c = d = 246.7mm

$
\alpha \,\,=\,\,\frac{1}{246.7}\left( \frac{246.7}{2}\,\,+\,\,\frac{1}{2}\frac{256.7 x 10^3}{26.8 x 255}\,\,-\,\,\left( 44.1{\,\,}+\,\,15.2 \right) \right) = 0.184$

since α < 0.5, class 1 limit is:

36ℇ/∝   = 36 x 0.960 /0.184 = 187.826

Since the actual value (9.21) < the limiting value (187.826), the web is Class 1

Cross-section Resistance

The column cross-section shall be verified for resistance to compression, and biaxial bending.

Compression Resistance

The expression below has to be satisfied for the section to be adequate in compression

$
\frac{N_{ED}}{N_{cRd}}\,\,\leqslant \,\,1
$

$
N_{cRd}\,\,=\frac{A\,\,f_y}{\gamma _{m}}\,\,
$ for class 1, 2, & 3 sections

$
N_{cRd}\,\,=\frac{360\,\,255}{\ 1 x 10^3}\,\,
$ = 9180KN

Since 256.7 < 9180, section is adequate

 

Resistance to Biaxial Bending

For biaxial bending the expression below should be satisfied.

$
\,\,\left[ \frac{M_{y,Ed}}{M_{Ny,Ed}} \right] ^{\alpha}\,\,+\,\,\left[ \frac{M_{z,Ed}}{M_{N,Z,Rd}} \right] ^{\,\,\beta}\leqslant \,\,1
$

M_Ny,Rd and M_Nz,Rd are the reduced plastic moment resistance reduced due to the axial force N_Ed

Check whether reduction of plastic moment about y axis is necessary

Reduction of plastic moment about y-axis shall be neglected if the two expressions below are satisfied.

N_Ed ≤ 0.25N_pl,Rd

$
N_{Ed}\,\,\leqslant \,\,\frac{\text{0.5}h_w\,\,t_w\,\,f_y}{\gamma _{mo}}
$

Check

0.25N_pl,Rd= 0.25 x 9180 = 2295KN

$
N_{Ed}\,\,\leqslant \,\,\frac{\text{0.5} x 365.3\,\,26.8\,\,255}{1 x 10^6}
$ = 946.85

256.7 ≤ 2295KN and 256.7 ≤ 946.85, reduction in plastic moment is unnecessary.

Since it is not necessary, plastic moment about y-axis is:

$
\,\,Mc,Rd\,\,=\,\,Mpl,Rd\,\,=\,\,\frac{W_{ply}.\,\,f_y\,\,}{\varUpsilon _{m0}}\,\,\,\,
$  (for class 1, and class 2 sections)

$
\,\,Mc,Rd\,\,=\,\,\frac{5110 x 10^3 .\,\,255\,\,}{1 x 10^6}\,\,\,\,
$  = 1303.05KNm

Check whether reduction of plastic moment about z axis is necessary

Reduction of plastic moment about z-axis shall be neglected if the expression below is satisfied.

$
N_{Ed}\,\,\leqslant \,\,\frac{\\,\,h_w\,\,t_w\,\,f_y}{\gamma _{mo}}
$

Check

$
N_{Ed}\,\,\leqslant \,\,\frac{\\,\,356.3\,\,26.8\,\,255}{1 x 10^6}
$ = 1893.7KN

256.7 ≤ 1893.7KN, reduction in plastic moment about z – axis is unnecessary.

$
\,\,Mc,Rd\,\,=\,\,Mpl,Rd\,\,=\,\,\frac{W_{plz}.\,\,f_y\,\,}{\varUpsilon _{m0}}\,\,\,\,
$  (for class 1, and class 2 sections)

$
\,\,Mc,Rd\,\,=\,\,\frac{2340 x 10^3.\,\,255\,\,}{1 x 10^6}\,\,\,\,
$  = 596.7KNm

 

Biaxial Check

α = 2

n = NEd/Npl,Rd = 256.7/9180 = 0.028

β = max (5 x 0.028, 1) = 1.0

$
\,\,\left[ \frac{20.8}{1303.05} \right] ^{2}\,\,+\,\,\left[ \frac{10.0}{596.7} \right] ^{\,\,1.0}\
$ = 0.63

0.017 ≤ 1, the section is satisfactory in biaxial bending

 

Buckling Resistance (Stability Check)

The buckling resistance of the section should be verified using the below expressions. It should also be noted that by using these expressions then the section would have been verified for local imperfection and second-order effects as these local effects have been allowed for in the expressions

$\frac{N_{E d}}{\frac{X_y N_{R k}}{Y_{m 1}}}+K_{y y} \frac{M_{y, E d}+\Delta M_{y, E d}}{\chi L T \frac{M_{y, R k}}{Y_{m 1}}}+K_{y z} \frac{M_{z, B d}+\Delta M_{z, E d}}{\frac{M_{z, R k}}{Y_{m 1}}} \leq 1$

$\frac{N_{E d}}{\frac{\chi_z N_{R k}}{Y_{m 1}}}+K_{z y} \frac{M_{y, E d}+\Delta M_{y, E d}}{M_{L T} \frac{M_{y, R k}}{Y_{m 1}}}+K_{z z} \frac{M_{z, E d}+\Delta M_{z, E d}}{\frac{M_{z, R k}}{Y_{m 1}}} \leq 1$

For simplification, the below expressions from the general expression should be determined so that their values are substituted back into the general expressions for buckling resistance.

$
\,\,\chi _{LT}\,\,\frac{M_{yRK}\,\,}{\varUpsilon _m}
$ is expression for lateral torsional buckling

$
\,\,\chi _{y}\,\,\frac{N_{RK}\,\,}{\varUpsilon _m}
$ is expression for flexural buckling about y-axis

$
\,\,\chi _{z}\,\,\frac{N_{RK}\,\,}{\varUpsilon _m}
$ is expression for flexural buckling about z-axis

 

Determine Lateral torsional buckling of the column

Lateral torsional buckling is evaluated using the expression below:

$$
M_{b,Rd}\,\,=\,\,\chi _{LT}\,\,\frac{W_y\,\,.\,\,f_y}{\varUpsilon _m}
$$

Where,

$
\,\,\chi _{LT}\,\,=\,\,\frac{1}{\phi _{LT}\,\,+\,\,\sqrt{\phi _{LT}^2\,\,+\,\,\ βλ _{LT}^2}}
$

Φ_LT = 0.5 (1 + ∝_LT( λ_LT – λ_LT,0) + βλ_LT ²)

$
\,\lambda _{LT}\,\,=\,\,\sqrt{\frac{W_y\,\,.\,\,f_y}{M_{cr}}}
$

Calculate the Elastic Critical Moment

$M_{C r}=C_1 \frac{\pi^2 E I_z}{L^2} \sqrt{\frac{I_w}{I_z}+\frac{L^2 G I_t}{\pi^2 E I_z}}$

C₁= 1.77

G = 80770 N/mm²

E = 21000cm³

Iw = 6.35dm⁶, It = 2030cm³

Iz = 24600

L = 3000

$M_{C r}= 1.77 \frac{3.142^2 x 210000 x 24600 x 10^4} {3000^2 x 10^3} \sqrt{\frac{6.35 x 10^12}{24600 x 10^4}+\frac{3000^2 x 81000 x 2030}{3.142^2 x 210000 x 24600}}$

M_cr= 23485.93

Calculate the Non-dimension Slenderness

$
\,\lambda _{LT}\,\,=\,\,\sqrt{\frac{5110 x 10^3\,\,.\,\,255}{23485.93 x 10^6}}
$ = 0.236

 

∝_LT = 0.35 (using Table 6.5)

β = 0.75

λ_LT,0 = 0.4

Φ_LT  = 0.5 (1 + 0.35(0.236– 0.4) + 0.75 x 0.236²) = 0.493

Calculate the reduction factor

$
\,\,\chi _{LT}\,\,=\,\,\frac{1}{0.493\,\,+\,\,\sqrt{0.493^2\,\,+\,\,\ 0.75 x 0.236^2}}
$ = 1.06

Since 1.06 ≥ 1 reduction factor is taken as 1

M_b,Rd = 1 x 1303.05 = 1303.05KNm

NB: It should be noted that the reduction factor is greater than one hence the cross-section is not susceptible to lateral torsional buckling

Check

$
\,\,\frac{M_{Ed}}{M_{bRd}}
$ = $
\,\,\frac{20.8}{1303.05}
$ = 0.016

Since 0.016 ≤ 1, the section is adequate

Determine the Flexural Buckling

The flexural buckling is determined both about the major (y-axis) and minor axis (z-axis). About any of the axis, the section is adequate for flexural buckling if the expression below is satisfied

$
\frac{N_{ED}}{N_{bRd}}\,\,\leqslant \,\,1
$

Determine the Flexural Buckling about y-axis

$
N_{bRd}\,\,=\chi \frac{A\,\,f_y}{\gamma _m}
$ for class 1, 2, & 3 sections

$
\,\,\chi \,\,=\,\,\frac{1}{\phi \,\,+\,\,\sqrt{\phi ^2\,\,-\,\,\lambda ^2}}\,\,\leqslant \,\,1
$

Φ = 0.5 (1 + α(λ – 0.2) + λ²)

$
\,\,\lambda \,\,=\,\,\sqrt{\frac{A\,\,f_y}{N_{cr}}}\,\,=\,\,\frac{L_{cr}}{i}\,\,\frac{1}{\lambda _1}
$ for class 1, 2 & 3 cross-sections

Determine the Non-dimension Slenderness

iy =78900

λ₁ = 93.9ℇ = 93.9 x 0.960 = 90.14

L_cr = 3000mm

$
\,\,\lambda \,\,=\,\,\frac{3000}{78900}\,\,\frac{1}{90.14}
$= 0.375

Determine the Reduction factor

Since tf = 44.1mm for S275, h/b ≤ 1.2, buckling curve b is to be used (Table 6.2 EN 1993-1-1)

α = 0.34 (Table 6.1 EN 1993-1-1)

Φ = 0.5 (1 + 0.34(0.375 – 0.2) + 0.375²) = 0.294

$
\,\,\chi \,\,=\,\,\frac{1}{0.294 \,\,+\,\,\sqrt{0.294^2\,\,-\,\,0.375 ^2}}
$ = 1.3

Since 1.3 ≥ 1, χ = 1

N_b,Rd = 1 x 9180

N_b,Rd = 9180KN

Check

$
\,\,\frac{N_{Ed}}{N_{bRd}}
$ = $
\,\,\frac{256.7}{9180}
$ = 0.028

Since 0.028 ≤ 1, the section is adequate

Determine the Flexural Buckling about z-axis

$
N_{bRd}\,\,=\chi \frac{A\,\,f_y}{\gamma _m}
$ for class 1, 2, & 3 sections

$
\,\,\chi \,\,=\,\,\frac{1}{\phi \,\,+\,\,\sqrt{\phi ^2\,\,-\,\,\lambda ^2}}\,\,\leqslant \,\,1
$

Φ = 0.5 (1 + α(λ – 0.2) + λ²)

$
\,\,\lambda \,\,=\,\,\sqrt{\frac{A\,\,f_y}{N_{cr}}}\,\,=\,\,\frac{L_{cr}}{i}\,\,\frac{1}{\lambda _1}
$ for class 1, 2 & 3 cross-sections

Determine the Non-dimension Slenderness

iz = 24600

λ₁ = 93.9ℇ = 93.9 x 0.960 = 90.14

L_cr = 3000mm

$
\,\,\lambda \,\,=\,\,\frac{3000}{24600}\,\,\frac{1}{90.14}
$= 0.67

Determine the Reduction factor

Since tf = 44.1mm for S275, h/b ≤ 1.2, buckling curve c is to be used (Table 6.2 EN 1993-1-1)

α = 0.49 (Table 6.1 EN 1993-1-1)

Φ = 0.5 (1 + 0.49(0.67 – 0.2) + 0.67²) = 0.399

$
\,\,\chi \,\,=\,\,\frac{1}{0.399 \,\,+\,\,\sqrt{0.399^2\,\,-\,\,0.67 ^2}}
$ = 0.848

N_b,Rd = 0.848 x 9180

N_b,Rd = 7781KN

Check

$
\,\,\frac{N_{Ed}}{N_{bRd}}
$ = $
\,\,\frac{256.7}{7781}
$ = 0.033

Since 0.33 ≤ 1, the section is adequate

 

Check for Buckling

We shall conservatively assume the interaction factors (Kyy, kyz, kzy and kzz) which accounts for non-linear effects to be 1.

Substitute the value of lateral torsional buckling resistance and flexural buckling about the two axis into the equations.

$\frac{256.7}{9180}+ \frac{20.8}{1303.05}+\frac{10}{596.7}=\leq 1$

0.06 ≤ 1

$\frac{256.7}{7781}+ \frac{20.8}{1303.05}+\frac{10}{596.7}=\leq 1$

0.066 ≤ 1

Since the value of both expressions are less that 1, then the section is adequate against buckling instability!