Eurocode 3. Example 6.7: Buckling resistance of a compression member

We compare the buckling resistance of a compression member according to SDC Verifier and Designer’s Guide to Eurocode3: Design of steel buildings EN 1993-1-1.

The results are generated with SDC Verifier 4.5 and calculated with FEMAP v11.3.1

Task:  A circular hollow section (CHS) member is to be used as an internal column in a multi-storey  building. The column has pinned boundary conditions at each end, and the inter-storey height is 4m. The critical combination of actions results in a design axial force of 2110 kN. Assess the suitability of a hot-rolled 244.5 x 10 CHS in grade S355 steel for this application.

Figure 6.21. Section properties for 244.5×10 CHS

Section Properties
\(d = 244.5 mm\)
\(t = 10.0 mm\)
\(A = 7370 mm^{2}\)
\(W_{el,y} = 415000 mm^{2}\)
\(W_{pl,y} = 550000 mm^{3}\)
\(I = 50730000 mm^{4}\)

Solution:

Model

Materials

TitleYoung Modulus [Pa]Shear Modulus [Pa]Poisson RatioShear [Pa]Mass Density [kg/m^3]Tensile Strength [Pa]Yield Stress [Pa]
1..S3552.10e+1100.3007850.00470.00e+6355.00e+6

Property “1..CHS 244.5 x 10”

PropertyValueProperty Shape
Type / ElementsBeam / 17
Material1..S355
Mass [kg]231.3
Gravity Center [m][0.00; 0.00; 2.00]
Area, [m^2]0.0074
I1, [m^4]5.073e-05
I2, [m^4]5.073e-05
I12, [m^4]0
Torsion Constant, [m^4]1.014e-04
Y Shear Area, [m^2]0.0039
Z Shear Area, [m^2]0.0039
Nonstructural Mass, [kg]0
Perimeter, [m]0.77
Warping Constant, [m^6]0
Y Neutral Axis Offset A, [m]0
Z Neutral Axis Offset A, [m]0
r [m]0.12225
t [m]0.01000

Boundary Conditions

Load “1..Axial -Z = 1630kN”

DefinitionLoad TypeApplied OnValues
Force on NodeForceNodes: 18(0; 0; -1630000)

Constraint “1..fixed”

DefinitionCountDOF
Constraint on NodeNodes: 1Tx Ty Tz
Constraint on NodeNodes: 18Tx Ty Rz

Cross-section classification

Check2..Circular Tube
PropertyShear AreaDiameterThicknessEpsilonLambdaClassBuckling CurveImperfection Factor
1..Tube0.00470.244500.010000.8124.451.002.000.21

Cross-section classification (clause 5.5.2)

\(\varepsilon \ = \sqrt{235/f_{y}} \ = \sqrt{235/355} \ = 0.81\)

Tubular sections (Table 5.2, sheet 3):

\(d/t \ = 244.5/10.0 \ = 0.81\)

Limit for Class 1 section = 50ε2 = 40.7

40.7 > 24.5 ∴ section is Class 1

Cross-section compression resistance (clause 6.2.4)

Standard1..Eurocode 3 Member ChecksCheck9..Axial Check
Individual Load1..Axial -Z = 1630kN.fixedSelection1..Eurocode3 Shapes
ExtremeAxial ForceDesign Tension ResistanceDesign Compression ResistanceDesign ResistanceUtilization Factor
Minimum-1630.0e+32615.3e+32615.3e+32615.3e+3-0.62
Maximum-1630.0e+32615.3e+32615.3e+32615.3e+3-0.62
Absolute-1630.0e+32615.3e+32615.3e+32615.3e+3-0.62

Cross-section compression resistance (clause 6.2.4)

\(N_{c,Rd} \ = \frac{Af_{y}}{\gamma M0} \)

for Class 1, 2 or 3 cross-sections

\(∴ N_{c,Rd} \ = \frac{7370 \times 355 }{1.00} \ = 2616 \times 10^{3} N \ = 2616 kN\)

2616 > 2110 kN    ∴ cross-section resistance is acceptable

Member buckling resistance in compression (clause 6.3.1)

Standard1..Eurocode 3 Member ChecksCheck12..Buckling Axial
Individual Load1..Axial -Z = 1630kN.fixedSelection1..Eurocode3 Shapes
ExtremeNcr_yNcr_zLambda YLambda ZFyFzXi_YXi_Z
Minimum6571.7e+36571.7e+30.630.630.740.740.880.88
Maximum6571.7e+36571.7e+30.630.630.740.740.880.88
Absolute6571.7e+36571.7e+30.630.630.740.740.880.88
ExtremeNb Rd YNb Rd ZUf_YUf_ZUf
Minimum2296.0e+32296.0e+3-0.71-0.71-0.71
Maximum2296.0e+32296.0e+3-0.71-0.71-0.71
Absolute2296.0e+32296.0e+3-0.71-0.71-0.71

Member buckling resistance in compression (clause 6.3.1)

\(N_{b,Rd} \ = \frac{\chi Af_{Y}}{\gamma M1}\)

for Class 1, 2 and 3 cross-sections

\(\chi \ = \frac{1}{\Phi + \sqrt{\Phi^{2} – \lambda^{2}}}\)   but  \(\chi \leq 1.0\)

where

\(\Phi \ = 0.5[1+\alpha(\lambda-0.2)+\lambda^{2}]\)

and

\(\lambda \ = \sqrt{\frac{Af_{y}}{N_{cr}}}\) for Class 1, 2 and 3 cross-sections

Elastic critical force and non-dimensional slenderness for flexural buckling

\(N_{cr} \ = \frac{\pi^{2}EI}{L^{2}_{cr}} \ = \frac{\pi^{2} \times 210000 \times 50730000}{4000^{2}} \ = \color{red}{6571 kN}\) \(∴\lambda \ = \sqrt{\frac{7370 \times 355}{6571 \times 10^{3}}} \ = \color{green}{0.63}\)

Buckling curves

\(\Phi \ = 0.5[1+0.21 \times (0.63 – 0.2)+0.63^{2}] \ = \color{orange}{0.74}\) \(\chi \ = \frac{1}{0.74 + \sqrt{0.74^{2} – 0.63^{2}}} \ = \color{blue}{0.88}\) \(∴N_{b,Rd} \ = \frac{0.88 \times 7370 \times 355}{1.0} \ = 2297 \times 10^{3} N \ = \color{yellow}{2297 kN}\)

Conclusion

The chosen cross-section, 244.5 × 10 CHS in grade S355 steel is acceptable.

Cross section compression and member buckling resistances in the SDC Verifier completely much with values in the example 6.7.

Check other comparation of Eurocode3 design examples standard and SDC Verifier.