# 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 $$d = 244.5 mm$$ $$t = 10.0 mm$$ $$A = 7370 mm^{2}$$ $$W_{el,y} = 415000 mm^{2}$$ $$W_{pl,y} = 550000 mm^{3}$$ $$I = 50730000 mm^{4}$$

Solution:

### Model

#### Materials

Title Young Modulus [Pa] Shear Modulus [Pa] Poisson Ratio Shear [Pa] Mass Density [kg/m^3] Tensile Strength [Pa] Yield Stress [Pa]
1..S355 2.10e+11 0 0.30 0 7850.00 470.00e+6 355.00e+6

#### Property “1..CHS 244.5 x 10”

Property Value Property Shape
Type / Elements Beam / 17 Material 1..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”

Definition Load Type Applied On Values
Force on Node Force Nodes: 18 (0; 0; -1630000) #### Constraint “1..fixed”

Definition Count DOF
Constraint on Node Nodes: 1 Tx Ty Tz
Constraint on Node Nodes: 18 Tx Ty Rz ### Cross-section classification

 Check 2..Circular Tube
Property Shear Area Diameter Thickness Epsilon Lambda Class Buckling Curve Imperfection Factor
1..Tube 0.0047 0.24450 0.01000 0.81 24.45 1.00 2.00 0.21

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)

 Standard 1..Eurocode 3 Member Checks Check 9..Axial Check Individual Load 1..Axial -Z = 1630kN.fixed Selection 1..Eurocode3 Shapes
Extreme Axial Force Design Tension Resistance Design Compression Resistance Design Resistance Utilization Factor
Minimum -1630.0e+3 2615.3e+3 2615.3e+3 2615.3e+3 -0.62
Maximum -1630.0e+3 2615.3e+3 2615.3e+3 2615.3e+3 -0.62
Absolute -1630.0e+3 2615.3e+3 2615.3e+3 2615.3e+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)

 Standard 1..Eurocode 3 Member Checks Check 12..Buckling Axial Individual Load 1..Axial -Z = 1630kN.fixed Selection 1..Eurocode3 Shapes
Extreme Ncr_y Ncr_z Lambda Y Lambda Z Fy Fz Xi_Y Xi_Z
Minimum 6571.7e+3 6571.7e+3 0.63 0.63 0.74 0.74 0.88 0.88
Maximum 6571.7e+3 6571.7e+3 0.63 0.63 0.74 0.74 0.88 0.88
Absolute 6571.7e+3 6571.7e+3 0.63 0.63 0.74 0.74 0.88 0.88
Extreme Nb Rd Y Nb Rd Z Uf_Y Uf_Z Uf
Minimum 2296.0e+3 2296.0e+3 -0.71 -0.71 -0.71
Maximum 2296.0e+3 2296.0e+3 -0.71 -0.71 -0.71
Absolute 2296.0e+3 2296.0e+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.