# 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 1630 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 mm2 Wel,y = 415000 mm2 Wpl,y = 550000 mm3 I = 50730000 mm4

## 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, [m2] 0.0074
I1, [m4] 5.073e-05
I2, [m4] 5.073e-05
I12, [m4] 0
Torsion Constant, [m4] 1.014e-04
Y Shear Area, [m2] 0.0039
Z Shear Area, [m2] 0.0039
Nonstructural Mass, [kg] 0
Perimeter, [m] 0.77
Warping Constant, [m6] 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 > 1630 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 comparisons of Eurocode3 design examples standard and SDC Verifier.