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.