EN 1994 Composite Column — Concrete-Filled Steel Tube per Eurocode 4
Complete guide to composite column design per EN 1994-1-1:2004. Concrete-filled tubes (CFT), fully and partially encased sections. Plastic resistance to compression N_pl,Rd, relative slenderness λ̄, buckling curves, shear connector design. Worked example with CHS 219.1×10 filled with C30/37 concrete.
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Composite Column Types — EN 1994-1-1 Clause 6.7
Concrete-Filled Steel Tube (CFT)
Steel tube filled with concrete. The tube acts as permanent formwork and reinforcement. Highest strength-to-weight ratio. No longitudinal reinforcement required in typical applications.
Fully Encased Section
Steel I-section fully encased in concrete with longitudinal and transverse reinforcement. Enhanced fire resistance.
Partially Encased Section
Steel I-section with concrete between flanges. Reduced fire protection requirement.
| Type | Strength | Fire Resistance | Formwork Cost | Construction Speed |
|---|---|---|---|---|
| CFT circular | Excellent | Good | None needed | Fast |
| CFT rectangular | Good | Good | None needed | Fast |
| Fully encased | Good | Excellent | Required | Slow |
| Partially encased | Moderate | Good | Minimal | Moderate |
Plastic Resistance to Compression — Clause 6.7.3.2
N_pl,Rd = A_a × f_yd + A_c × f_cd + A_s × f_sd
For concrete-filled tubes (typically no reinforcing steel A_s):
N_pl,Rd = A_a × f_y / γ_Ma + A_c × (0.85 × f_ck / γ_c)
Where:
- γ_Ma = 1.00 (steel partial factor)
- γ_c = 1.50 (concrete partial factor)
- 0.85 = coefficient for long-term effects
Relative Slenderness — Clause 6.7.3.4
λ̄ = √(N_pl,Rk / N_cr)
Where N_cr is the elastic critical force from:
N_cr = π² × (EI)_eff / L²
The effective flexural stiffness:
(EI)_eff = E_a × I_a + 0.6 × E_cd × I_c + E_s × I_s
The factor 0.6 accounts for concrete cracking and long-term effects.
Buckling Curves — Table 6.5
| Cross-Section | Buckling Curve | α |
|---|---|---|
| Concrete-filled CHS (hot-finished) | a | 0.21 |
| Concrete-filled CHS (cold-formed) | c | 0.49 |
| Fully/partially encased (y-y) | b | 0.34 |
| Fully/partially encased (z-z) | c | 0.49 |
Worked Example — CHS 219.1×10 CFT in S355, C30/37
| Parameter | Value |
|---|---|
| Steel tube | CHS 219.1×10, S355 |
| Concrete | C30/37 (f_ck = 30 MPa) |
| Column length | 4.0 m (pinned both ends) |
| A_a | 6570 mm² (from HSS table) |
| A_c | π/4 × (219.1 - 2×10)² = 31100 mm² |
Plastic Resistance
| Component | Calculation | Resistance |
|---|---|---|
| Steel | 6570 × 355 / 1.00 | 2332 kN |
| Concrete | 31100 × (0.85×30) / 1.50 | 529 kN |
| N_pl,Rd | Total | 2861 kN |
Buckling Check
| Parameter | Value |
|---|---|
| I_a (CHS) | 3586 × 10⁴ mm⁴ |
| I_c (concrete core) | π/64 × (199.1)⁴ = 7714 × 10⁴ mm⁴ |
| (EI)_eff | 210000 × 3586×10⁴ + 0.6 × 22000 × 7714×10⁴ = 7.53 × 10¹¹ + 1.02 × 10¹¹ = 8.55 × 10¹¹ N·mm² |
| N_cr | π² × 8.55×10¹¹ / 4000² = 5270 kN |
| λ̄ | √(2861 × 1.00 / 5270) = 0.74 |
| χ (curve a, α=0.21) | 0.87 |
| N_b,Rd | 0.87 × 2861 = 2489 kN |
The buckling capacity is 2489 kN, 13% less than the squash load.
Comparison — CHS 219.1×10 Hollow vs Filled
| Condition | N_pl,Rd (kN) | N_b,Rd, 4m (kN) | Gain |
|---|---|---|---|
| Steel only | 2332 | 1764 | — |
| Filled C30/37 | 2861 | 2489 | +41% |
| Filled C40/50 | 2975 | 2588 | +47% |
| Filled C50/60 | 3137 | 2729 | +55% |
Concrete filling provides 40-55% increase in axial capacity for the same steel tube.
Design Applications
Common Design Scenarios
This reference covers structural design scenarios commonly encountered in structural steel design practice:
- Strength verification: Check member or connection capacity against factored loads per the applicable design code
- Serviceability checks: Verify deflections, vibrations, and other serviceability criteria
- Code compliance: Ensure design meets all provisions of the governing standard
- Connection detailing: Verify weld sizes, bolt quantities, and edge distances
Related Design Considerations
- System behavior: consider the interaction between members and connections
- Load paths: verify that forces can be transferred through the structure to the foundations
- Constructability: check that the design can be fabricated and erected practically
- Cost optimization: evaluate alternative sections or connection types for economy
Worked Example
Problem: Verify a typical steel member for the following conditions:
Typical span: 6.0 m | Load: service loads per applicable code | Section: common section in this category
Design Check:
- Determine governing load combination (LRFD or ASD per applicable code)
- Calculate maximum internal forces (moment, shear, axial)
- Compute nominal capacity per code provisions
- Apply resistance/safety factors
- Verify interaction if combined forces exist
Result: Use the results from the Steel Calculator tool to verify design adequacy.
Frequently Asked Questions
What Australian Standard governs structural steel design?
AS 4100-2020 (Steel Structures) is the primary standard for structural steel design in Australia. It covers all aspects of design including member capacity, connections, serviceability, and fire resistance. The standard uses a limit states design philosophy with resistance factors (φ) applied to nominal capacities. Companion standards include AS/NZS 3679.1 for hot-rolled sections, AS/NZS 1554 for welding, and AS/NZS 4600 for cold-formed steel.
What are the common steel grades used in Australian construction?
The most common steel grades for Australian construction are Grade 300 and Grade 350 per AS/NZS 3679.1. Grade 300 (minimum yield 300 MPa for sections > 12 mm thick) is the standard for general structural applications. Grade 350 (minimum yield 340 MPa for sections > 12 mm) is used where higher strength reduces weight. Grade 400 and Grade 450 are available for specialized applications requiring higher strength-to-weight ratios.
How does AS 4100 compare to AISC 360?
Both AS 4100 and AISC 360 use limit states design (LRFD) principles. Key differences include: AS 4100 uses a single "capacity factor" φ approach rather than separate φ for different failure modes; AS 4100 specifies distinct buckling curves for hot-rolled and welded sections; the moment capacity formula in AS 4100 uses αm factor directly rather than Cb; and AS 4100 has more detailed provisions for slender sections and combined actions. Despite philosophical differences, both codes produce similar results for typical members.
Frequently Asked Questions
What is the advantage of concrete-filled steel tube columns over steel-only columns?
Concrete-filled tubes provide 40-55% higher axial capacity for the same steel tube, improved fire resistance (bare steel fails at 550°C, filled section continues for 60-120 minutes), no external formwork required (steel tube serves as permanent form), and enhanced ductility under seismic loading. The concrete core also prevents local buckling of the steel tube wall.
What buckling curve should be used for concrete-filled CHS columns?
Per EN 1994-1-1 Table 6.5, use buckling curve a (α = 0.21) for hot-finished CHS (EN 10210) filled with concrete. For cold-formed CHS (EN 10219), use curve c (α = 0.49). The difference is due to the higher residual stresses in cold-formed sections.
Related Pages
- Shear Stud Design — Shear connectors per EN 1994-1-1
- HSS Section Properties — CHS, RHS, SHS tables
- Column Design — Steel column per EN 1993-1-1
- All European References
Educational reference only. Design per EN 1994-1-1:2004 Clause 6.7. γ_Ma = 1.00, γ_c = 1.50. Verify concrete strength and steel grade. Results are PRELIMINARY — NOT FOR CONSTRUCTION without independent verification.
Design Resources
Calculator tools
- Column Capacity Calculator
- Steel Buckling Calculator
- Steel Column Base Design Calculator
- Composite Beam Design Calculator
- Composite Beam Calculator
Design guides
- Column Capacity Worked Example
- Column Buckling Guide
- Column Buckling Calculator Guide
- EN 1993-1-1 Column Buckling Worked Example
Reference pages