AS 4100 Compression Member Design Philosophy

AS 4100:2020 Clause 6 governs the design of steel columns, struts, and compression members. The standard uses a limit state approach with a distinction between:

• Section capacity (N_s): The maximum axial compression the cross-section can resist based on the yield strength (local buckling limit)
• Member capacity (N_c): The maximum axial compression the entire member can resist based on overall buckling (Euler buckling with initial imperfections)

The design condition is: N* ≤ φ_c × N_c (for overall member buckling) and N* ≤ φ_c × N_s (for local section capacity), where N* is the design axial compression force from the governing load combination.

Capacity Factors for Compression

The φ_c = 0.90 factor for compression matches the AISC 360 resistance factor of 0.90 for LRFD and EN 1993-1-1 γ_M1 = 1.0 (equivalent factor). This consistency across codes simplifies cross-border steel design.

Section Compression Capacity (N_s)

The section compression capacity is the squash load of the cross-section:

N_s = k_f × A_n × F_y

Where:

• k_f = form factor (accounts for local buckling of slender (Class 4) elements)
• A_n = net area of the section (gross area A_g if no holes)
• F_y = yield strength of the steel

Form Factor (k_f)

For sections with all elements classified as Class 1, 2, or 3 (compact or non-compact), k_f = 1.0 — the full section is effective in compression.

For sections with slender (Class 4) elements, k_f < 1.0 and is calculated as:

k_f = A_eff / A_g

Where A_eff is the effective area of the section considering local buckling of slender plate elements per AS 4100 Clause 6.2.2.

Section Classification for Compression (AS 4100 Clause 5.2)

For Australian sections in 300PLUS (F_y = 300 MPa):

• 310UC158: flange b/2t_f ≈ 9.5 → Class 3, web d_1/t_w ≈ 24.6 → Class 1
• 200UC52.2: flange b/2t_f ≈ 11.2 → Class 3, web d_1/t_w ≈ 20.0 → Class 1
• 150UC23.4: flange b/2t_f ≈ 8.7 → Class 2, web d_1/t_w ≈ 31.4 → Class 1

Most Australian UC sections have Class 2 or 3 flanges with Class 1 webs in pure compression — the flange slenderness is typically the governing classification.

Member Compression Capacity (N_c)

The member compression capacity for overall buckling is:

N_c = α_c × N_s ≤ N_s

Where α_c is the slenderness reduction factor determined from the non-dimensional slenderness λ_n and the column curve appropriate for the section type.

Non-Dimensional Slenderness (λ_n)

λ_n = (L_e / r) × √(k_f × F_y / 250)

Where:

• L_e = effective length (k_e × L) — see below
• r = radius of gyration about the buckling axis
• k_f = form factor (from above)
• F_y = yield strength in MPa

Note: AS 4100 uses a modified slenderness λ_n that normalises to F_y = 250 MPa (rather than the Euler stress as in AISC and EN 1993). The Australian λ_n is related to the Euler-based λ_e by:

λ_n = λ_e × √(F_y / 250) where λ_e = (L_e / r) × √(F_y / (π² × E))

Column Curves and α_b Factor

AS 4100 uses four column curves (a, b, c, d) selected based on the section type, the axis of buckling, and the method of manufacture:

The α_b factor modifies the slenderness reduction factor α_c through the column curve equations in AS 4100 Table 6.3.3(1):

For λ_n ≤ 0.422: α_c = 1.0 (no buckling reduction — the Euler stress exceeds yield)

For λ_n > 0.422: α_c = (ξ + λ_n² × (1 - ξ) + √((1 + ξ)² - 4 × λ_n²)) / (2 × λ_n²)

Where ξ = ((λ_n - 0.422) / 2.57)² + α_b × (λ_n - 0.422)² / (λ_n² + 0.422²)

The α_c factor reduces from 1.0 (at λ_n = 0.422) asymptotically to the Euler hyperbola (α_c = 250 / (λ_n² × F_y)) at high slenderness.

Column Curve Comparison

The difference between curves a and d is significant — at λ_n = 1.0, curve a gives 17.5% more capacity than curve d. The column curve selection has a material impact on column design economy. Hot-finished CHS columns (curve a) have a substantial advantage over cold-formed SHS (curve c) at intermediate slenderness.

Effective Length (L_e)

The effective length L_e = k_e × L accounts for end restraint conditions:

For Australian steel building frames:

• Braced frames (non-sway): k_e = 0.85-1.0 (typical for columns in braced bays)
• Unbraced frames (sway): k_e = 1.2-2.0 (depends on beam-to-column stiffness ratio — alignment chart method)
• Portal frame columns: k_e = 1.0-1.5 (pinned base) or 0.8-1.0 (fixed base) — use alignment chart

AS 4100 Clause 6.3.2 permits the use of k_e = 1.0 for braced frames without further refinement — a conservative approach that simplifies design office practice.

Worked Example: 310UC158 Column Design

Problem: Check a 310UC158 column in Grade 300PLUS for an axial compression load of N* = 3,200 kN.

Given:

• Section: 310UC158 (Australian Universal Column)
• Grade: 300PLUS (F_y = 300 MPa for t ≤ 20 mm)
• Gross area: A_g = 20,200 mm²
• Radii of gyration: r_x = 138 mm, r_y = 79.1 mm
• Flange: b_f = 327 mm, t_f = 21.7 mm, b/2t_f = 7.53
• Web: d = 308 mm, t_w = 12.6 mm, d_1/t_w = (308 - 2×21.7)/12.6 = 21.0
• Column height: L = 4.5 m
• End conditions: Pinned base, pinned top (braced frame)
• k_e = 1.0 (conservative for braced frame)

Step 1 — Section classification for compression:

Flange: b/2t_f = 7.53 < 8 → Class 1 (plastic) ✓ Web: d_1/t_w = 21.0 < 35 → Class 1 (plastic) ✓

All elements Class 1 → k_f = 1.0 (full section effective)

Step 2 — Section capacity N_s:

N_s = k_f × A_g × F_y = 1.0 × 20,200 × 300 / 1,000 = 6,060 kN

Section capacity check: N* / (φ_c × N_s) = 3,200 / (0.90 × 6,060) = 3,200 / 5,454 = 0.587 ✓

The section has adequate capacity at 59% utilisation.

Step 3 — Member capacity about minor axis (y-y):

The minor axis governs because r_y < r_x.

λ_n(y) = (L_e / r_y) × √(k_f × F_y / 250) = (1.0 × 4,500 / 79.1) × √(1.0 × 300 / 250) = 56.9 × √1.2 = 56.9 × 1.095 = 62.3

The non-dimensional slenderness is 62.3.

Step 4 — Determine α_b for UC section about minor axis:

From AS 4100 Table 6.3.3(2): UB and UC sections buckling about the minor (y-y) axis use Curve c → α_b = +0.5.

Step 5 — Calculate α_c:

For λ_n = 62.3, using the standard AS 4100 Table 6.3.3(1) interpolation:

λ_n = 60 → α_c = 0.877 (Curve c) λ_n = 70 → α_c = 0.821 (Curve c)

Interpolating for λ_n = 62.3: α_c = 0.877 - (62.3 - 60)/(70 - 60) × (0.877 - 0.821) = 0.877 - 0.23 × 0.056 = 0.877 - 0.013 = 0.864

Step 6 — Member capacity N_c:

N_c(y) = α_c × N_s = 0.864 × 6,060 = 5,236 kN

Member capacity (minor axis): φ_c × N_c(y) = 0.90 × 5,236 = 4,712 kN

Check: N* / (φ_c × N_c) = 3,200 / 4,712 = 0.679 ✓ (68% utilisation)

Step 7 — Check major axis (x-x):

λ_n(x) = (1.0 × 4,500 / 138) × √(1.0 × 300 / 250) = 32.6 × 1.095 = 35.7

For UB/UC sections about major axis (x-x): Curve b → α_b = 0.0

From AS 4100 Table 6.3.3(1): λ_n = 30 → α_c = 0.958 (Curve b) λ_n = 40 → α_c = 0.918 (Curve b)

Interpolating for λ_n = 35.7: α_c = 0.958 - (35.7 - 30)/(40 - 30) × (0.958 - 0.918) = 0.958 - 0.57 × 0.040 = 0.958 - 0.023 = 0.935

N_c(x) = 0.935 × 6,060 = 5,666 kN

φ_c × N_c(x) = 0.90 × 5,666 = 5,099 kN (not governing — minor axis governs as expected)

Step 8 — Conclusion:

The 310UC158 column in Grade 300PLUS is adequate for N* = 3,200 kN. The minor axis buckling governs with 68% utilisation. The column is well-sized for this load level.

Interaction of Compression and Bending (Beam-Columns)

AS 4100 Clause 8 governs combined compression and bending (beam-columns):

Section capacity (Clause 8.3):

(M_x* / φ × M_sx)^1.4 + (M_y* / φ × M_sy)^1.4 ≤ 1.0 for N* ≤ φ_c × N_s

Member capacity (Clause 8.4):

For in-plane buckling: M_x* ≤ φ × M_ix = φ × M_sx × (1 - N* / φ_c × N_cx)

For out-of-plane buckling (with LTB): M_x* ≤ φ × M_ox = φ × M_bx × (1 - N* / φ_c × N_cy)

These interaction equations are similar in form to AISC 360 Chapter H but use the Australian capacity factors and column curves.

Column Splices

AS 4100 Clause 6.4 requires column splices to:

• Transfer the full design compression N* through contact bearing if ends are prepared to bear (machined or sawn)
• Provide bolting or welding for at least 50% of the member capacity for tension (if tension can occur under uplift or overturning)
• Splices in ductile moment frames (seismic SMRF) must develop at least 1.2 × N_s for tension and compression

For a 310UC158 splice: minimum splice bolts = 50% × 300 MPa × 20,200 mm² = 3,030 kN in tension. This typically requires 8-12 M24 Grade 8.8 bolts per flange cover plate.

Frequently Asked Questions

What is the difference between section capacity N_s and member capacity N_c in AS 4100?

Section capacity N_s = k_f × A_n × F_y is the squash load of the cross-section, accounting for local buckling through the form factor k_f. Member capacity N_c = α_c × N_s accounts for overall buckling (Euler buckling with initial imperfections). For stocky columns (λ_n < 0.422), N_c = N_s — the column is limited by material yielding, not buckling. For slender columns (λ_n > 0.422), N_c < N_s. In typical building columns (λ_n = 30-80), the member capacity reduction ranges from 0-15% (stocky columns) to 30-50% (slender columns).

Which column curve should I use for an Australian UC section?

UC sections (Universal Columns) use Curve b (α_b = 0.0) for buckling about the major (x-x) axis and Curve c (α_b = +0.5) for buckling about the minor (y-y) axis. Since most columns are controlled by minor-axis buckling (r_y < r_x), the Curve c value typically governs. The Australian system has a 15-20% difference between Curve b and Curve c at λ_n = 60-80, so the column curve selection has a meaningful impact on design economy.

How does the AS 4100 column curve system compare with AISC 360 and EN 1993?

AISC 360 uses a single column curve (Fcr/Fy = 0.658^(Fy/Fe) for inelastic and 0.877Fe for elastic buckling). EN 1993-1-1 uses five curves (a0, a, b, c, d). AS 4100 uses four curves (a, b, c, d) that approximately align with EN 1993 curves a, b, c, and d respectively — the Australian system was intentionally harmonised with the European approach during the 1998 revision. Compared to AISC 360, the Australian Curve b is approximately 5-10% more conservative at intermediate slenderness (λ_n = 60-100), reflecting the different initial imperfection assumptions.

Related Pages

• AS 4100 Steel Design Overview — Australia — Full AS 4100 design reference
• AS 4100 Load Combinations — AS 1170.0 — Load combination guide for steel design
• Australian Steel Grades — AS/NZS 3678 & 3679.1 — Material properties
• AS 4100 Base Plate Design Guide — Column base plate design per AS 4100
• Australian Wind Load — AS 1170.2 — Wind load on steel structures
• Beam Capacity Calculator — Free multi-code beam calculator
• Column Compression Strength Tool — Free multi-code column calculator
• Section Properties — UB, UC, PFC — Australian section tables

Column Buckling Theory

Euler Buckling

The Euler buckling load represents the theoretical critical load for an ideal elastic column:

Pcr = π²EI / (KL)²

Where:

• E = modulus of elasticity (200 GPa for steel)
• I = moment of inertia about the buckling axis
• K = effective length factor
• L = unbraced length

Real Column Behavior

Real columns deviate from Euler theory due to:

• Initial out-of-straightness (typically L/1000)
• Residual stresses from manufacturing (hot-rolling or welding)
• Eccentricity of applied load
• Inelastic material behavior

These effects are accounted for through column strength curves that reduce the theoretical Euler capacity based on slenderness ratio (KL/r) and section type.

[object Object]

[object Object]

Frequently Asked Questions

What is the recommended design procedure for this structural element?

The standard design procedure follows: (1) establish design criteria including applicable code, material grade, and loading; (2) determine loads and applicable load combinations; (3) analyze the structure for internal forces; (4) check member strength for all applicable limit states; (5) verify serviceability requirements; and (6) detail connections. Computer analysis is recommended for complex structures, but hand calculations should be used for verification of critical elements.

How do different design codes compare for this calculation?

AISC 360 (US), EN 1993 (Eurocode), AS 4100 (Australia), and CSA S16 (Canada) follow similar limit states design philosophy but differ in specific resistance factors, slenderness limits, and partial safety factors. Generally, EN 1993 uses partial factors on both load and resistance sides (γM0 = 1.0, γM1 = 1.0, γM2 = 1.25), while AISC 360 uses a single resistance factor (φ). Engineers should verify which code is adopted in their jurisdiction.

Educational reference only. Compression member design per AS 4100:2020 Clause 6. Verify effective length factors, column curves, and section classification for your specific design. Results are PRELIMINARY — NOT FOR CONSTRUCTION without independent verification.

Design Resources

Calculator tools

• Column Compression Strength Tool
• Steel Buckling Calculator
• Steel Column Base Design Calculator

Design guides

• Column Capacity Worked Example
• Column Buckling Guide
• Column Buckling Calculator Guide
• EN 1993-1-1 Column Buckling Worked Example