AS 4100 Column Buckling Worked Example — 200UC46, ac Calculation, Clause 6
A steel column carrying axial compression can fail by buckling long before the steel reaches yield. AS 4100 Clause 6 handles this through the ac reduction factor, converting the section capacity Ns (based on area and yield) into the member capacity Nc (accounting for slenderness). Understanding the ac calculation is important because small changes in effective length produce large changes in capacity. This page works through the full ac calculation for a 200UC46.2 column, both axes.
PRELIMINARY — NOT FOR CONSTRUCTION. This is an educational worked example. All designs must be independently verified by a licensed Professional Engineer before use in any project.
Design Brief
| Parameter | Value |
|---|---|
| Column | 200UC46.2, Grade 300PLUS (fy = 300 MPa) |
| Section properties | Ag = 5,900 mm2, rx = 88.0 mm, ry = 51.3 mm, tf = 11.0 mm |
| Storey height | L = 3,500 mm (floor to floor) |
| Frame type | Braced frame (sway prevented) |
| End conditions | Pinned-pinned (ke = 1.0 both axes) |
| Form factor | kf = 1.0 (section is fully compact) |
Step 1 — Determine Form Factor kf
Check plate slenderness against yield limits in Table 6.2.4:
Flange: lambda-ef = (204 - 7.3)/(2 x 11.0) x sqrt(300/250) = 9.79 < 16 (yield limit) --> Not slender
Web: lambda-ew = (181/7.3) x 1.095 = 27.1 < 45 (yield limit) --> Not slender
kf = 1.0
Step 2 — Section Capacity Ns
Ns = kf x Ag x fy = 1.0 x 5,900 x 300 = 1,770 kN
Design Ns = 0.90 x 1,770 = 1,593 kN
This is the pure squash load capacity before buckling.
Step 3 — Effective Length and Modified Slenderness
For braced frame, pinned ends, ke = 1.0:
Le_x = Le_y = 1.0 x 3,500 = 3,500 mm
Major axis: Le_x/rx = 3,500/88.0 = 39.8
lambda-n_x = 39.8 x 1.0 x 1.095 = 43.6
Minor axis: Le_y/ry = 3,500/51.3 = 68.2
lambda-n_y = 68.2 x 1.0 x 1.095 = 74.7
The minor axis slenderness is substantially higher (68.2 vs 39.8), which is typical for UC sections. Minor axis will govern.
Step 4 — Select Buckling Curve
AS 4100 Table 6.3.3 for hot-rolled UC with tf = 11.0 mm (< 40 mm):
alpha-b = 0.0 (Curve b — most hot-rolled UB/UC sections)
Step 5 — Calculate ac (Slenderness Reduction Factor)
The ac formula from AS 4100 Clause 6.3.3 uses the Perry-Robertson formulation.
Major axis (lambda-n_x = 43.6):
eta = 0.0 x (43.6 - 13.5)/15.3 = 0.0
ksi = (43.6/90)^2 + 1 + 0.0 = 0.234 + 1 = 1.234
ksi x lambda-n = 1.234 x 43.6 = 53.8
Since 53.8 < 90, the column is on the plateau (stocky range):
ac_x = 1.0
Minor axis (lambda-n_y = 74.7) — GOVERNING:
eta = 0.0 x (74.7 - 13.5)/15.3 = 0.0
ksi = (74.7/90)^2 + 1 + 0.0 = 0.690 + 1 = 1.690
ksi x lambda-n = 1.690 x 74.7 = 126.3
Since 126.3 > 90, use full formula:
ac_y = 1.690 x [1 - sqrt(1 - (90/126.3)^2)]
= 1.690 x [1 - sqrt(1 - 0.7126^2)]
= 1.690 x [1 - sqrt(0.4922)]
= 1.690 x [1 - 0.7016]
= 1.690 x 0.2984
= 0.504
Step 6 — Member Capacity Nc
Nc_x = ac_x x Ns = 1.000 x 1,770 = 1,770 kN (major axis — not governing)
Nc_y = ac_y x Ns = 0.504 x 1,770 = 892 kN (minor axis — GOVERNING)
Design Nc = 0.90 x 892 = 803 kN
The minor axis buckling capacity governs at 803 kN.
Step 7 — Sensitivity: Fixed-Fixed (ke = 0.85)
If connections provided rotational restraint:
Le_y = 0.85 x 3,500 = 2,975 mm
Le_y/ry = 2,975/51.3 = 58.0
lambda-n_y = 58.0 x 1.095 = 63.5
ksi = (63.5/90)^2 + 1 = 1.498
ksi x lambda-n = 95.2 (> 90, use full formula)
ac_y = 1.498 x [1 - sqrt(1 - (90/95.2)^2)] = 0.829
Nc_y = 0.829 x 1,770 = 1,467 kN
Design Nc = 0.90 x 1,467 = 1,320 kN
A 15% reduction in effective length (ke from 1.0 to 0.85) increases capacity by 64% (803 to 1,320 kN).
Summary Table
| Parameter | Major Axis (x) | Minor Axis (y) | Governing |
|---|---|---|---|
| Le (mm) | 3,500 | 3,500 | — |
| r (mm) | 88.0 | 51.3 | Minor |
| Le/r | 39.8 | 68.2 | Minor |
| lambda-n | 43.6 | 74.7 | Minor |
| alpha-b | 0.0 | 0.0 | — |
| On plateau? | Yes (< 90) | No (126.3) | — |
| ac | 1.000 | 0.504 | Minor |
| Nc (kN) | 1,770 | 892 | Minor |
| Design Nc (kN) | 1,593 | 803 | Minor: 803 kN |
Frequently Asked Questions
Why does the minor axis always govern for UC sections?
UC sections have substantially more depth than width, so ry is much smaller than rx (approximately 0.55 x rx). Since lambda-n = (Le/r) x sqrt(fy/250), a smaller r means a larger lambda-n, and the ac reduction is steeper. Unless the column is braced at shorter intervals in the y-direction (e.g., intermediate girts), the minor axis always governs.
When can I use ke = 0.85 instead of ke = 1.0?
ke = 0.85 is appropriate for braced-frame columns where the column is nominally fixed at the base (base plate with four anchor bolts) and pinned at the top (simple connections). The alignment chart method in Figure 4.6.3.2 provides more precise values based on relative beam-to-column stiffness. For sway frames, ke > 1.0 applies.
What does kf < 1.0 mean in practice?
kf < 1.0 means some plate elements exceed the yield slenderness limit, and the effective area is less than the gross area. This occurs with fabricated sections or slender elements. For standard hot-rolled UB/UC sections, kf is almost always 1.0 because mills roll sections within compact limits.
How reliable is the ac calculation compared to physical testing?
The AS 4100 Perry-Robertson formulation is calibrated against extensive buckling tests on Australian steel. The five curves capture different imperfection sensitivities. Design is generally conservative by 5-15% at intermediate slenderness (lambda-n = 40-80) because the formulation assumes geometric imperfections grow elastically, while real columns have post-buckling reserve from plastic redistribution.
This worked example is for educational purposes only. Verify all section properties, material grades, effective length factors, and code clauses against the current editions of AS 4100 and AS/NZS 3679.1 before use in any project. All designs must be independently checked and certified by a licensed Professional Engineer.