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Steel Column Buckling

AISC 360 §E3 column compression capacity — Fe, Fcr, Pn, φPn for any KL/r including elastic and inelastic ranges.

Inputs

AISC 360 §E3
  • Fe = π²E / (KL/r)².
  • If KL/r ≤ 4.71√(E/Fy): Fcr = (0.658^(Fy/Fe))·Fy else Fcr = 0.877·Fe.
  • Pn = Fcr·A, φPn = 0.9·Pn.

Results

KL/r (slenderness)
Slenderness Limit
Fe (Euler)
Fcr
Pn
φPn
Mode

About this calculator

This tool computes the axial compressive strength of a steel column using the AISC 360 Chapter E compression-member provisions, covering both the inelastic (short-to-intermediate) and elastic (slender) buckling ranges. Given the member's effective length, radius of gyration, area, and yield strength, it returns the slenderness ratio, Euler buckling stress, critical stress, and factored design strength. It is intended for structural engineers doing preliminary axial-capacity checks on braced or unbraced steel columns.

Method & formulas

Slenderness ratio: KL/r, where K is the effective length factor set by the end restraint (e.g. 1.0 for pin-pin, 0.5 for fix-fix, 2.0 for fix-free), L is the unbraced length, and r is the governing (typically minimum) radius of gyration.

Elastic (Euler) buckling stress: Fe = π²E / (KL/r)². Slenderness limit separating the inelastic and elastic regimes: (KL/r)lim = 4.71·√(E/Fy).

Fe = π²E / (KL/r)²
if KL/r ≤ 4.71√(E/Fy): Fcr = 0.658(Fy/Fe)·Fy (inelastic)
else: Fcr = 0.877·Fe (elastic)

Nominal and design strength: Pn = Fcr·A; φPn = φc·Pn, with φc = 0.90.

Worked example

Inputs: A = 6000 mm², r = 50 mm, E = 200,000 MPa, Fy = 248 MPa (A36), L = 3500 mm, K = 1.0, φc = 0.90.

KL/r = 1.0×3500/50 = 70.
(KL/r)lim = 4.71×√(200,000/248) ≈ 133.8 → since 70 ≤ 133.8, inelastic buckling governs.
Fe = π²×200,000/70² ≈ 402.8 MPa.
Fcr = 0.658(248/402.8) × 248 = 0.6580.616 × 248 ≈ 191.7 MPa.
Pn = 191.7 × 6000 / 1000 ≈ 1150 kN.
φPn = 0.90 × 1150 ≈ 1035 kN.

Assumptions & limitations

FAQ

What's the physical difference between "inelastic" and "elastic" column buckling?
Short, stocky columns partially yield before they buckle, so their strength is governed by an inelastic curve tied to Fy; long, slender columns buckle elastically at a stress well below yield, governed purely by Euler's formula. The 4.71√(E/Fy) slenderness limit marks the transition between the two behaviors.
Why does doubling the unbraced length reduce capacity so much?
Fe is inversely proportional to (KL/r)², so doubling KL cuts the elastic buckling stress to roughly a quarter; Fcr follows that same steep reduction (moderated somewhat by the inelastic curve at lower slenderness), which is why column capacity is very sensitive to unbraced length.
Which radius of gyration should I use for r?
Use the smallest (typically weak-axis) radius of gyration unless bracing effectively restrains that axis at a shorter unbraced length than the strong axis — the governing check is whichever axis produces the larger KL/r.