Column Load Calculator

Axial capacity of a short concrete column (ACI 318) or a slender steel column (AISC Euler buckling). Pick the mode.

Column type

Mode
in
psi
Typical 3,000-5,000 psi residential / commercial.
psi
Grade 60 standard = 60,000 psi.
in²
4 × #8 bars = 3.16 in². 1-8 % of gross area allowed (ACI).

Axial capacity

Allowable axial load

— kip

Nominal load (P_n)

— kip

Cross-section area

— in²

—

—

Concrete mode uses ACI 318 short-column axial capacity (P_n) with the 0.80 reduction for tied columns. Steel mode uses Euler buckling capacity (no LTB or local-buckling adjustments — full AISC 360 method requires more inputs).

Informational only. Concrete mode assumes a short column (slenderness ratio kL/r ≤ 22). Steel mode does not check local buckling, slenderness limit (KL/r ≤ 200), or material yield governing. For any structural column, full design must follow ACI 318 / AISC 360 by a licensed engineer.

Formulas

Concrete short column (ACI 318)

Pn = 0.80 × [0.85 · f'c · (Ag − Ast) + fy · Ast]

The 0.80 factor accounts for accidental eccentricity in tied columns (use 0.85 for spiral columns). Result is the nominal capacity Pn. For LRFD, multiply by φ = 0.65 (tied) for design capacity. The calculator reports both.

Steel Euler buckling

Pcr = π² · E · I / (K · L)²

With E = 29,000 ksi, I = minimum (weak-axis) moment of inertia, K = effective length factor and L = unbraced length. For very stocky columns the material yield governs instead: Py = Fy × A.

Worked example — 12 × 12 concrete column, f'c = 4 ksi, 4 #8 rebars (3.16 in²)

  • Ag = 12 × 12 = 144 in²
  • Pn = 0.80 × [0.85 × 4 × (144 − 3.16) + 60 × 3.16]
  • = 0.80 × [0.85 × 4 × 140.84 + 189.6]
  • = 0.80 × [478.86 + 189.6]
  • = 0.80 × 668.46 = 534.8 kip nominal
  • Design φPn = 0.65 × 534.8 = 347.6 kip

Worked example — W8×31, 12 ft unbraced, K = 1.0

  • Iy = 37.1 in⁴ (weak axis governs)
  • (KL)² = (1.0 × 144)² = 20,736 in²
  • Pcr = π² × 29,000 × 37.1 / 20,736 = 512 kip
  • Material yield check: Py = 50 × 9.13 = 456 kip (governs!)
  • Allowable = min(Pcr, Py) / 1.67 = 273 kip (ASD)

Frequently asked questions

How much load can a 12 inch concrete column carry?
A 12×12 in tied column with f'c = 4,000 psi and 4 #8 bars (3.16 in² steel, ~2.2 % ratio) has nominal axial capacity ≈ 535 kip and design (φP_n) ≈ 348 kip. With less steel (1 % minimum), capacity drops to ~470 kip nominal.
What is the formula for concrete column axial capacity?
ACI 318 short tied column: P_n = 0.80 × [0.85·f'c·(A_g − A_st) + f_y·A_st]. The 0.80 factor accounts for accidental eccentricity. For spiral columns, use 0.85 instead of 0.80. Design strength is φP_n with φ = 0.65 (tied) or 0.75 (spiral).
When does Euler buckling govern a steel column?
When the column is slender — kL/r > 4.71·√(E/F_y) per AISC. For A992 steel that threshold is kL/r ≈ 113. Below that, material yield (or inelastic buckling) governs; above, elastic Euler buckling. Short stout columns yield; tall thin columns buckle.
What is K factor in column design?
Effective length factor — multiplies the actual unbraced length to account for end conditions. K = 1.0 for pinned-pinned (standard), K = 0.5 for fixed-fixed, K = 0.7 for fixed-pinned, K = 2.0 for fixed-free (cantilever). AISC Table C-A-7.1 lists all common cases.
How much reinforcement is required in a concrete column?
ACI 318 §10.6.1: minimum 1 % of gross cross-section area (A_st ≥ 0.01 A_g), maximum 8 %. Practical range 2-4 % for typical residential / commercial. Each face needs at least one longitudinal bar at each corner (4 bars minimum for square tied).
Does this calculator account for slenderness?
Concrete mode assumes a short column (kL/r ≤ 22). For slender concrete columns, ACI 318 requires a moment-magnification analysis or a P-Delta analysis — beyond this calculator. Steel mode does compute Euler buckling capacity, but does not check whether yield governs first — the worked example shows how to check both.