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RC Column Axial Capacity Calculator

Calculate the squash (concentric) axial load capacity of a short reinforced-concrete column. The tool returns the ultimate squash load Nuo, the design capacity φNuo and the longitudinal reinforcement ratio ρ, and — if you enter an applied axial load N* — the axial utilisation with an OK / over-capacity flag.

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How to use this calculator

Enter Values

Enter your values and press Calculate to see results.

Before you rely on this: First-pass guide only. Verify safety-critical or regulated work against the relevant standards, your project requirements and a qualified professional.

How to use this calculator

  1. Enter the gross cross-section area Ag (mm²) and the total longitudinal steel area As (mm²). As must be less than Ag.
  2. Optionally override the concrete strength f'c (default 32 MPa), the steel yield fy (default 500 MPa) and the squash capacity factor φ (default 0.6).
  3. Optionally enter the applied axial load N* (kN) to see the utilisation as a percentage of the design capacity φNuo.

How it works

At the squash load the concrete carries 0.85·f'c over the net area (Ag − As) and the steel carries fy over its area As, so Nuo = 0.85·f'c·(Ag − As) + fy·As, divided by 1000 to give kN. The design capacity is φNuo with φ ≈ 0.6 for a squash load. The reinforcement ratio is ρ = As / Ag × 100. Utilisation = N* / φNuo × 100.

Worked example

Worked example. Ag = 90000 mm² (a 300×300 column), As = 2400 mm², f'c = 32 MPa, fy = 500 MPa. Nuo = (0.85×32×(90000 − 2400) + 500×2400) / 1000 = (2 382 720 + 1 200 000) / 1000 = 3582.72 kN. φNuo = 0.6 × 3582.72 = 2149.63 kN. ρ = 2400 / 90000 × 100 = 2.67 %. With N* = 1500 kN the utilisation is 69.8 % (OK).

Common mistakes

  • Using the squash load for a slender column. Nuo is the capacity of a stocky (short) column only; slender columns buckle at a lower load and must be checked for slenderness.
  • Ignoring bending. Real columns carry moment as well as axial load; design uses an axial–moment interaction diagram and a minimum design eccentricity, not the pure squash load.
  • Exceeding the code steel-ratio limits. Practical columns keep ρ within roughly 1–4 %; this tool computes ρ but does not enforce the minimum or maximum limits.

Frequently asked questions

Why subtract As from Ag for the concrete term?

The steel physically displaces concrete, so the concrete only acts over the net area (Ag − As). The 0.85 factor accounts for the difference between the cylinder strength and the sustained in-situ compressive strength of the concrete in the member.

Why is φ only 0.6 for the squash load?

A concentrically loaded column fails in a sudden, brittle crushing mode with little warning, so AS 3600 applies a lower capacity factor (about 0.6) than the 0.85 used for ductile bending. φ increases toward the tension-controlled value as bending dominates on the interaction diagram.

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