CALC

Engineering

Column Axial Load Capacity Calculator

Calculate effective length, Euler load, squash load, reduced reference capacity, utilization, margin, radius of gyration, and slenderness.

Effective column length (mm)-
Ideal Euler buckling load (kN)-
Ideal axial yield load (kN)-
Lower idealized nominal reference (kN)-
Capacity after entered factor (kN)-
Applied load divided by reduced capacity-
Reduced capacity minus applied load (kN)-
Radius of gyration (mm)-
Effective slenderness-

Decision view

Column capacity engineering diagram

Column capacity engineering diagramThe member sketch and capacity hierarchy connect effective length and slenderness to Euler, squash, reduced, and applied loads.
Exact scenario comparisonApplied axial load (kN) changes while all other entered assumptions remain constant.
Applied axial load (kN)Effective column length (mm)Ideal Euler buckling load (kN)Ideal axial yield load (kN)Lower idealized nominal reference (kN)Capacity after entered factor (kN)Applied load divided by reduced capacityReduced capacity minus applied load (kN)Radius of gyration (mm)Effective slenderness

How to use Column Axial Load Capacity Calculator

  1. Enter column geometry, material values, K factor, reduction factor, and applied load.
  2. Compare Euler and squash references.
  3. Treat the result only as an idealized engineering screen.

Calculator guide

Understanding Column Axial Load Capacity Calculator

An idealized column check compares Euler buckling and axial yielding before applying an explicit reduction factor and load utilization.

Calculate effective length The default K factor leaves the physical length unchanged.
Calculate Euler load The division by 1,000 converts newtons to kilonewtons.
Calculate the axial yield reference The squash reference governs the two idealized limits.
Reduce and compare capacity The applied load is compared only after the entered reduction factor.

Calculation method

How the calculation works

Calculate ideal Euler and axial-yield references, select the lower value, apply an explicit reduction factor, and compare it with entered axial load. Use effective length in Euler's equation, compare with the axial yield reference, reduce the lower value, and compare with applied load.

Detailed calculation process

Build the idealized column load comparison

The default uses L=3,000 mm, K=1, E=200,000 MPa, I=8,000,000 mm⁴, A=2,400 mm², Fy=250 MPa, factor 0.65, and 180 kN applied.

General formula: L_e = KLP_E = pi^2EI/L_e^2/1000P_y = AF_y/1000P_n = min(P_E,P_y)P_r = phi P_nUtilization = 100P_applied/P_rMargin = P_r-P_appliedr_g = sqrt(I/A)lambda = L_e/r_g Euler load decreases with effective length squared, while the squash reference is area times yield strength. The lower idealized reference is reduced before applied load is compared.

What each symbol means

L, K, L_e Physical length, effective-length factor, and effective length (mm).
E Elastic modulus (MPa = N/mm²).
I Least second moment of area (mm⁴).
A Cross-sectional area (mm²).
F_y Entered yield strength (MPa).
P_E, P_y, P_n Euler, squash, and governing nominal references (kN).
phi, P_r Entered reduction factor and reduced reference capacity (no unit, kN).
r_g, lambda Radius of gyration (mm) and slenderness ratio (no unit).

Worked substitution with the default inputs

1. Calculate effective length L_e = 1(3,000)L_e = 3,000 mm The default K factor leaves the physical length unchanged.
2. Calculate Euler load P_E = pi^2(200,000)(8,000,000)/(3,000^2)/1,000P_E = 1,754.596 kN The division by 1,000 converts newtons to kilonewtons.
3. Calculate the axial yield reference P_y = 2,400(250)/1,000P_y = 600 kNP_n = min(1,754.596,600) = 600 kN The squash reference governs the two idealized limits.
4. Reduce and compare capacity P_r = 0.65(600) = 390 kNUtilization = 100(180/390) = 46.1538%Margin = 390-180 = 210 kN The applied load is compared only after the entered reduction factor.
5. Reconcile section slenderness r_g = sqrt(8,000,000/2,400) = 57.7350 mmlambda = 3,000/57.7350 = 51.9615 Radius of gyration converts area and inertia into a slenderness measure.

The default squash reference governs at 600 kN; the entered factor reduces it to 390 kN, with 46.154% utilization and 210 kN margin.

Purpose-built visual

Column section and capacity hierarchy

An engineering column sketch shows effective length and section properties beside a bullet hierarchy for Euler, squash, reduced, and applied loads.

Live The chart is regenerated from current inputs.
Units Every axis, marker, and endpoint retains its stated unit.
Check The chart reconciles to the displayed calculation.

Worked situations

Practical examples

  • The default uses L=3,000 mm, K=1, E=200,000 MPa, I=8,000,000 mm⁴, A=2,400 mm², Fy=250 MPa, factor 0.65, and 180 kN applied.
  • The default squash reference governs at 600 kN; the entered factor reduces it to 390 kN, with 46.154% utilization and 210 kN margin.

Better inputs

Useful tips

  • Change one input at a time and confirm both the result and visual move.
  • Keep the units stated beside every field.
  • Retain intermediate precision and round only the reported result.

Before relying on the result

Limitations and common mistakes

  • This is not a structural design check.
  • Residual stress, imperfections, eccentricity, bracing, connections, and local buckling are excluded.
  • Code equations and licensed engineering review govern real columns.

Reference

Key terms

Euler load
Ideal elastic buckling reference for a slender perfect column.
Squash load
Area times entered yield strength.
Slenderness
Effective length divided by radius of gyration.

Important note

Calculated from the entered values using the displayed engineering relationship. Confirm design values, load cases, safety factors, standards, and field conditions with a qualified professional.

Frequently asked questions

Is reduced capacity code-compliant?

Not by itself; the factor is user-entered.

Why use the least I?

Buckling tends to occur about the weaker axis.

Can squash load govern?

Yes, as in the default.

Does utilization include bending?

No.