Engineer's Calculator

Column Buckling Calculator

The Euler buckling load, effective length and slenderness of a column, with the Johnson limit for short columns and the AISC 360 strength of a steel column for comparison. The buckled shape and the column curve are drawn as you type. Free, no sign-up.

1. Pick the end conditions

Real connections are never perfectly fixed, so the AISC Commentary (Table C-A-7.1) recommends larger K values than theory for design. Uncheck the box to use the theoretical values.

2. Enter the column

Use the smaller moment of inertia unless the column is braced about that axis. The example is a W8×31 about its weak axis. Need I and A for a shape? Use the cross section calculator.

3. Buckling load

Steel column strength, AISC 360 section E3

Flexural buckling only. A full check also covers slender plate elements (section E7) and torsional buckling of angles, tees and other open shapes (section E4).

Critical stress against slenderness

Euler Johnson AISC 360 E3

Euler's formula is only right for slender columns. Below the transition slenderness the column yields before it buckles, and the Johnson parabola and the AISC curve take over. The AISC curve is lower because it allows for the residual stresses and the crookedness of real rolled shapes.

Sizing a real column? The WebStructural column designer picks steel W, HSS and pipe shapes or wood posts, combines axial load with bending, and checks them to AISC 360 and the NDS.

Open the column designer

The formulas

Pe= π² E I(K L)²Euler buckling load
r= √(I / A)radius of gyration
Fe= π² E(K L / r)²Euler buckling stress
Cc= √(2π² E / Fy)where Fe = Fy / 2
Fcr= Fy [1 − (KL/r)² / (2Cc²)]Johnson, KL/r < Cc
Fcr= 0.658Fy/Fe FyAISC, Fy/Fe ≤ 2.25
Fcr= 0.877 FeAISC, Fy/Fe > 2.25

Effective length factors

EndsTheoretical KRecommended K
Pinned at both ends1.01.0
Fixed base, pinned top0.70.8
Fixed at both ends0.50.65
Fixed base, free top (flagpole)2.02.1
Fixed at both ends, top free to sway1.01.2

Worked example: W8×31, 14 ft, pinned ends

About the weak axis, I = 37.1 in⁴ and A = 9.13 in², so r = 2.016 in and KL/r = 168 / 2.016 = 83.3. Euler gives Fe = π² × 29,000 / 83.3² = 41.2 ksi and Pe = 376 kips. That is less than Cc = 107, so the column is in the inelastic range: Johnson gives 34.8 ksi. With Fy = 50 ksi, AISC gives Fcr = 0.65850/41.2 × 50 = 30.1 ksi, Pn = 275 kips and φPn = 247 kips. The AISC Manual's column table lists about 248 kips for this shape and length.

Questions

Why is the Euler load higher than the AISC strength?

Euler assumes a perfectly straight, perfectly elastic column. Real columns are slightly crooked and have residual stresses from rolling and welding, so they fail sooner, especially at middle slenderness. The AISC curve is fitted to tests of real columns.

Which moment of inertia should I use?

The column buckles about the axis with the larger KL/r. If it is braced differently about each axis, check both and use the larger slenderness.

Does this work for wood or aluminum?

The Euler and Johnson results work for any material with the right E and yield stress. The AISC rows are for steel only. Wood columns are designed with the NDS column stability factor, which the WebStructural column designer handles.