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Steel · AISC 360-22

Steel column flexural buckling: an AISC 360-22 §E3 worked example

Classify the elements, find the governing slenderness, pick the right critical stress equation, then factor it. A W10x33 column worked through every step.

Axial compression strength of a steel column is a buckling problem long before it is a yielding problem. AISC 360-22 Chapter E covers it. For the common case, a rolled W-shape whose flanges and web are not slender, the governing limit state is usually flexural buckling under §E3, and the whole check fits in five equations. The equations are short; the mistakes happen in the setup: the wrong axis, the wrong effective length, or the wrong side of the inelastic/elastic boundary. This article walks through the procedure and then a complete example.

Step 1: confirm the section has no slender elements

§E3 applies to members without slender elements. Element slenderness for axial compression is classified under §B4.1 using Table B4.1a. For a rolled W-shape the two cases are:

If λ ≤ λr for every element, the section is nonslender for compression and §E3 (with §E4 where it applies) gives the strength. If any element exceeds λr, the member goes to §E7, which reduces the area through effective widths. The AISC Manual flags W-shapes that are slender for compression at Fy = 50 ksi with a footnote in Table 1-1.

Step 2: effective length and slenderness

Since the 2016 edition, and continuing in AISC 360-22, the Specification writes the effective length as Lc rather than KL:

Lc = K · L

K comes from the stability method in use. Under the direct analysis method of Chapter C, the effective length for flexural buckling is taken as the unbraced length (K = 1.0) unless a smaller value is justified by rational analysis. Under the effective length method of Appendix 7, K comes from the frame analysis or the alignment charts. A column pinned at both ends in a braced frame has K = 1.0 either way.

Compute Lc/r about both principal axes, each with its own unbraced length, and use the larger. For a W-shape braced equally about both axes, the weak axis (ry) governs. A user note in §E2 recommends Lc/r not exceed 200.

Step 3: elastic buckling stress

Fe = π²E / (Lc/r)²   (AISC 360-22 Eq. E3-4)

This is the Euler stress for the governing axis. For doubly symmetric members, §E4 also defines an elastic torsional buckling stress. For rolled W-shapes with the torsional unbraced length equal to the weak-axis length, flexural buckling about the weak axis governs; the example below checks this.

Step 4: critical stress, inelastic or elastic

When Lc/r ≤ 4.71√(E/Fy)  (or Fy/Fe ≤ 2.25):
  Fcr = 0.658Fy/Fe · Fy   (Eq. E3-2)

When Lc/r > 4.71√(E/Fy)  (or Fy/Fe > 2.25):
  Fcr = 0.877 · Fe   (Eq. E3-3)

The two forms of the limit are the same boundary: substituting Lc/r = 4.71√(E/Fy) into Eq. E3-4 gives Fe = 0.445 Fy, so Fy/Fe = 2.25. Eq. E3-2 is the inelastic range, where residual stresses cause partial yielding before buckling. Eq. E3-3 is the elastic range; the 0.877 factor reduces the Euler stress for initial out-of-straightness. For A992 steel (Fy = 50 ksi), the boundary is Lc/r = 113.4.

Step 5: nominal and available strength

Pn = Fcr · Ag   (Eq. E3-1)
LRFD: φcPn with φc = 0.90   (§E1)
ASD: Pn/Ωc with Ωc = 1.67   (§E1)

Worked example: W10x33, A992, Lc = 14 ft

A W10x33 column in a braced frame, pinned at both ends, 14 ft between supports, with no intermediate bracing about either axis. A992 steel: Fy = 50 ksi, E = 29,000 ksi. K = 1.0 about both axes, so Lc = 14 × 12 = 168 in.

W10x33 section properties (AISC Manual Table 1-1)
PropertyValuePropertyValue
Ag9.71 in²bf7.96 in
rx4.19 intf0.435 in
ry1.94 intw0.290 in
Ix171 in⁴bf/2tf9.15
Iy36.6 in⁴h/tw27.1
J0.583 in⁴Cw791 in⁶

Element slenderness (Table B4.1a)

√(E/Fy) = √(29,000 / 50) = 24.08
Flange: bf/2tf = 7.96 / (2 × 0.435) = 9.15 ≤ 0.56 × 24.08 = 13.49   nonslender
Web: h/tw = 27.1 ≤ 1.49 × 24.08 = 35.88   nonslender
→ no slender elements; §E3 applies

Governing slenderness

Lc/rx = 168 / 4.19 = 40.1
Lc/ry = 168 / 1.94 = 86.6   ← governs (weak axis)
86.6 ≤ 200   (§E2 user note)

Elastic buckling stress, Eq. E3-4

Fe = π² × 29,000 / 86.6² = 38.17 ksi

Which Fcr equation

4.71 × √(29,000 / 50) = 113.4
Lc/r = 86.6 ≤ 113.4   (check: Fy/Fe = 50 / 38.17 = 1.310 ≤ 2.25)
→ inelastic buckling, Eq. E3-2

Critical stress, Eq. E3-2

Fcr = 0.6581.310 × 50 = 0.5779 × 50 = 28.90 ksi

Nominal and available strength

Pn = 28.90 × 9.71 = 280.6 kips   (Eq. E3-1)
LRFD: φcPn = 0.90 × 280.6 = 252.5 kips
ASD: Pn/Ωc = 280.6 / 1.67 = 168.0 kips

Torsional buckling cross-check (§E4)

With the torsional unbraced length also 168 in and G = 11,200 ksi, the §E4 elastic torsional buckling stress for a doubly symmetric member is (π²ECw/Lcz² + GJ) / (Ix + Iy) = 70.1 ksi. That is well above the 38.17 ksi flexural value, so weak-axis flexural buckling governs, as expected for a W-shape.

Summary, W10x33, A992, Lc = 14 ft both axes
QuantityValueReference
Flange bf/2tf9.15 ≤ 13.49Table B4.1a, Case 1
Web h/tw27.1 ≤ 35.88Table B4.1a, Case 5
Governing Lc/r86.6 (y-axis)§E2, §E3
Fe38.17 ksiEq. E3-4
Fcr28.90 ksiEq. E3-2
Pn280.6 kipsEq. E3-1
φcPn (LRFD)252.5 kips§E1, φc = 0.90
Pn/Ωc (ASD)168.0 kips§E1, Ωc = 1.67

AISC Manual Part 4 (Table 4-1a) tabulates φcPn and Pn/Ωc for W-shapes against Lc about the y-axis, which is a quick independent check of a hand or software result.

The expensive mistake: using the elastic equation in the inelastic range. Here 0.877 × Fe = 0.877 × 38.17 = 33.47 ksi, which is 16% higher than the correct 28.90 ksi. Eq. E3-3 only applies past Lc/r = 113.4, which for this shape about its weak axis means Lc beyond 113.4 × 1.94 = 220 in, or about 18.3 ft.

Practical notes

How StructuralClaw handles it

The StructuralClaw steel column module checks W-shape columns for flexural buckling about both axes, with an effective-length case picker, and carries the result into the Chapter H interaction check. Its calculation trace lists each step with substituted values, so a reviewer can see which Fcr equation was used and why. The module currently cites AISC 360-16 Chapter E; the §E3 equations shown here are unchanged in AISC 360-22. It is part of the StructuralClaw desktop program, which is coming soon. For beams, the free steel beam calculator runs a W-shape flexure and shear check in the browser today.

Column checks with every equation named

The StructuralClaw desktop app runs AISC Chapter E flexural buckling on both axes and shows which critical stress equation governed. Coming soon. Need a beam check now? The steel beam calculator is free.

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