Steel Column Design Example — AISC 360-22 LRFD Compression Member

Complete step-by-step design of a steel column following AISC 360-22. This worked example covers load determination, member selection, effective length, slenderness classification, critical stress (Fcr) calculation per Chapter E, and the combined axial-plus-bending interaction check per Chapter H. All calculations are shown explicitly.

Problem statement

Design a W14 column for the first story of a 5-story office building.

Given:

• Column: W14 section (ASTM A992, Fy = 50 ksi, Fu = 65 ksi)
• Story height: 13 ft (floor-to-floor)
• Column effective length factor: K = 1.0 (braced frame, sidesway prevented)
• Factored axial load: Pu = 580 kips
• Factored moment (from eccentric beam connection): Mux = 85 kip-ft
• Lateral support: Column braced in both directions at each floor level
• Gravity framing: Braced frame (no moment connections to column)

Step 1 — Determine required capacity

Axial load

P_u = 580 kips (given, from LRFD load combinations)

Bending moment

M_ux = 85 kip-ft = 1,020 kip-in (minor axis bending from beam eccentricity)

This is a combined loading case. Chapter H interaction will be required.

Step 2 — Select trial section

Required area for axial-only

For a preliminary estimate, assume Fcr = 30 ksi (conservative for typical KL/r):

A_req = P_u / (phi x Fcr) = 580 / (0.90 x 30) = 21.5 in^2

Trial selection from AISC Table 4-1 (Available Strength in Axial Compression)

For KL = 1.0 x 13 = 13 ft, looking for phi*Pn >= 580 kips:

The W14x68 provides phi*Pn = 590 kips > 580 kips for axial only. But with bending, we need more capacity.

Select W14x82 for the combined loading check.

Step 3 — Section properties (W14x82, A992)

Step 4 — Compact section check (Table B4.1a)

Flange

For A992 (Fy = 50 ksi):

lambda_p = 0.56 x sqrt(E/Fy) = 0.56 x sqrt(29000/50) = 0.56 x 24.08 = 13.49
lambda = bf/(2tf) = 8.51 < 13.49 — compact

Web

lambda_p = 1.49 x sqrt(E/Fy) = 1.49 x 24.08 = 35.88
lambda = h/tw = 23.8 < 35.88 — compact

W14x82 is compact for Fy = 50 ksi.

Step 5 — Effective length and slenderness (AISC Chapter E)

Effective length

For a braced frame with sidesway prevented:

K = 1.0 (pinned-pinned, sidesway inhibited)
L_x = 13 ft = 156 in (buckling about strong axis, braced by floor beams)
L_y = 13 ft = 156 in (buckling about weak axis, braced by bracing connections)

Slenderness ratio

KL/r_x = 1.0 x 156 / 6.05 = 25.8
KL/r_y = 1.0 x 156 / 4.26 = 36.6  — GOVERNS (larger value controls)

Maximum KL/r = 36.6 < 200 — OK (Table D1.1 recommended limit).

Step 6 — Critical stress Fcr (AISC Section E3)

Elastic critical stress

F_e = pi^2 x E / (KL/r)^2
F_e = pi^2 x 29,000 / (36.6)^2
F_e = 286,159 / 1,340 = 213.5 ksi

Check if inelastic or elastic buckling

4.71 x sqrt(E/Fy) = 4.71 x sqrt(29000/50) = 4.71 x 24.08 = 113.4

Since KL/r = 36.6 < 113.4, use E3-2 (inelastic buckling):

F_cr = [0.658^(Fy/Fe)] x Fy
F_cr = [0.658^(50/213.5)] x 50
F_cr = [0.658^0.234] x 50
F_cr = 0.903 x 50 = 45.2 ksi

Nominal compressive strength

P_n = F_cr x A_g = 45.2 x 24.1 = 1,089 kips
phi x P_n = 0.90 x 1,089 = 980 kips

Axial check: phi*Pn = 980 kips > Pu = 580 kips (OK). Utilization = 59%.

Step 7 — Flexural capacity about weak axis (AISC Section F6)

The bending moment is about the weak axis (y-axis). For a compact I-shape bent about the weak axis:

phi x M_ny = phi x M_py = phi x Fy x Zy (Section F6.1)
phi x M_ny = 0.90 x 50 x 92.6 = 4,167 kip-in = 347 kip-ft

Flexural check: phi*Mny = 347 kip-ft > Mux = 85 kip-ft (OK). Utilization = 24%.

Step 8 — Combined loading check (AISC Chapter H, Section H1.1)

Interaction equation (H1-1a) — for Pu/Pn >= 0.2

P_u / (phi x P_n) = 580 / 980 = 0.592 >= 0.2

Use H1-1a:

P_u/(phi x P_n) + (8/9) x [M_ux/(phi x M_nx) + M_uy/(phi x M_ny)] <= 1.0

Since M_ux = 0 (no strong axis moment) and M_uy = 85 kip-ft:

0.592 + (8/9) x [0 + 85/347] = 0.592 + 0.889 x 0.245 = 0.592 + 0.218 = 0.810

Interaction check: 0.810 <= 1.0 (OK). Utilization = 81%.

The combined loading is the governing check at 81% utilization.

Step 9 — Summary of all limit states

Combined axial-plus-bending interaction governs at 81% utilization. W14x82 is adequate.

Step 10 — Design optimization

Can we use a lighter section?

W14x74 passes the interaction equation at 0.99, which is too close to the limit for practical design. W14x82 is the lightest practical section with adequate margin.

Cost comparison

W14x82: 82 lb/ft x 13 ft = 1,066 lb per story = $640 per column (at $0.60/lb erected)
W14x74: 74 lb/ft x 13 ft = 962 lb per story = $577 per column (savings: $63/column)
W14x90: 90 lb/ft x 13 ft = 1,170 lb per story = $702 per column

For a 16-column building, W14x82 costs $10,240 vs W14x74 at $9,232. The $1,008 savings on W14x74 is not worth the 99% utilization.

Connection detail summary

Column: W14x82, ASTM A992, 13 ft story height
K = 1.0 (braced frame)
KL/r_y = 36.6 (governing slenderness)
Fcr = 45.2 ksi
phi*Pn = 980 kips
phi*Mny = 347 kip-ft
Interaction = 0.81 (governs)
Capacity: 580 kips axial + 85 kip-ft bending

Design comparison — column sections for 580 kip load

W14 sections are preferred for columns because the wide flanges provide weak-axis stability and easy beam connections.

Cross-code comparison for column design

AISC uses a single column curve while AS 4100, EN 1993, and CSA S16 use multiple curves based on section type and axis. This can result in 5-15% capacity differences for the same section.

Common mistakes in column design

1. Using the wrong effective length factor K. In braced frames, K should be between 0.5 and 1.0 (alignment chart). In unbraced (moment) frames, K ranges from 1.0 to 3.0+. Using K=1.0 for all conditions is common but unconservative for sway frames.

2. Checking only axial capacity and ignoring interaction. Most building columns have some bending from beam eccentricity or lateral loads. The interaction equation (Chapter H) almost always governs. Always check it.

3. Forgetting weak-axis buckling. Strong-axis KL/r is usually smaller because rx > ry, but weak-axis buckling (KL/r using ry) typically controls. Always check both axes.

4. Not checking the bracing connection for shear transfer. The column must transfer floor diaphragm forces through the brace connection. This is a separate check from the column capacity itself.

5. Using the wrong phi factor in the interaction equation. Compression uses phi = 0.90 (AISC E3). Flexure uses phi = 0.90 (F2). The interaction equation (H1) uses the same phi factors as the individual checks. Don't use phi = 0.75 (that's for bolts and welds).

6. Ignoring splice requirements. Multi-story columns require splices every 2-3 stories. The splice must develop the full column capacity or at least the calculated demand. AISC Section J7 covers bearing on steel surfaces.

Frequently asked questions

What is the difference between braced and unbraced frame columns? In braced frames, lateral loads are resisted by diagonal braces or shear walls, not by column bending. Columns see only axial load plus minor moments from beam eccentricity. K = 0.5 to 1.0. In unbraced (moment) frames, columns resist lateral loads through bending. K = 1.0 to 3.0+. Braced frame columns are always more economical.

How do I determine the effective length factor K? Use the AISC alignment chart (Commentary Figure C-A-7-1) with GA and GB (relative stiffness factors at each end). For preliminary design: pinned-pinned = 1.0, fixed-fixed = 0.5, fixed-pinned = 0.7, cantilever = 2.0. In braced frames, K should never exceed 1.0.

When does the interaction equation govern over pure axial? When Pu/(phi*Pn) < 0.2, use equation H1-1b which is less restrictive. When Pu/(phi*Pn) >= 0.2, use H1-1a. For most building columns with beam connections, the ratio is 0.4-0.7, so H1-1a governs.

What Fcr should I use for preliminary sizing? For A992 (Fy = 50 ksi) columns with KL/r < 40: Fcr is approximately 43-46 ksi (close to Fy). For KL/r = 60-80: Fcr drops to 33-40 ksi. For KL/r = 100-120: Fcr is 20-30 ksi. For preliminary sizing, assume Fcr = 30 ksi and refine.

Do I need to check local buckling? Yes. Check bf/(2tf) and h/tw against the compact limits (Table B4.1a). Most W14 sections in A992 are compact. Noncompact sections have reduced Fcr per Section E7. Slender sections (like some W-shapes in high-strength steel) require further reduction.

What about column base plates? Column base plates transfer the axial load and moment to the concrete foundation. They are designed per AISC Design Guide 1. The base plate thickness depends on the bearing pressure distribution and the cantilever distance from the column flange to the edge of the plate.

AISC Chapter E design procedure — step-by-step summary

The following procedure applies to doubly symmetric I-shapes, HSS, and singly symmetric sections in compression per AISC 360-22 Chapter E.

Procedure checklist

1. Determine factored loads (Pu, Mux, Muy) from LRFD load combinations (ASCE 7-22).
2. Select trial section from AISC Manual Table 4-1 based on KL and required phiPn.
3. Verify compact section per Table B4.1a (flange and web slenderness).
4. Determine effective length KL for both axes using alignment chart or assumption.
5. Calculate slenderness ratio KL/r for each axis; the larger value governs.
6. Calculate Fe = pi^2 * E / (KL/r)^2.
7. Determine Fcr using E3-2 (inelastic, when KL/r <= 4.71*sqrt(E/Fy)) or E3-3 (elastic).
8. Calculate phiPn = 0.90 _ Fcr _ Ag.
9. Check interaction equation H1-1a (Pu/phiPn >= 0.2) or H1-1b (Pu/phiPn < 0.2).
10. Verify KL/r <= 200 (recommended maximum per AISC).

KL/r worked example — W14x61 column

Given: W14x61 (A992, Fy = 50 ksi), K = 1.0, L = 14 ft.

Slenderness calculation:

KL/rx = 1.0 x 14 x 12 / 5.98 = 28.1
KL/ry = 1.0 x 14 x 12 / 2.45 = 68.6  (GOVERNS)

Maximum KL/r = 68.6 < 200 (OK).

Fcr determination:

Fe = pi^2 x 29,000 / (68.6)^2 = 286,159 / 4,706 = 60.8 ksi
4.71 x sqrt(E/Fy) = 4.71 x 24.08 = 113.4

Since KL/r = 68.6 < 113.4, use inelastic buckling (E3-2):

Fcr = 0.658^(Fy/Fe) x Fy = 0.658^(50/60.8) x 50
    = 0.658^0.822 x 50 = 0.726 x 50 = 36.3 ksi
phiPn = 0.90 x 36.3 x 17.9 = 585 kips

Interaction equation H1-1a worked example — W14x61

Given: W14x61, Pu = 350 kips, Mux = 0, Muy = 55 kip-ft.

phiPn = 585 kips (from above)
phiMny = 0.90 x Fy x Zy = 0.90 x 50 x 55.0 = 2,475 kip-in = 206 kip-ft

Check ratio:

Pu / (phi x Pn) = 350 / 585 = 0.598 >= 0.2  --> Use H1-1a

H1-1a evaluation:

Pu/(phiPn) + (8/9) x [Mux/(phiMnx) + Muy/(phiMny)]
= 0.598 + (8/9) x [0 + 55/206]
= 0.598 + 0.889 x 0.267
= 0.598 + 0.237 = 0.835 <= 1.0  (OK)

Fcr summary table for common slenderness (Fy = 50 ksi)

The transition from inelastic to elastic buckling at KL/r = 113 marks where columns shift from yield-dominated to stability-dominated failure. Below this limit, Fcr is close to Fy. Above it, Fcr drops rapidly with increasing slenderness.

Run this calculation

• Column Capacity Calculator
• Base Plate Calculator

Related references

• Column K-Factor
• Column Buckling Equations
• Column Curve
• Combined Loading
• Column Splice Design
• Effective Length Factor
• Compact Section Limits
• Steel Grades
• How to Verify Calculations

Disclaimer

This page is for educational and reference use only. It does not constitute professional engineering advice. All design values must be verified against the applicable standard and project specification before use. The site operator disclaims liability for any loss arising from this information.