Column Capacity Worked Example — W10x49 per AISC 360-22 LRFD

Problem: Determine the available compressive strength of a W10x49 (A992 steel, Fy = 50 ksi) column with an unbraced length of 15 ft in both axes. Assume pinned supports at both ends (K = 1.0). Check flexural buckling about both principal axes and torsional buckling.

Step 1: Section Properties (W10x49)

From AISC Manual Table 1-1:

• A = 14.4 in²
• d = 10.0 in
• bf = 10.0 in
• tf = 0.560 in
• tw = 0.340 in
• rx = 4.35 in
• ry = 2.54 in
• J = 1.39 in⁴
• Cw = 1,700 in⁶ (warping constant)
• Ix = 272 in⁴
• Iy = 93.4 in⁴

Step 2: Effective Length and Slenderness

K = 1.0 (pinned-pinned) L = 15 ft = 180 in

X-axis slenderness:

KL/rx = (1.0 × 180) / 4.35 = 41.4

Y-axis slenderness:

KL/ry = (1.0 × 180) / 2.54 = 70.9 ← governs (larger slenderness)

Step 3: Elastic Buckling Stress (AISC 360 E3)

The Euler buckling stress is:

Fe = π² × E / (KL/r)²

Fe = π² × 29,000 ksi / (70.9)² Fe = π² × 29,000 / 5,027 Fe = 56.9 ksi

Compare with Fy = 50 ksi. Since Fe = 56.9 ksi > 0.44 × Fy = 22 ksi, use inelastic buckling (AISC E3-2).

Step 4: Critical Stress (AISC 360 E3)

Check slenderness limit:

4.71√(E/Fy) = 4.71 × √(29,000/50) = 4.71 × 24.08 = 113.4

KL/r = 70.9 < 113.4 → Inelastic buckling governs. Use Equation E3-2:

Fcr = [0.658^(Fy/Fe)] × Fy

Fcr = [0.658^(50/56.9)] × 50 Fcr = [0.658^0.879] × 50 Fcr = 0.701 × 50 Fcr = 35.1 ksi

Check the alternate method (E3-3): If Fe < 0.44Fy, use Fcr = 0.877 × Fe. 0.44 × 50 = 22 ksi. Fe = 56.9 > 22, so E3-2 governs (as used above).

Step 5: Nominal and Design Compressive Strength

Pn = Fcr × Ag = 35.1 ksi × 14.4 in² = 505 kips

For LRFD:

ϕcPn = 0.90 × 505 = 455 kips

For ASD:

Pn / Ωc = 505 / 1.67 = 302 kips

Step 6: Check Flexural-Torsional Buckling for Doubly-Symmetric Section

For doubly-symmetric W-shapes, the flexural and torsional buckling modes are uncoupled (the shear center coincides with the centroid). Per AISC E4, the torsional buckling stress for doubly-symmetric members is:

Fez = [π² × E × Cw / (Kz × L)² + G × J] × (1 / (Ix + Iy))

Where G = 11,200 ksi (shear modulus of steel)

π² × E × Cw / (Kz × L)² = π² × 29,000 × 1,700 / (1.0 × 180)² = 486,916,000 / 32,400 = 15,027 kip·in²

G × J = 11,200 × 1.39 = 15,568 kip·in²

Ix + Iy = 272 + 93.4 = 365.4 in⁴

Fez = (15,027 + 15,568) / 365.4 = 83.7 ksi

Since Fez = 83.7 ksi > Fcr_flexural = 56.9 ksi, flexural buckling about the weak axis governs. No reduction for torsional-flexural buckling is needed.

Step 7: Additional Length Check

Check if a longer unbraced length would change the mode:

At 25 ft, KL/r = 118.1 > 113.4 → Elastic buckling governs, using Fcr = 0.877 × Fe.

The column capacity is sensitive to unbraced length — increasing from 15 ft to 20 ft reduces capacity by 21%.

Step 8: Local Buckling Check (AISC 360 E7)

For the W10x49 section, verify compactness per AISC Table B4.1b:

Flange compactness: bf/2tf = 10.0/(2×0.560) = 8.93 λp (flange) = 0.38√(E/Fy) = 0.38√(29000/50) = 9.15 8.93 < 9.15 → Compact flange

Web compactness: h/tw = (d - 2tf)/tw = (10.0 - 2×0.560)/0.340 = 26.1 λp (web) = 3.76√(E/Fy) = 3.76√(29000/50) = 90.6 26.1 < 90.6 → Compact web

Since both flange and web are compact, no local buckling reduction applies. For slender elements, AISC E7 requires reducing Fcr using an effective area approach, which can reduce capacity by 10-40% depending on slenderness.

Step 9: Effects of End Restraint

The K = 1.0 assumption (pinned-pinned) is conservative. If the column has partial rotational restraint from beams framing into it at both ends, K could be reduced:

Using K = 0.8 instead of 1.0 increases design capacity by 12%, which is significant in design-build and optimization contexts. The actual K factor depends on the relative stiffness of beams and columns at each joint (alignment chart method per AISC Commentary Appendix 7).

Step 10: Design Summary and Recommendations

The W10x49 column at 15 ft unbraced length provides ϕcPn = 455 kips (LRFD). Key findings:

1. Governing mode: Flexural buckling about the weak axis (KL/ry = 70.9)
2. Failure type: Inelastic buckling (AISC E3-2, Fcr = 35.1 ksi)
3. Section efficiency: Utilization depends on applied load. At design load of 350 kips, utilization = 350/455 = 77%
4. Optimization: If the applied load is 300 kips or less, consider a W10x39 (A = 11.5 in², ry = 2.53 in) which would provide ϕcPn ≈ 365 kips — reducing beam weight by 20%.
5. Torsional check: Flexural-torsional buckling does not govern for this doubly-symmetric section.

For columns with biaxial bending (beam-columns), check interaction per AISC H1 (H2 for ASD). The interaction equation (H1-1a or H1-1b) combines axial and flexural utilization and typically governs over pure axial capacity for columns with significant frame action moments.

Step 11: AS 4100 and EN 1993 Comparison

The same column checked under international standards yields different capacities due to different design philosophies:

AS 4100-2020: Uses the member capacity factor αc, which is a function of member slenderness λn = (Ler/ry) × √(Fy/250). For the W10x49, λn = 70.9 × √(50×6.895/250) = 70.9 × 1.174 = 83.2. Using AS 4100 Table 6.3.3(1), αc ≈ 0.601 for a hot-rolled UB section. The nominal capacity Ns = Ag × Fy = 14.4 × 50 = 720 kips (3200 kN). The member capacity Nc = αc × Ns = 0.601 × 720 = 433 kips. With ϕ = 0.90 per AS 4100, ϕNc = 390 kips — approximately 14% lower than the AISC result due to different residual stress assumptions and slenderness formulations.

EN 1993-1-1: Uses buckling curves. For a hot-rolled I-section (buckling curve b for y-y axis), the imperfection factor α = 0.34. The slenderness λ̄ = √(Ag×Fy/Ncr) where Ncr = π²E/(Ler/ry)² × Ag. Ncr = π²×29000/(70.9)² × 14.4 = 56.9 × 14.4 = 819 kips. λ̄ = √(720/819) = 0.938. Using EN 1993-1-1 Table 6.1 and curve b, χ = 0.646. Nb,Rd = χ×Ag×Fy/γM1 = 0.646×720/1.0 = 465 kips — approximately 2% higher than the AISC LRFD result.

The differences between codes highlight the importance of using the correct standard for your jurisdiction.

Step 12: Applicability to Other Sections

For non-doubly-symmetric sections (channels, angles, tees), flexural-torsional buckling per AISC E4 must be checked explicitly using the full equations accounting for the shear center offset. The column capacity calculator handles all section types automatically.

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Use the Column Capacity Calculator to check your own column section, length, end conditions, and steel grade with AISC, AS 4100, EN 1993, or CSA S16.

Frequently Asked Questions

What is the difference between flexural buckling and torsional buckling? Flexural buckling involves lateral deflection of the member without twisting (Euler buckling). Torsional buckling involves twisting about the longitudinal axis. For doubly-symmetric W-shapes, these modes are uncoupled and flexural buckling about the weak axis typically governs. For unsymmetric sections, the modes interact.

How does end restraint affect column capacity? End restraint is captured by the effective length factor K. Practical ranges: K = 0.65 (fixed), K = 0.80 (partially restrained), K = 1.0 (pinned), K = 1.2 (pinned at one end, fixed at the other). Using K = 1.0 for a column with partial fixity underestimates capacity by 20-30%.

What limit state typically governs for typical building columns (10-15 ft)? For most W-shapes in the 10-15 ft range, inelastic flexural buckling about the weak axis governs. Local buckling (flange or web slenderness) rarely governs for compact sections with Fy ≤ 50 ksi.

How do I check local buckling in column design? For compact sections, local buckling is prevented by proportioning limits (check flanges per AISC Table B4.1b: bf/2tf ≤ 0.38√(E/Fy); web h/tw ≤ 3.76√(E/Fy)). If these limits are exceeded, AISC E7 applies an effective area reduction. Our W10x49 example has bf/2tf = 8.93 < 9.15 and h/tw = 26.1 < 90.6, confirming compactness and no local buckling reduction needed.

See Also

• Column Capacity Calculator
• Column Buckling Guide
• Column Design Guide
• Effective Length Reference