Combined Loading Worked Example — Beam-Column per AISC 360-22 LRFD

Problem: Check the adequacy of a W12x65 beam-column (A992 steel, Fy = 50 ksi) in a braced frame (no sway). The member is 14 ft long and supports a factored axial load of Pu = 200 kips combined with a factored end moment of M_nt = 180 kip·ft (strong-axis bending, single curvature, equal end moments). The member has no transverse loads between ends. Use AISC 360-22 Chapter H (combined forces) and Chapter C (stability).

Step 1: Section Properties (W12x65)

From AISC Manual Table 1-1:

Step 2: Required Compressive Strength (AISC 360 Chapter E)

KL = 1.0 × 14 = 14 ft for braced frame (K = 1.0, pinned-pinned)

Slenderness about weak axis (governs):

KL/ry = (1.0 × 14 × 12) / 3.02 = 168 / 3.02 = 55.6

Elastic buckling stress:

Fe = π² × E / (KL/r)² = π² × 29,000 / (55.6)² = 286,280 / 3,091 = 92.6 ksi

Critical stress:

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

KL/r = 55.6 < 113.4 → Inelastic buckling, Equation E3-2:

Fcr = [0.658^(Fy/Fe)] × Fy = [0.658^(50/92.6)] × 50

0.658^0.540 = 0.794

Fcr = 0.794 × 50 = 39.7 ksi

Nominal and design compressive strength:

Pn = Fcr × Ag = 39.7 × 19.1 = 758 kips

ϕcPn = 0.90 × 758 = 682 kips

Step 3: Required Flexural Strength (AISC 360 Chapter F)

Check Lb vs Lp for compact section:

Lb = 14 ft > Lp = 9.92 ft → Inelastic LTB, use AISC F2-2

Lb = 14 ft < Lr = 31.4 ft → Inelastic LTB zone

Mp = Fy × Zx = 50 × 96.8 / 12 = 403.3 kip·ft

0.7Fy × Sx = 0.7 × 50 × 87.9 / 12 = 256.4 kip·ft

Cb = 1.0 (conservative for equal end moments with no transverse load; actual Cb = 1.0 for M1/M2 = -1.0, single curvature)

Mn = Cb × [Mp - (Mp - 0.7FySx) × (Lb - Lp)/(Lr - Lp)] = 1.0 × [403.3 - (403.3 - 256.4) × (14.0 - 9.92)/(31.4 - 9.92)] = 1.0 × [403.3 - (146.9) × (4.08)/(21.48)] = 1.0 × [403.3 - 146.9 × 0.190] = 1.0 × [403.3 - 27.9] = 375.4 kip·ft

ϕbMn = 0.90 × 375.4 = 337.9 kip·ft

Step 4: Moment Amplification (AISC 360 Chapter C — Braced Frame)

For a braced frame (no sway), use AISC 360 Equation C2-1a:

Mu = B1 × M_nt

Where:

B1 = Cm / (1 - α × Pr / Pe1) ≥ 1.0

Calculate Pe1 (Euler buckling about the bending axis):

Pe1 = π² × E × Ix / (K1 × L)²

K1 = 1.0 (braced frame, pinned ends) L = 14 ft = 168 in

Pe1 = π² × 29,000 × 533 / (1.0 × 168)² = π² × 29,000 × 533 / 28,224 = 1,524,700 / 28,224 = 5,404 kips

Calculate Cm:

For a braced frame member with no transverse loads and equal end moments (M1/M2 = -1.0, single curvature):

Cm = 0.6 - 0.4 × (M1/M2) = 0.6 - 0.4 × (-1.0) = 0.6 + 0.4 = 1.0

Calculate B1:

α = 1.0 (LRFD) Pr = Pu = 200 kips

B1 = 1.0 / (1 - 1.0 × 200 / 5,404) = 1.0 / (1 - 0.037) = 1.0 / 0.963

B1 = 1.038

Check B1 ≥ 1.0 → 1.038 > 1.0 → OK

Amplified moment:

Mu = B1 × M_nt = 1.038 × 180 = 186.8 kip·ft

Step 5: Interaction Check — AISC 360 Chapter H (Equations H1-1a and H1-1b)

Calculate Pr/Pc:

Pc = ϕcPn = 682 kips Pr = 200 kips

Pr/Pc = 200 / 682 = 0.293

Since Pr/Pc > 0.20, use Equation H1-1a:

Pr/Pc + 8/9 × (Mrx/Mcx + Mry/Mcy) ≤ 1.0

Calculate Mrx/Mcx:

Mcx = ϕbMn = 337.9 kip·ft Mrx = Mu = 186.8 kip·ft

Mrx/Mcx = 186.8 / 337.9 = 0.553

For weak-axis bending (Mry/Mcy): None in this example (Mry = 0).

Interaction check:

0.293 + 8/9 × (0.553 + 0) = 0.293 + 0.491 = 0.784

0.784 ≤ 1.0 → OK (78% utilized)

Step 6: Check H1-1b for completeness

If Pr/Pc < 0.2 (which it is not in this case), use H1-1b:

Pr/(2Pc) + (Mrx/Mcx + Mry/Mcy) ≤ 1.0

For verification: 0.293/(2) + 0.553 = 0.147 + 0.553 = 0.700

This is lower than H1-1a result (0.784), confirming H1-1a governs for Pr/Pc > 0.2.

Step 7: Second-Order Effects — P-δ Check

The B1 factor accounts for P-δ (member-level second-order effects). The amplified moment includes the additional moment from axial load acting on the deflected shape of the member.

Verification of B1 approximation:

For members with end moments only (no transverse loads), the exact P-δ amplification can be checked:

B1_actual = (π² × EI / L²) / (π² × EI / L² - P) × Cm... this is already the B1 formula.

The B1 = 1.038 means second-order effects increase the first-order moment by 3.8%. This is modest because Pr/Pe1 = 200/5,404 = 0.037 (only 3.7% of the Euler load).

If the axial load were higher (e.g., Pu = 400 kips):

B1 = 1.0 / (1 - 400/5,404) = 1.0 / 0.926 = 1.080 → 8% amplification

Step 8: Check P-Δ (Sway Effects)

The frame is braced (no sway), so P-Δ effects are negligible by definition. For unbraced frames, the B2 factor from AISC C2-1b would also apply:

B2 = 1 / (1 - (α × ΣP_nt / ΣH) × (Δ_oh / L)) or B2 = 1 / (1 - (α × ΣP_nt / ΣPe2))

But since the frame is braced, B2 = 1.0.

Step 9: Check Slenderness Limits

Local buckling (AISC B4.1):

Flange: bf/(2tf) = 12.0/(2 × 0.605) = 9.92

λ_p (compact flange) = 0.38√(E/Fy) = 0.38 × 24.08 = 9.15

λ_r (noncompact flange) = 1.0√(E/Fy) = 24.08

9.92 > 9.15 but < 24.08 → Noncompact flange (requires AISC F2 for noncompact)

Actually, checking AISC Table B4.1b: For flanges of doubly-symmetric I-shaped sections, λ_p = 0.38√(E/Fy) = 9.15 and λ_r = 1.0√(E/Fy) = 24.1. Our bf/2tf = 9.92 > 9.15, so technically noncompact.

Web: h/tw for W12x65: d - 2tf = 12.1 - 2×0.605 = 10.89 in (clear distance between flanges)

h/tw = 10.89 / 0.390 = 27.9

λ_p (compact web) = 3.76√(E/Fy) = 3.76 × 24.08 = 90.5

27.9 < 90.5 → Compact web

The noncompact flange reduces the nominal flexural strength slightly. Recomputing Mn for noncompact flange per AISC F2-2 (which already accounts for LTB and is conservative) — the majority of the 9% reduction from Mp to Mn is from LTB rather than local buckling. The Mn = 375.4 kip·ft is still valid by F2-2.

Step 10: Summary

Check Value Limit Ratio Verdict
Pr/Pc 0.293 1.0 — —
Mrx/Mcx 0.553 1.0 — —
H1-1a interaction 0.784 1.0 0.78 OK
P-δ amplification (B1) 1.038 — — Negligible
Local buckling flange 9.92 9.15 (λp) 1.08 Noncompact (acceptable)
Local buckling web 27.9 90.5 (λp) 0.31 Compact

Final verdict: W12x65 is adequate for the combined loading of Pu = 200 kips and M_nt = 180 kip·ft. The beam-column is 78% utilized per the H1-1a interaction equation. The compression-flange slenderness is slightly noncompact, but this does not reduce the member capacity below the applied demand.

Try the Calculator

Use the Column Capacity Calculator to check beam-column interaction for your own sections, axial loads, and moments. The calculator applies AISC 360 Chapter H equations H1-1a and H1-1b with proper moment amplification per Chapter C.

Frequently Asked Questions

When do I use H1-1a versus H1-1b? Use H1-1a when Pr/Pc ≥ 0.20 (axial-dominated), and H1-1b when Pr/Pc < 0.20 (bending-dominated). The transition at Pr/Pc = 0.20 reflects the different interaction behavior: when axial load is significant, the interaction curve is convex (higher combined capacity), and when bending dominates, the interaction is more linear.

What is the difference between P-δ and P-Δ effects? P-δ (P-delta) refers to second-order effects within the member length — the axial load acting on the deflection between member ends. P-Δ (P-delta) refers to second-order effects at the frame level — the axial load acting on the relative lateral displacement between member ends. In AISC 360, B1 captures P-δ effects and B2 captures P-Δ effects. For this braced frame example, only B1 (P-δ) is needed.

How does Cb affect beam-column capacity? Cb (lateral-torsional buckling modification factor) accounts for the moment gradient along the member. Cb = 1.0 is the most conservative value. For non-uniform moment diagrams (e.g., transverse loads, unequal end moments, or reverse curvature), Cb > 1.0 and can increase Mn by up to 25-30%. Our example uses Cb = 1.0 with equal end moments in single curvature, which is correct (M1/M2 = -1.0 gives Cb = 1.0).

When is moment amplification not required? AISC 360 Chapter C exempts moment amplification when α × Pr ≤ 0.05 × Pe1 (i.e., B1 < 1.05 and the effect is less than 5%). Also, for members with very low axial load (Pr/Pc < 0.05), the interaction check reduces to a beam-only check with no P-δ interaction.

See Also