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 kipÃÂ÷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:
- A = 19.1 in²
- d = 12.1 in
- bf = 12.0 in
- tf = 0.605 in
- tw = 0.390 in
- Zx = 96.8 in³
- Sx = 87.9 in³
- Ix = 533 in⁴
- rx = 5.28 in
- ry = 3.02 in
- Lp = 9.92 ft (from Manual Table 3-2, Fy = 50 ksi)
- Lr = 31.4 ft
- Lb = 14 ft (unbraced length)
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 = ÃÂò × E / (KL/r)² = ÃÂò × 29,000 / (55.6)² = 286,280 / 3,091 = 92.6 ksi
Critical stress:
4.71âÃÂÃÂ(E/Fy) = 4.71 × Ã¢ÃÂÃÂ(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
ÃÂÃÂcPn = 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 kipÃÂ÷ft
0.7Fy × Sx = 0.7 × 50 × 87.9 / 12 = 256.4 kipÃÂ÷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 kipÃÂ÷ft
ÃÂÃÂbMn = 0.90 × 375.4 = 337.9 kipÃÂ÷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 - ÃÂñ × Pr / Pe1) ≥ 1.0
Calculate Pe1 (Euler buckling about the bending axis):
Pe1 = ÃÂò × E × Ix / (K1 × L)²
K1 = 1.0 (braced frame, pinned ends) L = 14 ft = 168 in
Pe1 = ÃÂò × 29,000 × 533 / (1.0 × 168)² = ÃÂò × 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:
ÃÂñ = 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 kipÃÂ÷ft
Step 5: Interaction Check — AISC 360 Chapter H (Equations H1-1a and H1-1b)
Calculate Pr/Pc:
Pc = ÃÂÃÂcPn = 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 = ÃÂÃÂbMn = 337.9 kipÃÂ÷ft Mrx = Mu = 186.8 kipÃÂ÷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-ÃÂô Check
The B1 factor accounts for P-ÃÂô (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-ÃÂô amplification can be checked:
B1_actual = (ÃÂò × EI / L²) / (ÃÂò × 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 - (ÃÂñ × ÃÂãP_nt / ÃÂãH) × (Δ_oh / L)) or B2 = 1 / (1 - (ÃÂñ × ÃÂãP_nt / ÃÂãPe2))
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
ÃÂû_p (compact flange) = 0.38âÃÂÃÂ(E/Fy) = 0.38 × 24.08 = 9.15
ÃÂû_r (noncompact flange) = 1.0âÃÂÃÂ(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, ÃÂû_p = 0.38âÃÂÃÂ(E/Fy) = 9.15 and ÃÂû_r = 1.0âÃÂÃÂ(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
ÃÂû_p (compact web) = 3.76âÃÂÃÂ(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 kipÃÂ÷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-ÃÂô amplification (B1) | 1.038 | — | — | Negligible |
| Local buckling flange | 9.92 | 9.15 (ÃÂûp) | 1.08 | Noncompact (acceptable) |
| Local buckling web | 27.9 | 90.5 (ÃÂûp) | 0.31 | Compact |
Final verdict: W12x65 is adequate for the combined loading of Pu = 200 kips and M_nt = 180 kipÃÂ÷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 Compression Strength Tool 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-ÃÂô and P-Δ effects? P-ÃÂô (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-ÃÂô effects and B2 captures P-Δ effects. For this braced frame example, only B1 (P-ÃÂô) 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 ÃÂñ × 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-ÃÂô interaction.
See Also
- Beam Capacity Calculator
- Beam Displacement and Sag Tool
- Steel Beam Sizes Reference
- Beam Design Guide
- Beam Span Reference
Disclaimer: This content is for educational purposes only. Results must be verified by a licensed professional engineer. Steel Calculator provides preliminary design tools — NOT a substitute for professional engineering judgment.