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LecturesModule 9

Beam-Columns

Beam-column design with second-order effects, B1/B2 factors.

AISC Reference Box
  • AISC 360-22 Chapter H — Design of Members for Combined Forces

Lecture Notes

This module introduces beam-columns. Lecture content here covers the governing physics, LRFD philosophy, and how the relevant AISC 360-22 chapter organizes the limit states.

Instructors can replace this text in Admin Mode. Each section is structured around: (1) behavior, (2) failure modes, (3) AISC limit-state equations, (4) design workflow, (5) detailing requirements.

A short comparison to ASD is included only where the resistance factor / safety factor relationship clarifies the LRFD design check.

Project case study — Cardinal Square — 4-story braced-frame office

Every chapter's worked example is one step in the design of the same building: Plan: 4 bays N–S × 3 bays E–W, each 30 ft × 30 ft. Stories: 4 @ 13 ft (52 ft roof). Composite floor: 4.5 in NW concrete on 3 VLI20 deck. Roof: 1.5 in B-deck + insulation + membrane. Materials: Wide-flange members A992 (Fy = 50 ksi, Fu = 65 ksi). Plates A572 Gr. 50. HSS bracing A500 Gr. C. Bolts A325-N 7/8 in dia. Welds E70XX. Concrete f'c = 4 ksi. Anchor rods F1554 Gr. 36.

Chapter 9 — Beam-columns w/ second-order effects
Same edge column with B1/B2 amplification
Demand carried forward
From Chapter 8: Pr, Mr Apply B1 (no sway) and B2 (sway) per App. 8.
This chapter contributes
Computes B1 for non-sway moments and B2 for sway moments, then re-runs the H1 interaction. Confirms whether the trial section still passes.
Mr / McPr / Pc1.00.21.0H1-1a: Pr/Pc + (8/9)·ΣMr/Mc ≤ 1H1-1b: Pr/(2Pc) + ΣMr/Mc ≤ 1
H1 envelope used for beam-columns with B1·Mnt + B2·Mlt applied to Mr.
P-δ (member)P-Δ (story)HB1 amplifies member moments; B2 amplifies story sway moments
B1 (member P-δ) and B2 (story P-Δ) amplification per AISC App. 8.

Formula Sheet

NameEquationAISC Ref
Interaction (Pr/Pc ≥ 0.2)Pr/Pc + (8/9)(Mrx/Mcx + Mry/Mcy) ≤ 1.0AISC §H1.1(a)
Interaction (Pr/Pc < 0.2)Pr/(2 Pc) + (Mrx/Mcx + Mry/Mcy) ≤ 1.0AISC §H1.1(b)

Worked Example

Beam-Columns

Given
Replace with project-specific given data (loads, geometry, material).
Load combination
Controlling LRFD load combination from ASCE 7.
Required strength
Compute required strength Ru from the controlling combination.
Limit states
  • Limit state 1
  • Limit state 2
AISC reference
AISC 360-22 — applicable chapter
Solution steps
  1. 1. Required strength
    Compute Ru.
  2. 2. Trial section
    Pick a trial from AISC shape tables Instructor should verify with official AISC Manual.
  3. 3. Check each limit state
    Apply φ Rn ≥ Ru for every governing limit state.
  4. 4. Iterate
    Resize until the most economical section satisfies all checks.
Final design decision
Select the lightest section that satisfies all LRFD limit states.
Common mistakes in this example
  • Skipping a limit state
  • Using the wrong φ factor
  • Forgetting serviceability checks

FE-Style Worked Examples (6)

Each example mirrors the NCEES FE Civil Reference Handbook style: brief givens, a labeled figure, AISC section reference, step-by-step numeric solution, and a single boxed answer.

Given
Pr=150 k, Pe1=900 k, Cm=0.85.
AISC Reference
AISC §C2.1 / App. 8
Step-by-step solution
  1. B1
    Cm/(1 − Pr/Pe1) = 0.85/(1 − 150/900) = 0.85/0.833 = 1.02 ≥ 1.0
Answer B1 = 1.02.
PuLcAxially loaded column

Textbook — Aghayere & Vigil (2009)(3 worked examples with figures + numerical answers)

Worked examples scanned directly from the CEGR 436 course textbook. Each card shows the original page (figure + full step-by-step solution) and adds an FE-style numerical multiple-choice prompt with answer key.

Chapter summary

Chapter 8 of the textbook develops beam-column design. AISC §H1 interaction equations combine the axial demand ratio Pr/Pc with bending demand ratios Mrx/Mcx, Mry/Mcy. Second-order effects use B1 (no-translation) and B2 (translation) amplifiers per Appendix 8, or solve via the Direct Analysis Method (Chapter C).

  • Use H1-1a when Pr/Pc ≥ 0.2 (axial dominates): Pr/Pc + (8/9)(Mrx/Mcx + Mry/Mcy) ≤ 1.0.
  • Use H1-1b when Pr/Pc < 0.2 (bending dominates): Pr/(2Pc) + (Mrx/Mcx + Mry/Mcy) ≤ 1.0.
  • B1 = Cm/(1 − αPr/Pe1) ≥ 1; Cm = 0.6 − 0.4(M1/M2) for end-moment-only loading.
  • B2 = 1/(1 − αΣPnt/ΣPe story) ≥ 1.
  • Direct Analysis Method (Chapter C): reduce EI, EA by 0.8·τb; apply notional loads Ni = 0.002 αYi.
Setup
W12×72, A992, Pu = 400 k, Mux = 200 k·ft, Muy = 0. φcPn = 818 k, φbMnx = 425 k·ft.
AISC Reference
AISC §H1.1
Numerical practice
Interaction ratio?
Textbook page 355 — Interaction check on a braced column
Aghayere & Vigil (2009), p. 355 — full worked solution & sketch.

FE Practice Bank (1)

Multiple-choice problems pulled from the instructor's CEGR 492 FE Review packet (EasyFEExam © 2025, Steve Efe, PhD). Pick an answer, then click Reveal solution.

P26D. RC Beam-Column AstEasyFEExam Structural Design – Problem 26
Tied column 16 × 20 in, Pu = 560 k, Mu = 3920 k·in, f′c = 4, fy = 60. Required Ast (ACI interaction chart)?

Interactive Calculator

Combined Axial + Bending

AISC §H1.1
Pr/Pc0.300
Equation H1-1a0.776 OK

Practice Problems

  1. [E] Define B1 (P-δ) and B2 (P-Δ) amplification factors.
  2. [E] State DAM notional load = 0.002·Yi.
  3. [E] When is B1 applied? When is B2 applied?
  4. [E] State Cm for an unbraced beam with end moments M1 and M2.
  5. [E] List the AISC chapter for stability (Chapter C) and appendix (App. 8).
  6. [M] Compute Cm for M1 = +50, M2 = +120 k-ft (single curvature).
  7. [M] B1 for an interior W14x90, Pr = 300 k, Pe1 = 5000 k, Cm = 0.85.
  8. [M] Use Table 6-2 to size W12 beam-column for Pu = 500 k, Mux = 180 k-ft.
  9. [M] Interaction for W14x120 with Pu = 850 k, Mux = 250 k-ft.
  10. [M] W10x60 beam-column, Pu = 200 k, Mux = 80 k-ft, KL = 14 ft. Compute interaction.
  11. [H] DAM: 3-story moment frame, 12 ft story, Pu = 420 k per col, ΣH = 80 k, ΣPstory = 1800 k. Compute B2.
  12. [H] M1/M2 = -0.4 (reverse curvature). Compare Cm vs Cm = 1.0.
  13. [H] Biaxial W14x120, Pu = 700 k, Mux = 200, Muy = 80 k-ft, KL = 16 ft.
  14. [H] Design perimeter MF column 14 ft, Pu = 350 k gravity, Mux = ±120 k-ft wind.
  15. [H] Compare 1st- vs 2nd-order moments: Pu = 400 k, M1st = 150 k-ft, B1 = 1.18, B2 = 1.05.
Structured Clues
  • B1 amplifies first-order moments for P-δ (no sway).
  • B2 amplifies moments for P-Δ (sway).
  • DAM: reduce EI to 0.8τb·EI, apply 0.002·Yi notional loads.
Code References
  • AISC 360-22 §C1, App. 8
  • AISC Manual Part 2 — Stability

Quiz (1 FE-bank + 2 concept)

1. [P26] Tied column 16 × 20 in, Pu = 560 k, Mu = 3920 k·in, f′c = 4, fy = 60. Required Ast (ACI interaction chart)?
2. Which AISC 360-22 chapter primarily governs beam-columns?
3. In LRFD, the basic design inequality is:

Common Student Mistakes

  • Mixing ASD and LRFD load combinations in the same problem.
  • Using nominal strength Rn instead of design strength φRn.
  • Forgetting to check every limit state listed in the AISC chapter.

"Professor Explains" Script

Today we're talking about beam-columns. Think of this topic as one step in the LRFD workflow: identify the demand, identify the limit states from the relevant AISC chapter, then check that φ·Rn is at least equal to Ru. We'll walk through the failure modes, the equations, and a worked example. Pay close attention to where the resistance factor changes — that's where students lose points on exams.