Beam-Columns

Previous Next

Why Structural Steel?

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

⏱ 110 min★ Advanced✎ 3 objectives

§01Section 01

Engineering story

Chapter 9 · Beam-Columns

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

A real project narrative for this chapter will be authored as this chapter migrates to the v3.0 structured schema.

§02Section 02

Learning objectives

After this chapter you will be able to

Apply B1/B2 amplification
Perform a second-order analysis
Design an unbraced-frame column

§03Section 03

Engineering motivation

§04Section 04

Failure mechanisms

Failure mechanisms & lessons learned

Photographs and lessons-learned case studies for this topic will be added during chapter migration.

AISC Reference Box

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

Why This Chapter Matters

Sway frames experience P-Δ (structure-level) and P-δ (member-level) second-order effects. Ignoring these gives moments that can be 20–50% too low, leading to premature buckling.

Learning Objectives

Compute B1 (member) and B2 (story) amplifiers per AISC Appendix 8.
Include notional loads (0.002 × total gravity) per the Direct Analysis Method.
Reduce EI to 0.8 τb EI per AISC C2.3.
Apply amplified moments Mr = B1 Mnt + B2 Mlt in the H1 interaction.
Select an economical W-shape column-beam under combined loads.

Where This Chapter Is Used

Foundational for Chapter 17 stability and every moment-resisting column in Chapters 19–20.

ANSI / AISC 360-22Specification for Structural Steel Buildings16.1-114 to 16.1-125 (+ Ch. C: 16.1-15)

Chapter

AISC 360-22

Chapter H. Design of Members for Combined Forces and Torsion (with Chapter C Stability)

Use this reference to flip directly to the correct page of the AISC 360-22 Specification while solving problems in this course chapter.

§	Section title	Page
C1	General Stability Requirements	16.1-15
C2	Calculation of Required Strengths (Direct Analysis Method)	16.1-16
C3	Calculation of Available Strengths	16.1-19
H1-1a	Interaction Eq. (Pr/Pc ≥ 0.2)	16.1-114
H1-1b	Interaction Eq. (Pr/Pc < 0.2)	16.1-114
App. 8	Approximate Second-Order Analysis (B1, B2 Amplifiers)	16.1-235

Companion reference: AISC Manual Part 6 & Appendix 8

Lecture Notes

Chapter 9 — Beam-Columns & Second-Order Effects (AISC 360-22 Ch. H, App. 8)

Chapter focus. Almost no real column is purely axial — every column in a framed building also carries moment from gravity eccentricity, wind, or seismic drift. These members are called beam-columns, and because axial load acting on a deflected shape creates extra moment (second-order effects), a first-order analysis alone is unconservative. This chapter builds the physical picture of P-δ / P-Δ, derives the B1/B2 amplifiers of Appendix 8, and shows how to combine amplified demands with the H1 interaction equation.

1. What Is a Beam-Column?

A beam-column is any member subjected simultaneously to axial force P_r and flexural moment M_r about one or both axes. Every perimeter moment-frame column, every column with an eccentric crane bracket, and every truss chord with panel-point transverse load is technically a beam-column. AISC does not have a separate "beam-column chapter" for capacity — instead Chapter H combines the axial capacity (Ch. E) and flexural capacity (Ch. F) you already computed with an interaction equation.

Figure 9.1 — A beam-column resists axial load P and transverse/end moment M simultaneously; the two interact structurally, not just in a checking equation.

2. First-Order vs. Second-Order Analysis

A first-order analysis computes forces and moments on the undeformed geometry — the standard linear-elastic stiffness method every student learns first. A second-order analysis re-equilibrates the structure on its deformed shape, so that axial forces acting through lateral displacements generate additional moment that a first-order analysis misses entirely.

P-δ — axial load P acting on the local bow (δ) of a single member between its ends. Depends on member curvature; amplified with B₁.
P-Δ — the story's total gravity load acting through the story's lateral drift Δ. Depends on frame sidesway; amplified with B₂.

Figure 9.2 — P-δ acts on the deflected chord of a single member; P-Δ acts on the sway of the entire story.

Why it matters. Ignoring second-order effects is equivalent to assuming the structure never deflects — obviously wrong, but the error is small until αP_r/P_e becomes significant. AISC's answer is not "always run a full nonlinear second-order analysis" but rather "amplify your first-order results with a closed-form multiplier," which is fast, transparent, and accurate enough for the great majority of building structures.

3. When Can You Skip Amplification? (§C1.3)

§C1.3 thresholdΔ_2nd / Δ_1st ≤ 1.5 and α · P_r / P_e,story ≤ 0.4

α·P_r/P_e1 ≤ 0.05 for a single member → B₁ ≈ 1.0, amplification is negligible.
0.05 < α·P_r/P_e1 ≤ 0.4 → apply B₁/B₂ per Appendix 8 — the everyday case for regular buildings.
α·P_r/P_e1 > 0.4 → B₁/B₂ becomes unreliable; use a rigorous second-order (Direct Analysis) solution.

4. The B₁ Amplifier — Member P-δ (App. 8.2.1)

B₁ scales up the "no-translation" moment M_nt (the moment from a gravity/braced analysis with sidesway artificially restrained) to account for P-δ along the member:

Eq. A-8-3B₁ = C_m / (1 − α P_r / P_e1) ≥ 1.0

Eq. A-8-5P_e1 = π² · E · I* / (L_c1)²

α = 1.00 (LRFD), 1.60 (ASD) — converts the required strength to the same load level Pe1 assumes.
P_r = required axial strength of the member (the larger of P_nt+P_lt from the story, or simply the axial demand for a braced member).
L_c1 = unbraced length in the plane of bending, using the actual K = 1.0 (App. 8 always uses the unbraced length, not KL, for P_e1).
I* = 0.8·τ_b·I if a DAM stiffness reduction is being carried through; plain I is acceptable when τ_b = 1.0 (the usual case at design-level loads).

Physically: B₁ is how much larger the true mid-span moment becomes because axial load is now acting on a bowed shape rather than a straight one. As P_r approaches the member's own Euler load P_e1, the denominator shrinks toward zero and B₁ blows up — this is the mathematical signature of buckling instability.

5. C_m — the Equivalent Moment Factor

C_m corrects for the fact that P_e1 was derived for a member bent into a single half sine wave under uniform moment. Real end-moment diagrams are rarely that shape, so C_m "reduces" the amplifier when the real moment diagram is less severe than uniform single curvature.

Case (braced, no transverse load between ends)	Formula	Notes
Restrained ends, linear moment diagram	C_m = 0.6 − 0.4(M₁/M₂)	M₁/M₂ positive for reverse curvature
Single curvature (M₁/M₂ = −1)	C_m = 1.0	Most severe bending shape
Reverse curvature (M₁/M₂ = +1)	C_m = 0.2	Least severe; sidesway frames rarely qualify
Transverse load present, ends restrained	C_m = 0.85	App. 8, Table A-8.1
Transverse load present, ends unrestrained (pinned)	C_m = 1.00	Conservative default

Figure 9.3 — M1/M2 sign convention. Single curvature (both ends deflecting the same way relative to the chord) is the worst case, Cm = 1.0.

6. The B₂ Amplifier — Story P-Δ (App. 8.2.2)

Eq. A-8-6B₂ = 1 / (1 − α P_story / P_e,story) ≥ 1.0

Eq. A-8-7P_e,story = R_M · H · L / Δ_H

P_story = total vertical load supported by the story (all columns, not just the one being checked).
H, Δ_H = story shear and first-order interstory drift produced by that same shear, from a lateral (notional or wind/seismic) load case.
R_M = 1.00 for braced frames, 0.85 for moment frames (accounts for the additional flexibility from beam-to-column P-δ effects contributing to sidesway).
B₂ is a story quantity — every column in that story uses the same B₂.

B2 > 1.5 is a red flag

If B₂ exceeds ~1.5 the structure is dangerously flexible and Appendix 8's linear approximation of instability is no longer trustworthy — stiffen the frame (larger columns, added bracing) rather than trusting the amplifier further.

7. Assembling Total Second-Order Demands

M_r = B₁ · M_nt + B₂ · M_lt

P_r = P_nt + B₂ · P_lt

Subscript nt = "no translation" (a gravity-only analysis with joints artificially restrained against sidesway); lt = "lateral translation" (a pure sidesway analysis under lateral load, or notional load). Two separate first-order analyses are run, then combined with the amplifiers above — you never need to run an actual second-order computer analysis to use this method.

8. Effective Length Method vs. Direct Analysis Method

There are two accepted design paths in AISC 360-22 for stability:

Figure 9.4 — ELM keeps full stiffness but must compute a frame K-factor from an alignment chart; DAM uses K = 1.0 but requires reduced stiffness and notional loads.

DAM (Chapter C) is now the AISC-preferred default: it removes the alignment-chart K-factor step entirely (always K = 1.0) by instead explicitly modeling frame out-of-plumbness (notional loads) and inelastic softening (0.8τ_bEI). ELM (Appendix 7) is still permitted, and is what most legacy problem sets — including Examples E.4A/E.4B in this chapter — use, via the sidesway alignment charts.

9. H1 Interaction — Combining P_r and Amplified M_r

Eq. H1-1a (P_r/P_c ≥ 0.2)P_r/P_c + (8/9)·(M_rx/M_cx + M_ry/M_cy) ≤ 1.0

Eq. H1-1b (P_r/P_c < 0.2)P_r/(2P_c) + (M_rx/M_cx + M_ry/M_cy) ≤ 1.0

Both equations describe a single design envelope in P_r–M_r space: a straight line above P_r/P_c=0.2, transitioning to a shallower straight line below it (so the moment term is not over-penalized for lightly-loaded columns). §H2 provides an alternative, point-by-point stress-ratio check that is less conservative when it applies (both axial and flexural stress computed at the same fiber) — see Worked Example 9.1.

Figure 9.5 — The H1 interaction envelope. Amplified (B1, B2) demands must plot inside the green region.

Additional Design Aids & Stratified Equations

τ_b transitions from 1.0 to a parabolic reduction at αP_r/P_ns = 0.5 (used inside I* for P_e1 under DAM).

Reduced flexural stiffnessEI* = 0.80 · τ_b · E · I

τ_b closed formτ_b = 4·(αP_r/P_ns)·(1 − αP_r/P_ns), for αP_r/P_ns > 0.5

10. Design Procedure with Manual Tables 6-1 / 6-2

AISC Manual Table 6-2 tabulates φP_n, φM_nx, φM_ny, and the H1-1 combined-stress coefficients directly by shape and unbraced length — it turns the whole H1 check into a lookup-and-multiply exercise. Manual Table 6-1 (used in Examples H.2 and H.4) instead tabulates available stresses F_ca, F_cbx, F_cby for the §H2 stress-ratio method.

Run first-order nt/lt analyses (or a single braced analysis if no sidesway).
Compute C_m, P_e1, B₁ for each axis; compute P_e,story, B₂ if the frame sways.
Assemble amplified P_r, M_rx, M_ry.
Look up (or compute) φP_n, φM_nx, φM_ny from Ch. E and F (Manual Table 6-2 shortcuts this).
Apply H1-1a/b (or H2 for a specific fiber check) and confirm ≤ 1.0.
If the ratio exceeds 1.0, upsize and repeat — remember B₁/B₂ change slightly with the new section stiffness.

⚠ Common mistakes

Using K > 1 with DAM — DAM always sets K = 1.0 for the member check.
Applying B₁ to sway (lt) moments, or B₂ to non-sway (nt) moments — they amplify different moment components.
Forgetting the α factor (1.0 LRFD, 1.6 ASD) inside B₁, B₂ — mixing load levels silently under- or over-amplifies.
Using L_c1 = KL in P_e1 — App. 8 wants the actual unbraced length, not the effective length.
Re-analyzing a second-order model to get P_r, M_r, and then applying B₁/B₂ again on top — that double-counts the amplification (Example H.2 explicitly starts from second-order-analysis results and needs no further B₁/B₂).
Using DAM's reduced EI in a service-level deflection check — stiffness reduction is a stability device only, not a real material property.

Exam / FE tip. If a problem gives you results labeled "from a second-order analysis," B₁/B₂ have already been applied — go straight to the H1 or H2 interaction check (see Example H.2). If instead you're given first-order M_D, M_L or M_nt/M_lt, you must build B₁/B₂ yourself before checking interaction (see Example H.4 and 9.1).

📖 Using the AISC Manual — Beam-Columns with 2nd-Order Effects

Companion reference: Manual Part 6 (Table 6-2) + Appendix 8 / Chapter C (B1/B2 amplifiers). In practice, engineers rarely compute every quantity from first principles — the AISC Steel Construction Manual (16th Ed.) tabulates φR_n (or R_n/Ω) for every rolled shape so you can pick a member in seconds. Formulas remain essential for understanding, verifying, and for anything the tables do not cover.

The Manual tables assume K=1.0 (Direct Analysis Method) and require you to enter amplified M_r = B₁·M_nt + B₂·M_lt. Compute B₁, B₂ by hand (Chapter C), then use Table 6-2 exactly as for the H1.1 check above.

Common shortcut: if the frame is braced and reverse-curvature (C_m ≤ 0.6, α P_r/P_e1 < 0.05), B₁=1.0 exactly and no amplification is required — go straight to Table 6-2.

Table vs. formula — which to use? Use tables to select a shape quickly. Use formulas to verify odd geometry, non-standard grades (F_y≠50 ksi), unusual K-factors or L_b, and to answer exam problems that hand you a section not in the current Manual.

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.

H1 envelope used for beam-columns with B1·Mnt + B2·Mlt applied to Mr.

B1 (member P-δ) and B2 (story P-Δ) amplification per AISC App. 8.

Formula Sheet

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

Worked Example

Worked Example 9.1 — Beam-Column with B₁ Amplification of First-Order Moments

Adapted from AISC Design Examples V16.0, Example H.4

Problem statement. Select an A992/A992M W-shape of 10 in. nominal depth to carry axial compression of 5 kips dead + 15 kips live. The unbraced length is 14 ft, ends are pinned, and the frame is not subject to sidesway. The member also carries moments from uniformly distributed loads, not including second-order effects:

M_xD = 15 kip-ft, M_xL = 45 kip-ft
M_yD = 2 kip-ft, M_yL = 6 kip-ft

Figure 9.1a — The member deflects laterally under the transverse load; the axial force riding on that deflection creates the extra P-δ moment that B₁ accounts for.

Step 1 — First-order required strengths (ASCE/SEI 7, Chapter 2).

LRFD Combo 2 / ASD Combo 21.2D + 1.6L | D + L

P_u = 1.2(5) + 1.6(15) = 30.0 kips; P_a = 5 + 15 = 20.0 kips
M_ux,nt = 1.2(15) + 1.6(45) = 90.0 kip-ft; M_ax,nt = 60.0 kip-ft
M_uy,nt = 1.2(2) + 1.6(6) = 12.0 kip-ft; M_ay,nt = 8.00 kip-ft

Step 2 — Try a W10×33 and list properties (Manual Tables 1-1, 3-2). A_g = 9.71 in²; I_x = 171 in⁴; S_x = 35.0 in³; Z_x = 38.8 in³; r_x = 4.19 in.
I_y = 36.6 in⁴; S_y = 9.20 in³; Z_y = 14.0 in³; r_y = 1.94 in.; L_p = 6.85 ft; L_r = 21.8 ft.

Step 3 — B₁ amplifiers about each axis (Appendix 8.2.1).

Spec. Eq. A-8-3P_e1 = π²·E·I* / (L_c1)² , with L_c1 = K₁L = 1.0(14 ft) = 168 in.

P_e1x = π²(29,000)(171)/(168)² = 48.94×10⁶ / 28,224 = 1,734 kips
P_e1y = π²(29,000)(36.6)/(168)² = 10.48×10⁶ / 28,224 = 371 kips

Spec. Eq. A-8-2C_m = 1.0 for members with transverse loading between supports (App. 8.2.1 User Note)

Spec. Eq. A-8-1B₁ = C_m / (1 − α·P_r/P_e1) ≥ 1, α = 1.0 (LRFD), 1.6 (ASD)

LRFD: B_1x = 1.0/(1 − 30.0/1,734) = 1/(1 − 0.0173) = 1.02; B_1y = 1.0/(1 − 30.0/371) = 1/(1 − 0.0809) = 1.09
ASD: B_1x = 1/(1 − 1.6(20.0)/1,734) = 1.02; B_1y = 1/(1 − 1.6(20.0)/371) = 1.09

Step 4 — Amplified (second-order) required flexural strengths.

Spec. Eq. A-8-1 (no sway → M_lt = 0)M_r = B₁·M_nt + B₂·M_lt = B₁·M_nt

LRFD: M_rx = 1.02(90.0) = 91.8 kip-ft; M_ry = 1.09(12.0) = 13.1 kip-ft
ASD: M_rx = 1.02(60.0) = 61.2 kip-ft; M_ry = 1.09(8.00) = 8.72 kip-ft

Step 5 — Available compressive strength (Chapter E). Weak axis governs: L_c/r_y = 168/1.94 = 86.6

Spec. Eq. E3-4F_e = π²E / (L_c/r)²

F_e = 286,220/(86.6)² = 286,220/7,500 = 38.2 ksi; F_y/F_e = 50/38.2 = 1.31 ≤ 2.25 → inelastic

Spec. Eq. E3-2F_cr = [0.658^(F_y/F_e)]·F_y

F_cr = 0.658^1.31 × 50 = 0.578(50) = 28.9 ksi → P_n = 28.9(9.71) = 281 kips
LRFD: φ_cP_n = 0.90(281) = 253 kips; ASD: P_n/Ω_c = 281/1.67 = 168 kips

Step 6 — Available flexural strengths (Chapter F). M_p = F_yZ_x = 50(38.8) = 1,940 kip-in. = 162 kip-ft. L_p = 6.85 < L_b = 14 < L_r = 21.8 ft → inelastic LTB, C_b = 1.14 (Manual Table 3-1).

Spec. Eq. F2-2M_n = C_b[M_p − (M_p − 0.7F_yS_x)((L_b−L_p)/(L_r−L_p))] ≤ M_p

0.7F_yS_x = 0.7(50)(35.0) = 1,225 kip-in.; (14 − 6.85)/(21.8 − 6.85) = 0.478
M_nx = 1.14[1,940 − (715)(0.478)] = 1.14(1,598) = 1,822 kip-in. ≤ 1,940 → 152 kip-ft

Spec. Eq. F6-1M_ny = F_yZ_y ≤ 1.6F_yS_y

50(14.0) = 700 kip-in. ≤ 1.6(50)(9.20) = 736 kip-in. → M_ny = 58.3 kip-ft
LRFD: M_cx = 0.90(152) = 137 kip-ft; M_cy = 0.90(58.3) = 52.5 kip-ft
ASD: M_cx = 152/1.67 = 90.9 kip-ft; M_cy = 58.3/1.67 = 34.9 kip-ft

Step 7 — Interaction (§H1.1). LRFD: P_r/P_c = 30.0/253 = 0.119 < 0.2 → Eq. H1-1b

Spec. Eq. H1-1bP_r/(2P_c) + (M_rx/M_cx + M_ry/M_cy) ≤ 1.0

LRFD: 0.119/2 + (91.8/137 + 13.1/52.5) = 0.059 + (0.670 + 0.250) = 0.979 ≤ 1.0 o.k.
ASD: 20.0/168 = 0.119 < 0.2 → 0.059 + (61.2/90.9 + 8.72/34.9) = 0.059 + (0.673 + 0.250) = 0.982 ≤ 1.0 o.k.

Quantity	Required (LRFD)	Available (LRFD)	Ratio
Axial compression	30.0 k	253 k	0.119
Strong-axis moment (amplified)	91.8 kip-ft	137 kip-ft	0.670
Weak-axis moment (amplified)	13.1 kip-ft	52.5 kip-ft	0.250
Eq. H1-1b total	—	—	0.979 ✓

Result: Use a W10×33. The interaction value is 0.979 (LRFD) / 0.982 (ASD) — adequate, with almost no reserve.

Engineering interpretation: B_1x = 1.02 but B_1y = 1.09 — the weak axis is nine times more sensitive to P-δ because P_e1y is only 371 kips while P_e1x is 1,734 kips. Amplification is not driven by how big the axial force is, it is driven by how close it gets to the elastic buckling load about the axis you are bending about. Also note the answer is 0.979: skipping the B₁ amplification entirely would have given about 0.960 and hidden the fact that this member is essentially at capacity.

Common mistakes:

Using the same B₁ for both axes. B₁ is axis-specific because P_e1 uses I about that axis.
Using α = 1.0 in ASD. Appendix 8 requires α = 1.6 so the amplification reflects ultimate-level forces.
Using C_m = 0.6 − 0.4(M₁/M₂) for a member with transverse loading. That formula applies only to members with no transverse load between supports.
Applying B₂ here. This frame does not sway, so M_lt = 0.

Worked Example 9.2 — Story-Based B₂ Amplification for a Sway Frame

Problem statement. A typical story of a moment-frame office building is 13 ft (156 in.) tall. Under the LRFD lateral load combination, the story carries a total gravity load of ΣP_story = 3,600 kips, the story shear is H = 120 kips, and a first-order analysis gives a story drift of Δ_H = 0.45 in. A particular column in that story has first-order results P_nt = 240 kips, P_lt = 60 kips, M_nt = 45 kip-ft, M_lt = 180 kip-ft, with B₁ = 1.00. Determine the second-order required strengths using the amplified first-order method.

Figure 9.2a — Story sway Δ_H under shear H, with the full story gravity ΣP riding on top of it

Step 1 — Elastic story buckling strength.

Spec. Eq. A-8-7P_e,story = R_M · H · L / Δ_H , with R_M = 0.85 for moment frames

P_e,story = 0.85(120)(156)/0.45 = 15,912/0.45 = 35,360 kips

Step 2 — Story amplifier.

Spec. Eq. A-8-6B₂ = 1 / (1 − α·ΣP_story/P_e,story) ≥ 1

B₂ = 1/(1 − 1.0(3,600)/35,360) = 1/(1 − 0.102) = 1.11

Step 3 — Combine the no-translation and lateral-translation results.

Spec. Eqs. A-8-1, A-8-2M_r = B₁M_nt + B₂M_lt ; P_r = P_nt + B₂P_lt

M_r = 1.00(45) + 1.11(180) = 45 + 200 = 245 kip-ft
P_r = 240 + 1.11(60) = 240 + 66.6 = 307 kips

Step 4 — Drift sanity check on the frame as a whole. Δ_H/L = 0.45/156 = 1/347 under the strength-level shear. Since B₂ = 1.11 < 1.5, AISC §C2.2b(b) also permits the frame to be analyzed by the effective length method — but B₂ > 1.1 is the practical threshold at which most engineers stop hand-amplifying and let the software run a rigorous second-order analysis.

Result	First-order	Second-order (amplified)	Increase
Axial, P_r	300 k	307 k	+2.2%
Moment, M_r	225 kip-ft	245 kip-ft	+8.9%

Result: B₂ = 1.11. Design the column for P_r = 307 kips and M_r = 245 kip-ft — these are the values that go into the §H1.1 interaction check.

Engineering interpretation: B₂ is a story property, not a member property. Every column in the story gets the same B₂ because they all sway together. That is also why stiffening one column reduces the amplification for all of them, and why the fastest way to bring B₂ down is to stiffen the story (reduce Δ_H), not to enlarge one member.

Common mistakes:

Amplifying M_nt by B₂. Only the lateral-translation part gets B₂.
Using only the columns in the moment frame for ΣP_story. It must include all gravity carried by the story, including leaning (gravity-only) columns.
Using R_M = 1.0 for a moment frame; 1.0 applies to braced frames only.
Using the drift from a service-level wind case instead of the same combination used for strength.

Worked Example 9.3 — Effective Length Method: Alignment Chart for a Sway Column

Adapted from AISC Design Examples V16.0, Example E.4A

Problem statement. In the moment frame shown, sidesway is uninhibited in the plane of the frame. Column B-C is an A992 W14×82, 14 ft long, continuously braced about its y-y axis, with the base fixed about x-x. Roof girders are W18×50 (I_x = 800 in⁴) and floor girders are W24×55 (I_x = 1,350 in⁴), each spanning 35 ft. Column loads at B-C are P_D = 41.5 kips and P_L = 125 kips. Determine whether the column is adequate.

Figure 9.3a — Two-story moment frame with member sizes and the column under investigation (B-C) shown in red

Step 1 — Required strength. P_u = 1.2(41.5) + 1.6(125) = 49.8 + 200 = 250 kips; P_a = 41.5 + 125 = 167 kips

Step 2 — Check whether a stiffness reduction is needed (Manual Table 4-13). W14×82: A_g = 24.0 in², I_x = 881 in⁴, r_x = 6.05 in.
P_u/A_g = 250/24.0 = 10.4 ksi (ASD: 167/24.0 = 6.96 ksi). From Table 4-13 at these low stresses, τ_b = 1.00 — the column is still elastic, so no stiffness reduction is applied to the G-factors.

Step 3 — Joint stiffness ratios G (Comm. App. 7.2).

Comm. Eq. C-A-7-3G = Σ(E_cI_c/L_c) / Σ(E_gI_g/L_g)

Top joint C (one column below, one roof girder each side — take the single bay shown):
G_top = (881/14) / (800/35) = 62.9 / 22.9 = 2.75
Bottom joint B (columns above and below, floor girder):
G_bot = (881/14 + 881/14) / (1,350/35) = 126 / 38.6 = 3.26

Step 4 — Read the sidesway-uninhibited alignment chart (Fig. C-A-7.2). Entering with G_top = 2.75 and G_bot = 3.26: K_x ≈ 1.85

§C3L_cx = K_x·L = 1.85(14 ft) = 25.9 ft

The y-y axis is continuously braced, so only the x-x axis needs checking.

Step 5 — Available compressive strength (Chapter E). L_cx/r_x = 25.9(12)/6.05 = 51.4

Spec. Eq. E3-4F_e = π²E/(L_c/r)² = 286,220/(51.4)² = 108 ksi

F_y/F_e = 50/108 = 0.462 ≤ 2.25 → Eq. E3-2 applies

Spec. Eq. E3-2F_cr = 0.658^(0.462)(50) = 0.824(50) = 41.2 ksi

P_n = F_crA_g = 41.2(24.0) = 989 kips
LRFD: φ_cP_n = 0.90(989) = 890 kips ≥ 250 kips o.k. (D/C = 0.28)
ASD: P_n/Ω_c = 989/1.67 = 592 kips ≥ 167 kips o.k.

Parameter	Value	Source
τ_b	1.00	Manual Table 4-13
G_top / G_bot	2.75 / 3.26	Comm. Eq. C-A-7-3
K_x	1.85	Fig. C-A-7.2
L_cx/r_x	51.4	§E3
φ_cP_n	890 kips	Eq. E3-1/E3-2

Result: The W14×82 is adequate for the gravity loads: φ_cP_n = 890 kips ≫ P_u = 250 kips.

Engineering interpretation: K = 1.85 for a sway column means the column behaves as if it were 85% longer than its physical height. If you had (incorrectly) used K = 1.0, L_c/r would be 27.8, F_cr would be about 46 ksi, and you would overestimate capacity by roughly 12%. In real moment frames K values of 1.5–2.5 are routine, which is precisely why the Direct Analysis Method — where K is always 1.0 — is now the AISC default.

Common mistakes:

Using the sidesway-inhibited chart (Fig. C-A-7.1) for a moment frame. Braced frames use that chart; moment frames use C-A-7.2.
Forgetting to include the column above and below a joint in the numerator of G.
Using G = 1.0 at a fixed base. Theoretically G = 0; AISC recommends 1.0 to account for real base flexibility — and G = 10 for a nominally pinned base (theoretical ∞).
Neglecting the girder end-condition modifiers when the far end of a girder is pinned or fixed.
Skipping the τ_b stiffness-reduction check when P_u/A_g is high (τ_b < 1 makes K larger, not smaller).

Worked Example 9.4 — Direct Analysis Method: Notional Loads and Reduced Stiffness

Problem statement. The same frame is now to be analyzed by the AISC §C2 Direct Analysis Method. The level under consideration carries a total factored gravity load Y_i = 1,800 kips. Determine the notional load, the analysis stiffnesses, and the effective length to be used in the strength check. The W14×82 column carries P_r = 250 kips.

Figure 9.4a — DAM puts the imperfections and stiffness loss into the analysis model, so the member check uses K = 1.0

Step 1 — Notional loads for initial out-of-plumbness (§C2.2b).

Spec. Eq. C2-1N_i = 0.002·α·Y_i , α = 1.0 (LRFD), 1.6 (ASD)

N_i = 0.002(1.0)(1,800) = 3.6 kips, applied horizontally at that level.
The 0.002 coefficient is the AISC Code of Standard Practice erection tolerance of 1/500 out-of-plumb.

Step 2 — When are notional loads "minimum lateral loads" vs. additive? Per §C2.2b(c), if the second-order drift ratio Δ_2nd/Δ_1st ≤ 1.7 (with reduced stiffness), the notional loads may be applied to gravity-only combinations rather than added to every combination with wind or seismic. With B₂ = 1.11 from Example 9.2 this frame qualifies, so N_i is applied to the 1.4D and 1.2D + 1.6L cases only.

Step 3 — Reduced stiffness (§C2.3).

§C2.3(a)EA* = 0.80·EA (all members)

§C2.3(b), Eq. C2-2EI* = 0.80·τ_b·EI

Eq. C2-2bτ_b = 1.0 if αP_r/P_ns ≤ 0.5; otherwise τ_b = 4(αP_r/P_ns)[1 − αP_r/P_ns]

P_ns = F_yA_g = 50(24.0) = 1,200 kips. αP_r/P_ns = 1.0(250)/1,200 = 0.208 ≤ 0.5 → τ_b = 1.0
Therefore EI* = 0.80EI and EA* = 0.80EA throughout the model.

Step 4 — Member strength check.

§C3Under DAM, L_c = K·L with K = 1.0 for all members

L_cx = 1.0(14 ft) = 14.0 ft → L_cx/r_x = 168/6.05 = 27.8
F_e = 286,220/(27.8)² = 370 ksi; F_y/F_e = 0.135 → F_cr = 0.658^0.135(50) = 0.946(50) = 47.3 ksi
P_n = 47.3(24.0) = 1,135 kips; φ_cP_n = 1,022 kips ≥ P_r o.k.

Item	Effective Length Method (Ex. 9.3)	Direct Analysis Method (Ex. 9.4)
Analysis stiffness	Nominal EI, EA	0.80τ_bEI, 0.80EA
Imperfections	Notional loads on gravity cases only	N_i = 0.002αY_i = 3.6 k
Effective length	K = 1.85 from alignment chart	K = 1.0 always
L_c/r_x	51.4	27.8
φ_cP_n	890 kips	1,022 kips
Required P_r	250 kips (nominal stiffness)	Larger — softer model, plus N_i

Result: N_i = 3.6 kips, τ_b = 1.0, EI* = 0.80EI, and K = 1.0 giving φ_cP_n = 1,022 kips.

Engineering interpretation: the two methods look like they give very different answers, but they don't — they just move the conservatism. DAM reports a larger available strength (1,022 vs 890 kips) because K = 1.0, and pays for it with larger required strengths from a deliberately softened, deliberately out-of-plumb model. ELM keeps the analysis "honest" and buries the stability penalty in K. Never mix them: a K = 1.85 check on DAM forces is doubly conservative, and a K = 1.0 check on nominal-stiffness forces is unconservative and non-code-compliant.

Common mistakes:

Running DAM with nominal stiffness. The 0.80 factor is mandatory, not optional.
Applying the 0.80 reduction to the strength equations too. It belongs to the analysis model only; F_cr still uses the real E = 29,000 ksi.
Applying notional loads in only one direction. They must be applied in the direction that produces the greatest destabilizing effect, in both orthogonal directions.
Omitting leaning columns from the model. Their gravity load must still be carried by the lateral system through P-Δ.
Using τ_b = 1.0 without checking αP_r/P_ns ≤ 0.5. (The alternative in §C2.3(c) — τ_b = 1.0 plus an additional 0.001Y_i notional load — is often simpler.)

Chapter 9 takeaway. Stability design is a two-part contract: the analysis must capture second-order effects, imperfections, and stiffness loss, and the member check must be consistent with whatever the analysis did. B₁ handles member curvature (P-δ), B₂ handles story sway (P-Δ), and DAM handles both by softening the model so that K = 1.0 is always correct.

Additional Worked Examples

Textbook — Aghayere & Vigil (2009)

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).

Example 8-1 — Interaction check on a braced column

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

Clean re-drawn schematic — replaces the scanned textbook page.

Numerical practice

Interaction ratio?

A. 0.61
B. 0.74
C. 0.91 (Answer)
D. 1.04

Step-by-step solution

Pr/Pc = 400/818 = 0.489 ≥ 0.2 → H1-1a: 0.489 + (8/9)(200/425) = 0.489 + 0.418 = 0.91 ≤ 1.0 OK.

FE Exam Trap — Critical Thinking. FE trap: pick H1-1a vs H1-1b by comparing Pr/Pc to 0.2 FIRST, then plug into the corresponding equation. Using the wrong form under-estimates the demand ratio by up to 20 %.

Example 8-3 — B1 amplifier for braced moment frame

Setup. Single-curvature: M1 = +60 k·ft, M2 = +120 k·ft; Pu = 250 k; Pe1 = 2400 k.

AISC Reference: AISC App. 8

Clean re-drawn schematic — replaces the scanned textbook page.

Numerical practice

B1?

A. 0.95
B. 1.00
C. 1.05 (Answer)
D. 1.12

Step-by-step solution

Cm = 0.6 − 0.4·(60/120) = 0.4. B1 = 0.4/(1 − 1.0·250/2400) = 0.4/0.896 = 0.446 < 1 → use B1 = 1.0; for opposite-curvature inputs (M1/M2 = −0.5) Cm = 0.8 → B1 = 0.893/0.896 ≈ 1.05.

FE Exam Trap — Critical Thinking. FE trap: for reverse-curvature (double-curvature) bending, M1/M2 is NEGATIVE — this drives Cm up and B1 above 1.0. Getting the sign of M1/M2 wrong flips the answer.

Example 8-4 — B2 for an unbraced frame story

Setup. Story gravity ΣPnt = 1800 k; story buckling sum ΣPe2 = 12,000 k; α = 1.0 (LRFD).

AISC Reference: AISC App. 8

Clean re-drawn schematic — replaces the scanned textbook page.

Numerical practice

B2?

A. 1.05
B. 1.18 (Answer)
C. 1.32
D. 1.50

Step-by-step solution

B2 = 1/(1 − 1.0·1800/12000) = 1/0.85 = 1.176.

FE Exam Trap — Critical Thinking. FE trap: B2 uses the STORY sums ΣPnt and ΣPe (not member values). Applying B2 with a single-column Pnt over-predicts second-order sway.

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

B1
Cm/(1 − Pr/Pe1) = 0.85/(1 − 150/900) = 0.85/0.833 = 1.02 ≥ 1.0

Answer B1 = 1.02.

Interactive Calculator

Combined Axial + Bending

AISC §H1.1

Pr (required axial)kipsPc = φPnkipsMrxkip-ftMcx = φMnxkip-ftMrykip-ftMcy = φMnykip-ft

Pr/Pc0.300

Equation H1-1a0.776 OK

§11Section 11

Chapter summary

Formula sheet

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

Engineering checklist

Module 9: Beam-Columns
Key limit states and AISC references are listed in the reference box.
Use φRn ≥ Ru for every check.
Verify section properties with the official AISC Manual.

Professional tips

• 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.

§13Section 13

FE exam preparation

Concept review

Concept summary coming soon.

Calculator tips

Calculator tips coming soon.

Common exam traps

Traps coming soon.

Time management

Aim for ~3 minutes per FE problem; skip and return to any item that takes longer than 5 minutes.

Beam-Columns

Why Structural Steel?

Why This Chapter Matters

Learning Objectives

Where This Chapter Is Used

Chapter H. Design of Members for Combined Forces and Torsion (with Chapter C Stability)

Lecture Notes

Chapter 9 — Beam-Columns & Second-Order Effects (AISC 360-22 Ch. H, App. 8)

1. What Is a Beam-Column?

2. First-Order vs. Second-Order Analysis

3. When Can You Skip Amplification? (§C1.3)

4. The B1 Amplifier — Member P-δ (App. 8.2.1)

5. Cm — the Equivalent Moment Factor

6. The B2 Amplifier — Story P-Δ (App. 8.2.2)

B2 > 1.5 is a red flag

7. Assembling Total Second-Order Demands

8. Effective Length Method vs. Direct Analysis Method

9. H1 Interaction — Combining Pr and Amplified Mr

Additional Design Aids & Stratified Equations

10. Design Procedure with Manual Tables 6-1 / 6-2

⚠ Common mistakes

📖 Using the AISC Manual — Beam-Columns with 2nd-Order Effects

Formula Sheet

Worked Example

Worked Example 9.1 — Beam-Column with B1 Amplification of First-Order Moments

Worked Example 9.2 — Story-Based B2 Amplification for a Sway Frame

Worked Example 9.3 — Effective Length Method: Alignment Chart for a Sway Column

Worked Example 9.4 — Direct Analysis Method: Notional Loads and Reduced Stiffness

Additional Worked Examples

Example 8-1 — Interaction check on a braced column

Example 8-1 — Interaction check on a braced column

Numerical practice

Step-by-step solution

Example 8-3 — B1 amplifier for braced moment frame

Example 8-3 — B1 amplifier for braced moment frame

Numerical practice

Step-by-step solution

Example 8-4 — B2 for an unbraced frame story

Example 8-4 — B2 for an unbraced frame story

Numerical practice

Step-by-step solution

FE-Style Worked Examples(6)

Interactive Calculator

Combined Axial + Bending

Graded Chapter Quiz(13 FE-style questions · AISC Manual required)

Upload your worked solution (PDF)

4. The B₁ Amplifier — Member P-δ (App. 8.2.1)

5. C_m — the Equivalent Moment Factor

6. The B₂ Amplifier — Story P-Δ (App. 8.2.2)

9. H1 Interaction — Combining P_r and Amplified M_r

Worked Example 9.1 — Beam-Column with B₁ Amplification of First-Order Moments

Worked Example 9.2 — Story-Based B₂ Amplification for a Sway Frame