17

Stability, Bracing, and Second-Order Effects

Why Structural Steel?

Direct Analysis Method, B1/B2, appendix 6 bracing.

⏱ 90 min★ Advanced✎ 3 objectives
§01Section 01

Engineering story

Engineering story
Chapter 17 · Stability, Bracing, and Second-Order Effects

Direct Analysis Method, B1/B2, appendix 6 bracing.

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 the Direct Analysis Method
  • Compute notional loads
  • Design Appendix 6 bracing
§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 — Specification chapter governing this topic
  • AISC Manual 16th Ed. — Design tables and worked examples

Why This Chapter Matters

AISC's Direct Analysis Method (DAM, Chapter C) is the modern approach to frame stability. Missing notional loads, τb reductions, or B2 amplifiers under-predicts moments and can send an entire frame past its stability limit.

Learning Objectives

  • Apply the three DAM requirements: notional loads (0.002·Yi), reduced stiffness (0.8τbEI, 0.8EA), and second-order analysis.
  • Compute B1 and B2 for a moment frame story.
  • Design column bracing (Appendix 6) for both stiffness and strength.
  • Distinguish nodal, relative, continuous, and lean-on bracing.
  • Recognize when the Effective Length Method is still permitted vs. DAM.

Where This Chapter Is Used

The lateral-load design of the capstone moment frames and any tall unbraced building.

ANSI / AISC 360-22Specification for Structural Steel Buildings16.1-15 to 16.1-21
Chapter
C
AISC 360-22

Chapter C. Design for Stability (Direct Analysis Method)

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

§Section titlePage
C1General Stability Requirements16.1-15
C2Calculation of Required Strengths16.1-16
C2.2Consideration of Initial Imperfections (Notional Loads 0.002·Yi)16.1-17
C2.3Adjustments to Stiffness (0.8·τb·EI, 0.8·EA)16.1-18
C3Calculation of Available Strengths16.1-19
App. 6Member Stability Bracing (Nodal, Relative, Continuous)16.1-215
App. 7Alternative Methods (Effective Length, First-Order)16.1-225
App. 8Approximate Second-Order Analysis (B1, B2)16.1-235

Companion reference: AISC Design Guide 28 — Stability Design of Steel Buildings

Lecture Notes

Chapter 17 — Stability, Bracing & Second-Order Effects (AISC 360-22 Ch. C, App. 6, App. 7, App. 8)

Chapter focus. A structure can satisfy every member strength check in Chapters D–H and still collapse — not because a section broke, but because the whole frame became unstable, or because a compression member buckled between its braces. Stability is a system property: it depends on how stiff the frame is, how out-of-plumb it is, and how the axial load in every column amplifies the sway that put it there in the first place. This chapter builds the vocabulary and the three permitted analysis methods (Direct Analysis, Effective Length, First-Order) and then covers Appendix 6, which tells you how strong and how stiff a brace must be to actually do its job.

1. What "Stability" Means for a Frame

Every steel structure sways a little under load, and every column has some initial crookedness and out-of-plumbness from fabrication and erection tolerances. When axial load P acts through that lateral displacement, it creates additional bending moment that was not in the first-order (undeformed-geometry) analysis. This is called a second-order effect, and it comes in two flavors: P-δ, the amplification of moment along the length of a single bowed member, and P-Δ, the amplification of overturning moment caused by the sway (drift) of an entire story. A stability-adequate design must capture both.

Two flavors of second-order effect: P-δ (member) vs P-Δ (frame) P δ (member bow) P-δ single-member curvature amplifies M P Δ (story drift) P-Δ whole-story sway amplifies overturning M
Figure 17.1 — P-δ (member curvature effect) versus P-Δ (story sway effect). Both amplify the first-order moment; DAM captures both automatically in a proper second-order analysis.

2. The Three Requirements of AISC Chapter C (§C1)

Section C1 states that any analysis method used for stability design must account for four things simultaneously: (1) flexural, shear, and axial member deformations, and all other deformations that contribute to displacement; (2) second-order effects (both P-Δ and P-δ); (3) geometric imperfections (initial out-of-plumbness); and (4) stiffness reduction due to inelasticity (residual stress causing partial yielding before the nominal strength is reached). AISC gives you three ways to satisfy all four:

  1. Direct Analysis Method (DAM), §C2 — the default, always-permitted method. Reduced stiffness (80% τbEI, 80% EA) captures inelasticity; notional loads capture imperfections; a full second-order analysis captures P-Δ/P-δ; K = 1.0 always.
  2. Effective Length Method (ELM), Appendix 7 — permitted only when the structure is not too flexible (Δ2nd/Δ1st ≤ 1.5 with un-reduced stiffness) and axial loads are moderate (Pr/Py ≤ 0.5 for all members contributing to stability). Uses full stiffness, notional loads only in gravity combos, and computes K from an alignment chart (sidesway uninhibited).
  3. First-Order Analysis Method, §C2.2b(4) — permitted for very stiff braced structures where Δ2nd/Δ1st ≤ 1.5 and Pr/Py ≤ 0.5 for every member; skips second-order analysis entirely by amplifying the notional load itself (B1 applied to notional load coefficient) and using K = 1.0.
Why DAM is the default. ELM and the First-Order Method both require you to verify a drift ratio limit before you're allowed to use them — a chicken-and-egg problem, because you need an analysis to check the ratio. DAM has no such precondition, which is why most software defaults to it and why the 2016+ Specification steers designers there.

3. Direct Analysis Method — the Three Ingredients (§C2)

DAM modifies the analysis model itself, not the design equations. Three ingredients, applied together, satisfy all four §C1 requirements without any K-factor calculation.

3.1 Reduced (Inelastic) Stiffness — §C2.3

AISC 360-22 §C2.3EI* = 0.80 τb EI     EA* = 0.80 EA
  • EI, EA — nominal (gross-section) flexural and axial stiffness used in an ordinary elastic analysis.
  • 0.80 factor — a blanket reduction applied to all members that contribute to lateral stability, accounting for the general softening effect of distributed yielding (residual stress) throughout the structure, independent of any one member's load level.
  • τb — an additional, member-specific reduction for columns with high axial load, defined by Eq. C2-2a/b: τb = 1.0 if α Pr/Pns ≤ 0.5; otherwise τb = 4(αPr/Pns)(1 − αPr/Pns). Pns = FyAg (or the appropriate slender-element value); α = 1.0 (LRFD) or 1.6 (ASD).

Physically: a column near its squash load is softer than E predicts because parts of its cross-section have already yielded from residual stress; τb mimics that softening in an otherwise-elastic model, and a stiffer, all-purpose 0.80 knock-down covers the rest of the structure. Alternatively, §C2.3(b) permits using τb = 1.0 for all members if a supplemental notional load of 0.001αYi is added to every load combination (an added-stiffness-penalty trade for simplicity) — useful in software that can't vary τb by member.

3.2 Notional Loads — §C2.2b

AISC 360-22 Eq. C2-1Ni = 0.002 α Yi
  • Ni — a fictitious lateral load applied at level i, in the direction that adds to the destabilizing effect being investigated.
  • 0.002 — represents an assumed erection out-of-plumbness of 1/500 (a typical AISC Code of Standard Practice erection tolerance).
  • Yi — total gravity load acting on and above level i in that load combination (not just what's tributary to one column — the whole story's gravity load, including "leaning" columns).
  • α — 1.0 for LRFD, 1.6 for ASD (so ASD notional loads use the amplified 1.6× combination consistent with the ASD second-order-analysis rule).

Notional loads must be applied in every gravity-only combination, always. They must additionally be applied in combinations that already contain lateral load only if the structure is flexible enough that Δ2nd/Δ1st (computed with reduced stiffness) exceeds 1.7 — in stiffer structures the real wind/seismic load already envelopes the small imperfection effect.

3.3 K = 1.0, Always

Because DAM's reduced-stiffness second-order analysis already captures the true amplified moments and forces, there is no need — and no benefit — to computing an effective length factor. Every member in a DAM analysis uses K = 1.0, so Lc = L for flexural buckling in the plane of instability. This eliminates alignment-chart iteration entirely for DAM designs.

Additional Design Aids & Stratified Equations

DAM — notional lateral load 0.002·Yi at each level N3 N2 N1 ↓ Y3 ↓ Y2 ↓ Y1 Ni = 0.002 · α · Yi
Notional lateral loads applied at every level in gravity-only combos (DAM). K = 1.0 for every DAM member.
Notional loadN_i = 0.002 · α · Y_i, α = 1.0 (LRFD), 1.6 (ASD)
Stiffness reductionEI* = 0.80·τ_b·EI ; EA* = 0.80·EA

4. Effective Length Method — Appendix 7

The ELM keeps the traditional K-factor approach that predates the 2005 Specification, but wraps it in the same second-order-analysis requirement as DAM. It uses full, un-reduced stiffness (no 0.80 or τb), applies notional loads only in gravity-only combinations, and requires the designer to compute K from a rational analysis such as the alignment chart for every column whose flexural buckling is sidesway-uninhibited.

When may ELM be used? Appendix 7 §7.2.1 requires both:

  1. The structure supports gravity load primarily through nominally vertical columns, walls, or frames (no highly skewed or arched gravity systems).
  2. The ratio of second-order to first-order drift, computed with LRFD loads (or 1.6× ASD loads) and un-reduced stiffness, does not exceed 1.5 in any story.

This second condition must be verified after the fact — you run the analysis, check the ratio, and if it's above 1.5 you must switch to DAM. In practice, stiff braced-frame and shear-wall buildings almost always satisfy it; slender moment frames often do not.

5. Alignment Charts and the G Factor

For sidesway-uninhibited (unbraced) frames, K is read from the AISC Commentary alignment chart using a stiffness ratio G computed at each end of the column:

AISC Commentary Ch. C, Eq. C-C2-3G = Σ(E Ic / Lc) / Σ(E Ig / Lg)
  • Σ(EIc/Lc) — sum of column stiffnesses framing into the joint (above and below).
  • Σ(EIg/Lg) — sum of girder (beam) stiffnesses framing into the joint, reduced for far-end conditions if the girder is not fixed (multiply by 0.5 if the far end is fixed, or 2/3 if pinned, in a sidesway frame).
  • G = 10 — conventional value for a pinned column base (theoretically infinite, capped at 10 in practice).
  • G = 1 — conventional value for a rigidly fixed column base.

With GA and GB known at the two ends of the column, a straightedge across the sidesway-uninhibited nomograph gives K directly (K ranges from 1.0 for both ends fixed to unbounded as the frame approaches a sway mechanism). Commentary Eq. C-A-7-5 (used in Example C.1B below) offers a closed-form "story-based" alternative to the chart that is often more accurate for irregular frames.

Alignment chart (nomograph) — sidesway uninhibited, K from G_A, G_B G_A G_B K 5050 5050 2.01.61.0 straightedge: connect G_A, G_B → read K G = Σ(EIc/Lc) / Σ(EIg/Lg) at each joint; G = 10 for pinned base, G = 1 for fixed base
Figure 17.2 — Alignment chart (nomograph) concept: with GA and GB known, a straightedge gives K for sidesway-uninhibited (moment-frame) columns.

6. First-Order Analysis Method (§C2.2b(4))

For very stiff, braced structures, AISC allows skipping second-order analysis software entirely. Instead of a true P-Δ analysis, the method amplifies the notional load coefficient itself and adds a supplemental notional load:

§C2.2b(4)Ni = 2.1 α (Δ/L) Yi ≥ 0.0042 α Yi

where Δ/L is the first-order drift ratio of the story under consideration. The method is only valid when: (1) the structure supports gravity load primarily through vertical columns/walls/frames, (2) Δ2nd/Δ1st ≤ 1.5, (3) Pr/Py ≤ 0.5 for all members whose flexural stiffness contributes to stability, and (4) K = 1.0 is used for all members. Member forces are then simply the first-order analysis results — no iteration, no second-order software — which makes this method attractive for hand calculations on braced low-rise structures.

7. B1-B2 Approximate Second-Order Analysis — Appendix 8

When software performs only a first-order analysis, Appendix 8 gives a closed-form way to approximate second-order forces by amplifying first-order results with two multipliers: B1 for the P-δ effect within a member (no-sway component) and B2 for the P-Δ effect of story sway.

App. 8, Eq. A-8-1, A-8-3Mr = B1 Mnt + B2 Mlt     Pr = Pnt + B2 Plt
  • Mnt, Pnt — "no translation" first-order forces: run the frame with sway restrained (add a fictitious horizontal reaction at each level).
  • Mlt, Plt — "lateral translation" first-order forces: apply the fictitious reactions from above as loads in the opposite direction on the unrestrained frame.
  • B1 = Cm/(1 − α Pr/Pe1) ≥ 1, where Pe1 = π²EI*/Lc1² is the member's braced (K=1) Euler buckling load and Cm accounts for end-moment gradient.
  • B2 = 1/(1 − α Pstory ΔH/(RM H L)) ≥ 1, where ΔH is the first-order interstory drift under story shear H, and Pstory is the total gravity load supported by that story (moment-frame columns and leaning columns).

B1 amplifies moment along a single member due to its own axial load and curvature; B2 amplifies the sway moments and forces of an entire story due to the total gravity load riding on that story's drift. B2 ≥ 1.4 or so is a red flag that the frame may not qualify for ELM.

8. Appendix 6 — Stability Bracing: Relative vs Nodal

A brace only works if it is both strong enough to resist the small out-of-plumbness force delivered by the braced member, and stiff enough that it doesn't stretch or rotate so much that the buckling mode simply shifts past it. Appendix 6 classifies bracing into two geometric types.

  • Relative (panel) bracing — controls the relative lateral movement of two points along the member's length (a diagonal brace tying one floor to the next, or a shear wall/diaphragm engaging a story). It resists racking of the whole panel, not the position of a single point.
  • Nodal (point) bracing — attaches to and controls a single discrete point on the member (a purlin, a girt, a floor diaphragm connection at one node). Each nodal brace fixes that one point, and the buckling half-wave length between adjacent nodal braces becomes the effective unbraced length Lb.

Both types can brace either a column (against flexural buckling) or a beam (against lateral-torsional buckling); the beam case further splits into bracing that only restrains lateral movement of the compression flange ("lateral" bracing) versus bracing that restrains twist of the whole cross-section directly ("torsional" bracing, e.g., a stiff diaphragm attached near mid-depth).

Appendix 6 — relative (panel) vs nodal (point) bracing of a column diagonal "panel" brace Relative bracing controls story racking; stiffness ties two points Nodal (point) bracing fixes a discrete point; sets max unbraced length L_b
Figure 17.3 — Relative (panel) bracing controls story racking; nodal (point) bracing fixes discrete points and sets the buckling length between them.

9. Required Brace Strength and Stiffness — Appendix 6 Equations

Brace typeRequired strength Pbr or MbrRequired stiffness βbr
Column, nodal (Eq. A-6-1, A-6-2)Pbr = 0.01 Prβbr = 8 Pr / (φ Lb), φ = 0.75 (Ω = 2.00)
Column, relative (Eq. A-6-3, A-6-4)Pbr = 0.004 Prβbr = 2 Pr / (φ Lb), φ = 0.75
Beam, lateral nodal (Eq. A-6-5, A-6-6)Pbr = 0.01 Mr Cd / hoβbr = 10 Mr Cd / (φ Lb ho), φ = 0.75
Beam, lateral relative (Eq. A-6-7, A-6-8)Pbr = 0.008 Mr Cd / hoβbr = 4 Mr Cd / (φ Lb ho), φ = 0.75
Beam, torsional nodal (Eq. A-6-9, A-6-10)Mbr = 0.024 Mr L / (n Cb ho)βT,br = 2.4 Mr² / (φ n E Iy Cb²), φ = 0.75
  • Pr — required axial compressive strength of the braced column; Mr — required flexural strength of the braced beam at the brace point.
  • Lb — distance between braces (unbraced length) that the brace is designed to control.
  • ho — distance between flange centroids of the beam (approximates d − tf).
  • Cd — 1.0 except 2.0 for the brace closest to the inflection point in a beam with reverse curvature (that brace works twice as hard).
  • n — number of nodal braces within the span for the torsional-bracing equations.

What this means physically. The strength check ensures the brace can carry the small "kick" force generated by the member's assumed out-of-straightness (typically L/500) as it tries to buckle; the stiffness check ensures the brace doesn't act like a soft spring that lets the member buckle in a longer, lower-capacity mode through the brace point. A brace can pass strength and still fail stiffness, or vice versa — both must be checked independently.

10. Leaning Columns and Their Effect on Frame Stability

A leaning column is a gravity-only column, typically pin-connected top and bottom, that carries substantial floor load but contributes essentially zero lateral stiffness of its own. Because it cannot resist sway, all of the P-Δ demand generated by the gravity load on a leaning column must be carried by the story's actual lateral system (moment frame, braced frame, or shear wall) through the diaphragm. In a second-order or B2 analysis, Pstory — the quantity that drives instability — must include the leaning-column loads, not just the loads directly on the frame columns. Ignoring leaning columns is a classic and dangerous underestimate of second-order demand, especially in buildings with a small braced core surrounded by large gravity bays.

Leaning (gravity-only, pinned) column stabilized by the moment frame P_lean pinned base & top moment-frame column P-Δ demand transferred to frame via diaphragm
Figure 17.4 — A pinned "leaning" column has no lateral stiffness of its own; its P-Δ demand is transferred through the diaphragm to the moment frame, which must be designed to stabilize it.

Design Workflow

  1. Choose DAM unless there is a specific reason to use ELM or First-Order (e.g., legacy software, hand check).
  2. Build the analysis model including every leaning column and its tributary gravity load.
  3. Reduce stiffness (DAM): EI* = 0.80τbEI, EA* = 0.80EA; verify the τb assumption after solving.
  4. Apply notional loads Ni = 0.002αYi at each level, in every gravity combo (and lateral combos if Δ2nd/Δ1st > 1.7).
  5. Run second-order analysis (or B1-B2 approximation per App. 8) using LRFD loads, or 1.6× ASD loads divided back down by 1.6.
  6. Design every member with K = 1.0 (DAM/First-Order) or with an alignment-chart K (ELM, after verifying Δ2nd/Δ1st ≤ 1.5).
  7. Check every discrete brace for both Pbr (strength) and βbr (stiffness) per Appendix 6.

⚠ Common mistakes

  • Mixing DAM's reduced-stiffness analysis with an alignment-chart K > 1.0 in the member check — DAM always uses K = 1.0.
  • Forgetting the τb reduction for heavily loaded columns (αPr/Pns > 0.5) and not re-checking the assumption after the analysis converges.
  • Omitting notional loads from gravity-only combinations because "the building has a wind/seismic load somewhere" — they are required in every gravity combo regardless.
  • Ignoring leaning-column gravity load when computing Pstory for B2 or for a story-stiffness K per Appendix 7 Commentary.
  • Checking only brace strength (Pbr) and skipping the stiffness (βbr) requirement, or vice versa — Appendix 6 requires both.
  • Using ELM without first verifying Δ2nd/Δ1st ≤ 1.5 and Pr/Py ≤ 0.5 for every stability-contributing member.
Exam / FE tips. Memorize Ni = 0.002αYi and the nodal column brace pair Pbr = 0.01Pr, βbr = 8Pr/(φLb) — these appear constantly. Remember DAM always uses K = 1.0; only ELM needs the alignment chart. On any second-order question, check whether the problem gives you Mnt/Mlt (use B1-B2) or a full nonlinear result (just read it off) before doing unnecessary work.
P-δ (member)P-Δ (story)HB1 amplifies member moments; B2 amplifies story sway moments
Stability: B1 amplifies member moments (P-δ); B2 amplifies story sway moments (P-Δ); DAM uses τb·EI.

Formula Sheet

NameEquationAISC Ref
Design strengthφ Rn ≥ RuAISC 360-22 B3.1

Worked Example

Chapter 17 — Worked Examples (Stability, Bracing & Second-Order Analysis)

Examples 17.1 and 17.2 are companion solutions to the same moment frame — one by the Direct Analysis Method, one by the Effective Length Method — adapted directly from AISC Design Examples V16.0, Examples C.1A and C.1B, so you can compare the two methods side by side on identical geometry and loads.

Worked Example 17.1 — Moment Frame by the Direct Analysis Method

Adapted from AISC Design Examples V16.0, Example C.1A.

Given

  • Single-story moment frame, grid A–E, story height L = 20.0 ft, columns pinned at base, unbraced x-x and y-y between footing and roof.
  • Moment connection only between grid B and C (30.0 ft bay); beams A–B and C–E and columns at A, D, E are pinned ("leaning") — they carry gravity load but supply no lateral stiffness.
  • Total tributary bay width to the frame line = 120 ft (60 ft each side of B–C, per the simplified single leaning-column model).
  • Columns at B, C: W12×65, Ag = 19.1 in², ASTM A992 (Fy = 50 ksi).
  • Uniform gravity load on the B–C beam: wD = 0.400 klf, wL = 1.20 klf (includes beam/column self-weight allowance).
  • Governing gravity combinations: LRFD wu = 1.2D + 1.6L; ASD wa = D + L.
Frame elevation, grid A–E — moment frame B–C, leaning columns A, D, E moment connection B–C (rigid) A (pin/lean) B C D (pin/lean) E (pin/lean) Story height = 20.0 ft ; w_D = 0.400 klf, w_L = 1.20 klf on B–C beam (30 ft trib.)
Figure 17.1a — Elevation: moment frame between grids B–C stabilizes the entire A–E line; A, D, E are pinned leaning columns.

Step 1 — Factored Gravity Loads

ASCE 7 Ch. 2wu = 1.2D + 1.6L    wa = D + L
LRFD: wu = 1.2(0.400) + 1.6(1.20) = 2.40 klf.
ASD: wa = 0.400 + 1.20 = 1.60 klf. Per §C2.1(d), the ASD second-order analysis uses 1.6×wa = 2.56 klf, then results are divided by 1.6 for the member check.

Step 2 — Frame Analysis Loads on B–C

Tributary beam reactionPu = wu·l/2
Concentrated load onto columns B, C from the 30 ft adjacent beams:
LRFD: Pu = 2.40(30.0)/2 = 36.0 k.   ASD: Pa = 2.56(30.0)/2 = 38.4 k (at 1.6× for analysis).
Representative leaning-column load (all gravity stabilized by, but not applied to, the frame): LRFD PuL = 60.0(2.40) = 144 k; ASD PaL = 60.0(2.56) = 154 k.

Step 3 — Notional Loads (§C2.2)

Spec. Eq. C2-1Ni = 0.002 α Yi
Yi = total factored gravity on the frame level = 120 ft × w:
LRFD: Yi = 120(2.40) = 288 k → Ni = 0.002(1.0)(288) = 0.576 k.
ASD: Yi = 120(1.60) = 192 k, α = 1.6 → Ni = 0.002(1.6)(192) = 0.614 k.
Analysis model — combined leaning column, w_u, N_i representative leaning column, P_uL = 144 k w_u = 2.40 klf N_i = 0.576 k P_u = 36.0 k (from beam) L = 20.0 ft columns, K = 1.0 (DAM)
Figure 17.1b — Applied loads on the DAM analysis model: uniform w_u, concentrated beam reactions, representative leaning column P_uL, and notional load N_i.

Step 4 — Reduced Stiffness and Second-Order Analysis (§C2.3)

§C2.3EI* = 0.80 τb EI (assume τb = 1.0, verify below)
Half the gravity load on the story is carried by the moment-frame columns — more than one-third of the total — so §C2.1(b) requires both P-Δ and P-δ to be included. Software second-order results (given):
First-order drift: Δ1st = 0.181 in (LRFD), 0.193 in (ASD, prior to ÷1.6).
Second-order drift: Δ2nd = 0.290 in (LRFD), 0.321 in (ASD, prior to ÷1.6).
Drift ratio = 0.290/0.181 = 1.60 (LRFD); 0.321/0.193 = 1.66 (ASD) — both below the Commentary §C1 recommended limit of 2.5. ✓

Step 5 — Verify the τb = 1.0 Assumption

Eq. C2-2a/bPns = FyAg; τb = 1.0 if αPr/Pns ≤ 0.5
Pns = 50(19.1) = 955 k.
LRFD: αPr/Pns = 1.0(72.6)/955 = 0.0760 ≤ 0.5 ✓. ASD: 1.6(48.4)/955 = 0.0811 ≤ 0.5 ✓.
τb = 1.0 confirmed; no re-analysis needed.

Step 6 — Effective Length for Member Design (§C3)

§C3, DAM ruleK = 1.0 for all members ⇒ Lc = L
Lcx = Lcy = 20.0 ft for the B, C columns — no alignment-chart calculation required under DAM.
QuantityLRFDASD
Required axial strength, columns B/C72.6 k48.4 k
Required moment (base → top, linear)0 → 127 k-ft0 → 84.8 k-ft
Lcx = Lcy (K = 1.0)20.0 ft20.0 ft
By the Direct Analysis Method: Pr = 72.6 k (LRFD) / 48.4 k (ASD), Mr linear to 127 k-ft (LRFD) / 84.8 k-ft (ASD) at the top, K = 1.0 (Lcx = Lcy = 20.0 ft) for both moment-frame columns. Drift ratio 1.60/1.66 < 2.5, τb = 1.0 verified.

Engineering interpretation. Even though the sway multiplier (drift ratio) is a substantial 1.60, the actual moment increase is modest because the only source of sway in this pure-gravity combination is the small notional load — real wind/seismic combinations with larger lateral loads will show a much bigger absolute moment amplification for the same ratio.

⚠ Common mistakes

  • Forgetting to include the leaning-column gravity load in Yi when computing the notional load — it must include ALL gravity supported by that level, not just the load directly on frame columns.
  • Using K > 1.0 with DAM's reduced-stiffness analysis — double-counts the P-Δ effect.
  • Skipping the τb verification step after solving — it is an assumption made before running the analysis and must be checked against the actual Pr.

Worked Example 17.2 — Same Frame by the Effective Length Method

Adapted from AISC Design Examples V16.0, Example C.1B — repeats Example 17.1 with identical geometry and loads.

Given

  • Identical frame, loads, and W12×65 columns as Example 17.1; additionally Ix = 533 in⁴ for the W12×65.
  • Design by the Effective Length Method (Appendix 7) instead of DAM.

Step 1 — Check ELM Applicability (App. 7, §7.2.1)

App. 7 §7.2.1(a) gravity carried by nominally vertical elements; (b) Δ2nd/Δ1st ≤ 1.5 (full stiffness, LRFD or 1.6×ASD)
Condition (a) is satisfied by inspection. Condition (b) must be verified after running the analysis (Step 4) — proceed provisionally.

Step 2 — Loads (identical to Example 17.1)

wu = 2.40 klf, wa = 1.60 klf; Pu = 36.0 k, Pa = 38.4 k; PuL = 144 k, PaL = 154 k — same as Step 2 of Example 17.1.
Notional load per Spec. Eq. C2-1 (same formula, App. 7 §7.2.2 requires it only in gravity-only combinations since Δ2nd/Δ1st ≤ 1.7 is guaranteed whenever ELM applies): Ni = 0.576 k (LRFD), 0.614 k (ASD).

Step 3 — Analysis with FULL (Un-reduced) Stiffness

App. 7 §7.2.2Use 100% of nominal E, I, A — no 0.80 or τb reduction
First-order results: Δ1st = 0.145 in (LRFD), 0.155 in (ASD, before ÷1.6).
Second-order results: Δ2nd = 0.204 in (LRFD), 0.223 in (ASD, before ÷1.6).
Drift ratio = 0.204/0.145 = 1.41 (LRFD); 0.223/0.155 = 1.44 (ASD) — both ≤ 1.5. ELM is confirmed permitted.

Step 4 — Story-Based Kx per Commentary Appendix 7

Comm. App. 7, Eq. C-A-7-5Kx = √[ π²EI/L² / (RM Pstory) ] , using ΔH/(H·L) from the drift results
Pstory = w(120 ft): LRFD 2.40(120) = 288 k; ASD 1.60(120) = 192 k.
RM = 1 − 0.15(Pmf/Pstory) (Eq. C-A-7-6): Pmf = 144/288 → RM = 1 − 0.15(144/288) = 0.925 (both LRFD, ASD).
Euler load of one frame: Pe = π²EIx/L² = π²(29,000)(533)/(20.0×12)² = 2,650 k.
Solving Eq. C-A-7-5 with the analysis drift/shear results gives Kx = 3.45 (LRFD), 3.46 (ASD).

Step 5 — Effective Lengths for Member Design

§C3Lcx = Kx Lx; Lcy = Ky Ly (Ky = 1.0, braced out-of-plane)
Lcx = 3.46(20.0) = 69.2 ft.   Lcy = 1.00(20.0) = 20.0 ft (out-of-plane braced, Ky = 1.0 by definition when not part of the sway mechanism in that direction).
QuantityDAM (Ex. 17.1)ELM (Ex. 17.2)
Stiffness used in analysis0.80τbEI, 0.80EAFull EI, EA
Drift ratio Δ2nd/Δ1st1.60 (LRFD)1.41 (LRFD)
Kx for in-plane buckling1.00 (always)3.45 (LRFD) / 3.46 (ASD)
Lcx20.0 ft69.2 ft
By the Effective Length Method: ELM is permitted (Δ2nd/Δ1st = 1.41–1.44 ≤ 1.5). Using full stiffness and Commentary Eq. C-A-7-5, Kx = 3.46, giving Lcx = 69.2 ft versus Lcy = 20.0 ft — a dramatically larger effective length than the DAM answer's flat Lc = 20.0 ft, illustrating why K-factor bookkeeping in ELM is easy to get wrong.

Engineering interpretation. DAM and ELM must give member designs of comparable adequacy for the same frame — but they get there differently. DAM folds all instability effects into the analysis and member forces, then uses the simplest possible K = 1.0. ELM keeps a first-order-like member-force philosophy but pushes all of the instability penalty into a large K-factor. Both are valid; DAM is preferred because it needs no applicability check up front.

⚠ Common mistakes

  • Using reduced (0.80) stiffness in an ELM analysis — App. 7 requires full nominal stiffness.
  • Forgetting to multiply column loads by 1.6 in the ASD version of Eq. C-A-7-5 before computing RM and Kx.
  • Applying Ky = Kx when the frame is only a moment frame in one direction — out-of-plane K is usually 1.0, not the sidesway value.
  • Proceeding with ELM without confirming the 1.5 drift-ratio limit after the analysis — it is not optional.

Worked Example 17.3 — Nodal (Point) Column Bracing per Appendix 6

Given

  • W10×49 column, unbraced length 66.7 ft, braced at the two third-points by nodal braces (steel struts to an adjacent braced bay), giving Lb = 22.2 ft per segment.
  • Required axial compressive strength at the brace location: Pr = 500 k (LRFD).
  • φ = 0.75 for brace stiffness per Appendix 6 (Ω = 2.00 for ASD).
Example 17.3 — W10×49 column, two nodal braces at third points nodal brace 1 nodal brace 2 P_r = 500 k L_b = 66.7 ft/3 = 22.2 ft each Total column length = 66.7 ft, 3 equal segments
Figure 17.3a — W10×49 column braced at the third points by two nodal (point) braces; each controls an unbraced length L_b = 22.2 ft.

Step 1 — Required Brace Strength

App. 6, Eq. A-6-1Pbr = 0.01 Pr
Pbr = 0.01(500) = 5.00 k, applied perpendicular to the column axis at the brace point, in the direction that produces the worst-case buckled shape.

Step 2 — Required Brace Stiffness

App. 6, Eq. A-6-2βbr = 8 Pr / (φ Lb)
Lb = 22.2 ft = 266 in.
βbr = 8(500) / (0.75 × 266) = 4,000 / 199.5 = 20.1 k/in.

Step 3 — Select the Brace Member (Strength)

Ch. D, tension braceφPn = φFyAg ≥ Pbr
Try a single-angle strut, Ag = 1.19 in² (L3×3×¼), Fy = 50 ksi.
φPn = 0.90(50)(1.19) = 53.6 k ≫ Pbr = 5.00 k ✓ (strength governs by a wide margin — typical for stability bracing).

Step 4 — Check the Brace Member Axial Stiffness

Axial spring stiffnesskbrace = Ag E cos²θ / Lbrace
Diagonal strut length Lbrace = 14.0 ft = 168 in, angled at θ = 35° to the column axis (cos θ = 0.819).
kbrace = 1.19(29,000)(0.819)²/168 = 1.19(29,000)(0.671)/168 = 137 k/in ≥ βbr = 20.1 k/in ✓.
CheckRequiredProvidedResult
Brace strength5.00 k53.6 k (φPn, L3×3×¼)✓ OK
Brace stiffness20.1 k/in137 k/in✓ OK
A single L3×3×¼ diagonal strut satisfies both the Appendix 6 nodal-brace strength (5.00 k) and stiffness (20.1 k/in) requirements for the W10×49 column at each third-point, with wide margin on both checks.

Engineering interpretation. Stability braces are almost always governed by the stiffness requirement architecturally (the member needs to be stiff/short/steep enough), while the strength requirement is easily satisfied by nearly any reasonable section — but both must always be checked, because for very long or very shallow-angle braces the stiffness demand can dominate and force a larger member than strength alone would suggest.

⚠ Common mistakes

  • Using the column's overall unbraced length (66.7 ft) instead of the segment length Lb = 22.2 ft in the βbr formula.
  • Forgetting the cos²θ term when the brace is diagonal rather than perpendicular to the member — only the stiffness component perpendicular to the column counts.
  • Checking only member strength and skipping the connection stiffness/slip, which can dominate for bolted brace connections.

Worked Example 17.4 — B1/B2 Amplification of a Braced-Frame Beam-Column

Given

  • W12×65 column in a braced bay (X-braced), L = 14.0 ft, Ag = 19.1 in², Ix = 533 in⁴, ASTM A992.
  • First-order "no translation" (nt) results: Pnt = 210 k, Mnt = 95 k-ft (gravity load, no sway).
  • First-order "lateral translation" (lt) results: Plt = 18 k, Mlt = 22 k-ft (from notional/lateral load with sway permitted).
  • Story data: Pstory = 3,600 k, RM = 0.925 (approximate, per Comm. Eq. C-A-7-6), first-order interstory drift ΔH = 0.14 in under story shear H = 32 k, story height L = 168 in.
  • Cm = 0.85 (member subject to transverse loading between supports, conservative default).
Example 17.4 — Braced-frame beam-column, B1/B2 amplification X-brace (lateral system) P_nt = 210 k, P_lt = 18 k M_nt = 95 k-ft (gravity), M_lt = 22 k-ft (notional) W12×65 column, L = 14.0 ft, first-order results given
Figure 17.4a — Braced-frame column: first-order "no translation" (gravity) and "lateral translation" (notional-load) analyses are combined via B1 and B2.

Step 1 — B1 (Member P-δ Amplifier)

App. 8, Eq. A-8-1, A-8-4Pe1 = π²EI*/Lc1²;   B1 = Cm/(1 − αPr/Pe1) ≥ 1
Using DAM-consistent reduced stiffness EI* = 0.80(29,000)(533) = 12,370,000 kip-in² (τb = 1.0, verified below), K = 1.0, Lc1 = 168 in.
Pe1 = π²(12,370,000)/168² = 4,330 k.
Pr = Pnt + B2Plt — but B2 must be found first (Step 2) since Pr in B1 depends on it (iteration is normally trivial since B2 ≈ 1 for braced frames).
Using Pr ≈ Pnt + Plt = 210 + 18 = 228 k as a first pass: αPr/Pe1 = 1.0(228)/4,330 = 0.0527.
B1 = 0.85/(1 − 0.0527) = 0.85/0.947 = 0.898 → use B1 = 1.0 (App. 8 requires B1 ≥ 1).

Step 2 — B2 (Story P-Δ Amplifier)

App. 8, Eq. A-8-6B2 = 1 / [1 − α Pstory ΔH / (RM H L)] ≥ 1
α Pstory ΔH = 1.0(3,600)(0.14) = 504 kip-in.
RM H L = 0.925(32)(168) = 4,973 kip-in.
B2 = 1/(1 − 504/4,973) = 1/(1 − 0.1014) = 1/0.8986 = 1.113.

Step 3 — Amplified Required Strengths

App. 8, Eq. A-8-1, A-8-3Mr = B1Mnt + B2Mlt;   Pr = Pnt + B2Plt
Mr = 1.0(95) + 1.113(22) = 95 + 24.5 = 119.5 k-ft.
Pr = 210 + 1.113(18) = 210 + 20.0 = 230 k.
Re-check B1 with the refined Pr = 230 k: αPr/Pe1 = 230/4,330 = 0.0531 (negligible change) — B1 = 1.0 stands.
QuantityFirst-order (nt + lt)Amplified (B1/B2)
Axial Pr228 k230 k
Moment Mr117 k-ft119.5 k-ft
Design the W12×65 beam-column for Pr = 230 k and Mr = 119.5 k-ft using the H1 interaction equation with K = 1.0 (DAM-consistent), Lcx = Lcy = 14.0 ft. B2 = 1.11 shows the braced frame is only mildly sway-sensitive; B1 = 1.0 confirms the member's own curvature amplification is negligible at this load level.

Engineering interpretation. In a well-braced frame, B2 stays close to 1.0 (here 1.11) — most of the "second order" effect comes from the story sway captured by B2, while B1 commonly floors out at 1.0 for members with modest axial load relative to their Euler buckling load. This is the everyday hand-calculation alternative to running a full nonlinear second-order analysis when your software only performs first-order runs with and without sway.

⚠ Common mistakes

  • Forgetting Pstory and RM must include all gravity load on the story (including leaning columns), not just the load on the member being checked.
  • Applying B1 to the axial force — B1 amplifies moment only; B2 amplifies both Plt and Mlt.
  • Using B1 or B2 < 1.0 — App. 8 requires both floored at 1.0.
  • Mixing full-stiffness Pe1 with a DAM (reduced-stiffness) member design, or vice versa — be consistent about which analysis method's stiffness convention is in use.

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
Second-order/first-order drift ratio = 1.7.
AISC Reference
AISC §C1
Step-by-step solution
  1. Pick
    DAM mandatory (always allowed) Effective length method only when Δ2/Δ1 ≤ 1.5.
Answer Use the Direct Analysis Method.
Choose analysis method
Problem statement image
Loads (ASCE 7)LRFD CombosRequired RuTrial SectionLimit StatesφRn ≥ Ru?Serviceability
DIMDimensions from the problem statement
ratio = 1.7.
LRFD workflow (ASCE 7 → AISC 360)
  • Loads (D, L, L_r, S, W, E) → LRFD combos → P_u, M_u, V_u
  • Nominal capacity R_n from AISC 360-22 → apply φ
  • Require φR_n ≥ R_u for every controlling limit state

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

These questions reference AISC Steel Construction Manual (16th ed.) — sections, equations, and tables are cited explicitly. Use a calculator. Each question offers a clue you may reveal before answering. Submissions are recorded to your account once signed in.

C17-01AISC 360-22 §C1
1. AISC 360-22 default stability method is:
Braced frame · X-brace bay V
C17-02AISC 360-22 Eq. C2-1
2. Notional lateral load in DAM (all combos):
C17-03AISC 360-22 §C2.2b(d)
3. When can notional load be applied ONLY in gravity-only combos?
C17-04AISC 360-22 Eq. C2-2a
4. τb reduction (Eq. C2-2a/b) when αPr/Py ≤ 0.5:
C17-05AISC 360-22 §C1
5. Under DAM, K for member design =
C17-06AISC 360-22 App. 7
6. Effective Length Method (App. 7) allowed when:
C17-07AISC 360-22 §C2
7. Second-order elastic analysis captures:
C17-08AISC 360-22 App. 8
8. Alternative approximate B1–B2 method uses:
C17-09AISC 360-22 §C
9. Sensitivity check α·Δ2nd/Δ1st = B2. If B2 > 1.7:
Braced frame · X-brace bay V
C17-10AISC 360-22 §C2.2b Comm.
10. Global initial out-of-plumbness assumed:
C17-11AISC 360-22 §C2.3
11. Direct Analysis Method requires stiffness reduction on:
C17-12AISC 360-22 §C
12. In DAM, LOAD combinations use:
C17-13AISC 360-22 §C
13. For simple gravity braced frame, DAM simplifies to:

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§11Section 11

Chapter summary

Formula sheet
  • Design strength
    φ Rn ≥ Ru
    AISC 360-22 B3.1
Engineering checklist
  • Module 17: Stability, Bracing, and Second-Order Effects
  • 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

FE exam preparation
Concept review
Concept summary coming soon.
Calculator tips

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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.