19

Steel Framing Systems

Why Structural Steel?

Braced frames, moment frames, dual systems, gravity systems.

⏱ 75 min★ Core✎ 3 objectives
§01Section 01

Engineering story

Engineering story
Chapter 19 · Steel Framing Systems

Braced frames, moment frames, dual systems, gravity systems.

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
  • Compare braced vs moment frames
  • Lay out a gravity system
  • Select a lateral system
§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

The choice of lateral system — braced frame, moment frame, dual, or shear-wall-with-steel-gravity — sets the cost, the erection sequence, and the seismic performance of the whole building.

Learning Objectives

  • Compare concentrically braced frames (CBF), eccentrically braced frames (EBF), and buckling-restrained braced frames (BRBF).
  • Compare special (SMF), intermediate (IMF), and ordinary (OMF) moment frames.
  • Understand response-modification factor R and how it drives design forces.
  • Layout a floor framing plan with beams, girders, columns, and lateral system.
  • Estimate steel tonnage per square foot for common systems.

Where This Chapter Is Used

This chapter directly precedes the capstone (Chapter 20), where students choose and design a full lateral system.

Lecture Notes

Chapter 19 — Steel Framing Systems (Gravity, Lateral, and Load Path)

Chapter focus. Every member you've designed so far (beams, columns, connections) is a piece of a larger puzzle: a coherent framing system that gets every pound of dead, live, wind, and seismic load safely from where it's applied to the ground. This chapter is about the "big picture" decisions engineers make before a single member is sized: how the gravity framing hierarchy works, which lateral system to choose, how forces actually travel through a diaphragm, and why bay size, expansion joints, and erection sequence matter as much as any code equation.

1. Gravity Framing Systems

Gravity framing carries dead and live load down through a hierarchy of elements, each spanning to the next, until the load reaches a column and finally a foundation. The most common arrangement is one-way framing: a metal deck spans a short distance between closely spaced filler beams, the filler beams span onto girders at the column lines, and the girders frame directly into columns. Two-way framing (a slab or deck spanning in both directions onto a denser grid, or waffle/flat-plate concrete systems) is less common in steel because a corrugated deck has an inherently strong and weak direction, but two-way steel framing appears in heavy industrial or transfer-level construction where load must be shared among many supports.

  • Metal deck spans perpendicular to the closely spaced filler beams, typically 3 in deep composite deck spanning 8–12 ft unshored, or non-composite deck for roofs spanning less.
  • Filler (infill) beams spaced 8–10 ft o.c., spanning 30–45 ft between girders; usually composite with the slab via shear studs.
  • Girders at column lines carry the concentrated reactions of the filler beams; often composite, sometimes deeper to control deflection over longer spans.
  • Open-web steel joists replace filler beams on long-span (≥ 40 ft), lighter live-load floors/roofs — cheaper per square foot but limited to lighter concentrated loads and larger deflections.
Gravity framing plan — deck / beam / girder / column hierarchy girder (grid line) filler beams @ 10 ft o.c. column line metal deck spans perpendicular to filler beams; beams span onto girders; girders onto columns
Figure 19.1 — Gravity framing hierarchy: deck spans onto filler beams, filler beams onto girders, girders onto columns.

2. Composite Floors and Deck Spans

In a composite floor, shear studs welded through the metal deck to the beam top flange bond the concrete slab to the steel beam so they act as one T-shaped section, dramatically increasing stiffness and strength for the same beam weight. Deck profiles are typically 1.5 in, 2 in, or 3 in deep; a 3 in deep, 20-gauge composite deck can span roughly 10–12 ft unshored with normal-weight topping. Total slab thickness (deck depth + topping) is usually 4.5–6.5 in, chosen to satisfy both strength and the 2-hour fire rating typical of office construction.

3. Lateral Framing Systems

A lateral system resists wind and seismic loads and controls drift. The major steel families are:

  • Concentric braced frames (CBF) — diagonal members whose centerlines intersect at a common work point, so the system behaves as an efficient, stiff truss. Configurations include X-bracing (two diagonals per bay, very stiff, blocks door/window openings), chevron (V or inverted-V) bracing (diagonals meet at mid-span of the beam, allowing a door on either side but inducing an unbalanced vertical force on the beam if a brace buckles), and single-diagonal bracing (one diagonal per bay, most architecturally open but must resist tension and compression alternately, requiring symmetric detailing across the building).
  • Eccentrically braced frames (EBF) — the diagonal intentionally misses the beam-column joint, leaving a short "link" beam segment that yields in shear and provides ductility (similar to a fuse) while keeping the brace itself close to elastic. Combines the stiffness of a CBF with much of the ductility of a moment frame.
  • Moment frames (MF) — rigid beam-column connections resist lateral load through frame bending rather than diagonal bracing; the most architecturally open system (no diagonals blocking space) but also the most flexible, often governed by drift rather than strength.
  • Dual systems — a moment frame combined with a braced frame or shear wall; ASCE 7 requires the moment frame alone to resist at least 25% of the design story shear, providing redundancy if the stiffer bracing yields or degrades.
  • Shear-wall interaction — in composite steel/concrete buildings, concrete or CMU shear walls (often at stair/elevator cores) can act as the primary lateral system, with the steel gravity frame designed only for gravity loads and "riding along" with the wall's lateral drift; compatibility of steel beam connections to the wall's drift must be checked.

Additional Design Aids & Stratified Equations

Steel lateral systems — visual comparison SCBF (X-brace) Chevron EBF (with link) SMF (moment)
Four common steel lateral systems compared: SCBF (X), chevron, EBF (with link), SMF.
Seismic base shearV = C_s · W, C_s = min( S_DS/(R/I_e), S_D1 / (T·R/I_e) )
SCBF capacity designP_expected = R_y · F_y · A_g; R_y = 1.1 for A992

4. Load Path: Cladding → Diaphragm → Lateral System → Foundation

Every lateral load must have a complete, continuous load path. Wind pressure first strikes the exterior cladding, which spans (typically vertically) to floor-level girts or the slab edge; that force enters the floor diaphragm (the composite slab acting as a deep horizontal beam/plate); the diaphragm distributes the force to the vertical lateral elements (braced frames, moment frames, or shear walls) in proportion to their relative stiffness; those elements carry the load down as shear and overturning moment to the foundation, which must resist sliding, overturning, and — for braced frames especially — net uplift at the base.

Lateral load path — cladding → diaphragm → braced frame → foundation cladding rigid floor diaphragm braced frame foundation V collected by diaphragm, delivered to braced bay, resolved to base shear + overturning
Figure 19.2 — Complete lateral load path from cladding pressure through the diaphragm to the braced frame and down to the foundation.

5. Diaphragm Behavior — Rigid vs Flexible

A diaphragm is classified as rigid if its in-plane deflection is small compared to the story drift of the vertical elements it connects (ASCE 7 §12.3.1) — typically a reinforced concrete slab or a well-connected composite steel deck slab. A rigid diaphragm distributes story shear to the vertical lateral elements in proportion to their relative rigidity (stiffness), and also transfers torsional shear if the center of rigidity and center of mass don't coincide. A diaphragm is flexible (bare steel deck without adequate topping, or a wood diaphragm) if its own deflection is large relative to the story drift — a flexible diaphragm instead distributes shear to vertical elements by tributary area, like a simple beam spanning between the lines, ignoring relative stiffness and torsion.

6. Transfer Forces, Collectors, and Drag Struts

Where the vertical lateral elements are not uniformly distributed under the diaphragm (e.g., a braced bay is offset from a re-entrant corner, or a lateral line stops at one level and must "transfer" to a different line below), the diaphragm must collect the distributed shear along a wall or frame line and deliver it as a concentrated axial force to the vertical element. A collector (also called a drag strut) is the beam or added steel member designed to carry this axial "dragged" force in tension and compression, in addition to its normal gravity bending. Collector forces are frequently amplified by the seismic overstrength factor Ω0 (ASCE 7 §12.10.2.1) because collector failure is a non-ductile, brittle failure mode that must be protected against yielding of the ductile lateral system elsewhere.

Collector / drag strut at a re-entrant (L-shaped) diaphragm corner collector (drag strut) re-entrant corner to braced bay diaphragm shear collected along wing
Figure 19.3 — At a re-entrant corner the diaphragm's distributed shear must be collected along the wing and dragged, as a concentrated axial force, into the braced bay.

7. Bay-Size Economics and Typical Spans

Bay size / spanTypical systemNotes
20–30 ftFiller beam + composite girderMost economical for standard office loading; deck spans directly without intermediate beams if ≤ 12 ft.
30–45 ftComposite beam + girder, deeper sectionsCommon parking-structure and retail bay; deflection/vibration often governs over strength.
40–60 ftOpen-web steel joist + joist girderWarehouse/light-industrial; low cost per lb but larger deflections, less concentrated-load capacity.
60–150+ ftLong-span truss, staggered truss, transfer girderColumn-free space (arenas, atria); heavier connections, erection sequencing critical.

The classic "sweet spot" for typical office/commercial steel is roughly a 30 ft × 40–45 ft bay: wide enough to minimize the number of columns (and their foundations), narrow enough that beam and deck sizes stay light and deflection-controlled rather than strength-controlled.

8. Expansion Joints

Buildings longer than roughly 200–300 ft (IBC/AISC rule-of-thumb, varies by climate and cladding type) are typically split by a full structural expansion joint — two independent structural frames placed close together with a gap — to accommodate thermal expansion/contraction and to limit differential movement demands on cladding, MEP, and finishes. Expansion joints also often mark a natural break for lateral system redundancy: each independent structure needs its own complete, self-sufficient lateral system, since forces cannot reliably transfer across the joint.

9. System Selection Matrix

PriorityFavorsDisfavors
Open floor plan / large windowsMoment frame (SMF), EBF with link away from openingsX-braced CBF blocking bays
Minimum steel tonnage / costCBF (X or chevron) — most efficient axial-only load pathSMF (heavier for equivalent drift control)
Drift control on a tall buildingCBF, BRBF, dual system, shear wall coreOMF (flexible, height-limited)
High seismic ductility demandSCBF, BRBF, EBF, SMF (all AISC 341 detailed)OCBF, OMF above code height limits

10. Seismic Force-Resisting System R Factors (ASCE 7 Table 12.2-1)

SystemRΩ0CdDetailing reference
Steel Special Moment Frame (SMF)835.5AISC 341 Ch. E, AISC 358 pre-qualified connections
Steel Ordinary Moment Frame (OMF)3.533AISC 341 Ch. E; height-limited in SDC D–F
Special Concentrically Braced Frame (SCBF)625AISC 341 Ch. F2; capacity-designed connections/columns
Ordinary Concentrically Braced Frame (OCBF)3.2523.25AISC 341 Ch. F1; low-seismic/low-rise only
Eccentrically Braced Frame (EBF)824AISC 341 Ch. F3; link beam ductile fuse
Buckling-Restrained Braced Frame (BRBF)825AISC 341 Ch. F4; symmetric hysteresis
Steel systems not specifically detailed for seismic333AISC 360 only; SDC B/low-risk use

Higher R means lower design seismic force but demands more ductile detailing (capacity design of connections and adjoining members per AISC 341) — R is a "discount" for guaranteed ductile behavior, not a free reduction in actual earthquake demand.

11. Constructability and Erection Sequence

A framing system's cost and schedule depend heavily on how it goes together in the field, not just its steel weight. Considerations include: connection standardization (repeating a handful of connection types across the whole building speeds shop fabrication and field erection); temporary bracing/stability of columns before the permanent bracing or diaphragm is installed (governed by AISC's Code of Standard Practice and OSHA erection requirements); crane reach and lift weight limiting maximum piece size, especially for long-span trusses; and pour sequencing for composite floors so that shoring/re-shoring, if used, does not overload lower floors.

Design Workflow

  1. Establish bay geometry and gravity framing hierarchy (deck → beam → girder → column) based on architectural grid and occupancy loads.
  2. Determine seismic design category and wind exposure; select candidate lateral systems from the selection matrix and R-factor table.
  3. Locate lateral elements to minimize torsional eccentricity and provide a short, direct load path from every diaphragm area.
  4. Classify each diaphragm rigid or flexible; distribute story shear accordingly (relative rigidity vs tributary area).
  5. Identify collector/drag-strut lines at re-entrant corners, transfers, and discontinuities; size for Ω0-amplified force where required.
  6. Check drift against ASCE 7 Table 12.12-1 (seismic) or H/400–H/500 (wind); iterate system choice if drift governs.
  7. Coordinate erection sequence, temporary bracing, and expansion-joint locations with the construction manager before finalizing member sizes.

⚠ Common mistakes

  • Assuming a bare steel deck diaphragm is rigid without checking its actual flexibility relative to the vertical elements.
  • Forgetting the Ω0 overstrength amplification on collector/drag-strut design forces in seismic systems.
  • Using OCBF or OMF above their SDC height limits (ASCE 7 Table 12.2-1) — often prohibited entirely in SDC D and above.
  • Ignoring the 25% independent-strength rule for the moment-frame portion of a dual system.
  • Sizing a long building without an expansion joint, leading to unanticipated thermal stress in cladding and diaphragm connections.
  • Choosing bay sizes purely by architectural preference without checking that deck span and beam depth stay in the deflection/vibration "sweet spot."
Exam / FE tips. Know the rigid-vs-flexible diaphragm distinction and how each distributes shear (stiffness vs tributary area) — it's tested constantly. Memorize the R-factor ordering: OMF (3.5) < OCBF (3.25, comparable) < SCBF (6) < SMF = EBF = BRBF (8). Remember collectors need Ω0, and dual systems need the moment frame to carry ≥25% of story shear alone.
Braced (CBF)Moment (SMF)Dual system
Lateral systems: concentrically braced frames, moment frames, and dual systems (AISC 341).

Formula Sheet

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

Worked Example

Worked Example 19.1 — Lateral System Selection for Cardinal Square

Given

  • 6-story office building, plan 120 ft × 90 ft, hsx = 13 ft, roof 78 ft.
  • Site: Cincinnati, OH. SDS = 0.18, SD1 = 0.08 → SDC B.
  • Basic wind V = 115 mph (ASCE 7); Exposure B.
  • Base shear (approx): Vwind = 380 k, Vseismic = 210 k.
Cardinal Square — chevron-braced-frame plan (2 bays per direction) V_wind 120 ft (6 × 20) 90 ft (5 × 18) ■ N-S chevron CBF (2 bays) ■ E-W chevron CBF (2 bays)
Figure 19.1a — Plan showing chevron braced-bay locations (2 per direction)

Step 1 — SDC Constraints

ASCE 7 Table 12.2-1Vdesign = V/R (R = response modification factor);  Governing lateral = max(Vwind, Vseismic/R)
SDC B → most systems permitted with no height limit. AISC 341 seismic detailing not required for R ≤ 3 systems (R = 3 "steel systems not specifically detailed for seismic").

Step 2 — Candidate Systems

SystemREst. VdesignPros / Cons
R = 3 (steel not detailed)3.0Vwind governs = 380 kSimplest; wind governs anyway.
OCBF3.25~194 k seismic; wind still governsOCBF permitted, but SDC B doesn't reward it.
OMF3.5180 k seismicOpen plan; larger members; costly.

Step 3 — Drift Check (Wind, H/400)

ServiceabilityΔstory ≤ H/400 (wind, ASCE 7 CC.1.2);  Δx ≤ Cd·Δxe/Ie ≤ 0.020 hsx (seismic)
Allowable Δ = 13·12/400 = 0.39 in/story.
R = 3 chevron CBF at each end + interior gravity columns → estimated Δstory = 0.25 in ✓.
OMF alternative Δstory ≈ 0.55 in → NG without column upsizing.

Step 4 — Decision

Adopt a chevron braced frame (R = 3) along each principal axis. Simpler than OCBF, meets wind drift, and passes seismic by inspection since SDC B. Deliver as 4 braced bays total (2 per direction) with HSS diagonals.
Reflection: Had the site been in high-seismic California (SDC D+), an SCBF or BRBF with capacity design under AISC 341 would be required and OMF would be prohibited above 35 ft.

Worked Example 19.2 — Gravity Column Load Takedown with Live-Load Reduction

Problem statement. Design the ground-floor segment of an interior gravity column in Cardinal Square. Tributary area per floor is 30 ft × 30 ft = 900 ft². Floor dead load D = 75 psf, floor live load Lo = 50 psf (office). Roof dead Dr = 40 psf, roof live Lr,o = 20 psf. The column supports the roof plus five floors. Story height 13 ft, pinned top and bottom, braced in both directions by the CBF system. Use ASTM A992 (Fy = 50 ksi).

Interior gravity column takedown — 900 ft² tributary per level, K_LL = 4 Roof: D = 36.0 k, L_r = 10.8 k (R1 = 0.6) L6: D = 67.5 k, L = 22.5 k (red. 0.500) L5: D = 67.5 k, L = 19.2 k (red. 0.427) L4: D = 67.5 k, L = 18.0 k (floor at 0.4L_o) L3: D = 67.5 k, L = 18.0 k L2: D = 67.5 k, L = 18.0 k A_T = 9001,8002,700 3,6004,500A_T = 4,500 ft² P_u = 598 kips at the base
Figure 19.2a — Accumulating tributary area drives the live-load reduction factor down level by level until it hits the 0.4Lo floor
Step 1 — Live-load reduction (ASCE/SEI 7 §4.7).
ASCE 7 Eq. 4.7-1L = Lo·(0.25 + 15/√(KLL·AT)) ≥ 0.40Lo (members supporting two or more floors)
KLL = 4 for an interior column (ASCE 7 Table 4.7-1). Reduction is permitted because KLLAT > 400 ft².
Floors supportedAT (ft²)KLLAT0.25 + 15/√(KLLAT)L (psf)
19003,6000.50025.0
21,8007,2000.42721.3
32,70010,8000.394 → 0.400 floor20.0
43,60014,4000.375 → 0.400 floor20.0
5 (base)4,50018,0000.362 → 0.400 floor20.0
Step 2 — Roof live-load reduction (ASCE/SEI 7 §4.8).
ASCE 7 Eq. 4.8-1Lr = Lr,o·R1·R2, 12 ≤ Lr ≤ 20 psf
AT = 900 ft² > 600 ft² → R1 = 0.6; flat roof → R2 = 1.0
Lr = 20(0.6)(1.0) = 12.0 psf (at the 12 psf floor)
Step 3 — Accumulate the loads at the base. D = roof + 5 floors = 40(900)/1000 + 5[75(900)/1000] = 36.0 + 337.5 = 373.5 kips
L = 20.0(900)(5)/1000 = 90.0 kips (using the fully reduced 20 psf on all 5 floors)
Lr = 12.0(900)/1000 = 10.8 kips
Step 4 — Load combinations (ASCE/SEI 7 §2.3).
CombinationArithmeticPu (kips)
1.4D1.4(373.5)523
1.2D + 1.6L + 0.5Lr1.2(373.5) + 1.6(90.0) + 0.5(10.8)598 ← governs
1.2D + 1.6Lr + 1.0L448.2 + 17.3 + 90.0556
ASD: D + L = 373.5 + 90.0 = 464 kips
Step 5 — Select the column (§E3). Lc = KL = 1.0(13 ft) = 156 in. in both directions (pinned, fully braced by the CBF).
Try W14×68: Ag = 20.0 in², ry = 2.46 in.
Lc/ry = 156/2.46 = 63.4; Fe = π²E/(Lc/r)² = π²(29,000)/(63.4)² = 71.2 ksi
Fy/Fe = 50/71.2 = 0.702 ≤ 2.25 → inelastic buckling
Spec. Eq. E3-2Fcr = [0.658Fy/Fe]Fy = (0.658)0.702(50) = 0.746(50) = 37.3 ksi
Spec. Eq. E3-1φcPn = 0.90(37.3)(20.0) = 672 kips ≥ 598 kips o.k. (D/C = 0.89)
Check W14×61 (Ag = 17.9, ry = 2.45): φcPn ≈ 601 kips — D/C = 0.995, too tight for a column that may pick up future load.
Result: Use W14×68 for the ground-floor segment. φcPn = 672 kips ≥ Pu = 598 kips.

Engineering interpretation: live-load reduction cut the base live load from 50 psf to 20 psf — a 60% reduction that removed 135 kips from the column. That is not a loophole; it is a statistical statement that 4,500 ft² of office floor is never simultaneously loaded to its design intensity. Note the 0.40Lo floor kicked in at three floors and never released, so levels 2–4 all use the same 20 psf. In practice you standardize column sizes over two-story lifts (splices every other floor) to keep the erection simple, so this W14×68 would run from the base to Level 3.

Common mistakes:
  • Using KLL = 2 (an edge/interior beam value) for a column. Interior columns use KLL = 4; exterior columns without cantilever slabs use 4; corner columns use 4.
  • Applying the 0.50Lo floor. That floor applies to members supporting one floor; multi-floor members get 0.40Lo.
  • Reducing roof live load with the §4.7 formula. Roofs use §4.8 with R1 and R2.
  • Reducing live load in an assembly area, a garage, or where Lo > 100 psf — §4.7.3–4.7.6 prohibit or restrict it.
  • Forgetting to check the 1.4D-only combination on a heavily dead-loaded column.

Worked Example 19.3 — Rigid-Diaphragm Distribution with Accidental Torsion

Problem statement. The roof-level lateral force in the E-W direction is Fx = 95 kips. The 120 ft × 90 ft concrete-on-metal-deck diaphragm is rigid. Two identical E-W chevron frames sit on grid lines at y = 0 and y = 90 ft; two identical N-S frames sit at x = 0 and x = 120 ft. All four frames have equal rigidity R = 1.0 (relative). Determine the design shear in one E-W frame including ASCE/SEI 7 accidental torsion.

Rigid diaphragm: direct shear + torsional shear from the 5% accidental eccentricity E-W frame 1 (d = 45 ft) E-W frame 2 (d = 45 ft) N-S frame (d = 60 ft) N-S frame (d = 60 ft) C.R. = C.M. displaced C.M. (e_a = 6 ft) F_x = 95 k M_t = 570 k-ft 120 ft e_a = 0.05(120 ft) = 6.0 ft · J = Σ R·d² = 2(45)² + 2(60)² = 11,250 ft² V_frame = 47.5 + 2.28 = 49.8 kips
Figure 19.3a — Direct shear splits equally; torsional shear is distributed in proportion to R·d/J and picked up by all four frames
Step 1 — Direct shear.
Rigid diaphragmVdirect,i = Fx · Ri / ΣR
Vdirect = 95(1.0)/2.0 = 47.5 kips per E-W frame
Step 2 — Accidental eccentricity (ASCE/SEI 7 §12.8.4.2). ea = 0.05 × (plan dimension perpendicular to the applied force) = 0.05(120) = 6.0 ft
Mt = Fx · ea = 95(6.0) = 570 kip-ft
Step 3 — Torsional rigidity of the frame layout.
Polar rigidityJ = Σ Ri·di² (both directions participate)
E-W frames at d = ±45 ft: 2(1.0)(45)² = 4,050
N-S frames at d = ±60 ft: 2(1.0)(60)² = 7,200
J = 4,050 + 7,200 = 11,250 ft²
Step 4 — Torsional shear in the critical E-W frame.
Torsional distributionVt,i = Mt · Ri · di / J
Vt = 570(1.0)(45)/11,250 = 2.28 kips
Step 5 — Total design shear.
Vframe = Vdirect + Vt = 47.5 + 2.28 = 49.8 kips
Check torsional irregularity (ASCE 7 Table 12.3-1): with symmetric frames and only accidental eccentricity, δmax/δavg ≈ 1.05 < 1.2 → no torsional irregularity, so the Ax amplification of §12.8.4.3 does not apply.
ContributionValue% of total
Direct shear47.5 k95.4%
Accidental torsion2.28 k4.6%
Design shear per frame49.8 k100%
Result: Design each E-W chevron frame for V = 49.8 kips at the roof level, a 4.8% increase over the direct-shear-only value.

Engineering interpretation: the torsional penalty is small here because the frames are pushed to the perimeter. Move the braced bays inboard — say to d = ±20 ft — and J collapses to 800 + 7,200 = 8,000, the torsional shear jumps to 570(20)/8,000 = 1.43 k but the N-S frames now soak up an outsized share and the building becomes torsionally soft. The design lesson is geometric, not numerical: put lateral stiffness as far from the centre of rigidity as the architecture allows. A flexible diaphragm would change everything — force would distribute by tributary area, not rigidity, and torsion would not be transferred at all.

Common mistakes:
  • Omitting the perpendicular (N-S) frames from J. They resist torsion even though they carry no direct E-W shear.
  • Using the plan dimension parallel to the force for ea. It is the perpendicular dimension.
  • Subtracting torsional shear on the "relieved" side. ASCE 7 requires ea applied in the direction that produces the worst effect on each frame.
  • Distributing by tributary area (flexible-diaphragm logic) when the diaphragm is a concrete slab.
  • Forgetting the Ax torsional amplification when a torsional irregularity does exist (δmax/δavg > 1.2 in SDC C and above).

Worked Example 19.4 — Chevron Brace Design

Problem statement. Design the diagonal braces for a chevron (inverted-V) braced bay carrying a design story shear V = 100 kips. Bay width 30 ft, story height 13 ft, so each half-bay diagonal spans from a base corner to the beam midpoint. Braces are ASTM A500 Gr. C round/square HSS (Fy = 50 ksi for round, 50 ksi for square Gr. C — use Fy = 46 ksi for A500 Gr. B square as noted). Pinned ends, K = 1.0.

Chevron braced bay — 30 ft bay × 13 ft story, V = 100 kips C = 66.1 k T = 66.1 k V = 100 k θ = 40.9° 15 ft 13 ft Brace length L = √(15² + 13²) = 19.85 ft = 238 in. · P_brace = V / (2 cos θ) = 100 / (2 × 0.756) = 66.1 kips
Figure 19.4a — Chevron geometry: half the story shear goes to each diagonal, amplified by 1/cos θ
Step 1 — Brace geometry and force. Horizontal projection = 30/2 = 15 ft; vertical = 13 ft.
θ = tan−1(13/15) = 40.9°; cos θ = 0.756
Lbrace = √(15² + 13²) = 19.85 ft = 238 in.
Statics of the chevronPbrace = V / (2 cos θ) = 100 / (2 × 0.756) = 66.1 kips
One brace is in compression (66.1 k), the other in tension (66.1 k). Compression governs the section.
Step 2 — First trial: HSS5×5×¼ (A500 Gr. B, Fy = 46 ksi). Ag = 4.30 in², r = 1.90 in.
Lc/r = 1.0(238)/1.90 = 125.3; Fe = π²(29,000)/(125.3)² = 18.2 ksi
Fy/Fe = 46/18.2 = 2.53 > 2.25 → elastic buckling
Spec. Eq. E3-3Fcr = 0.877Fe = 0.877(18.2) = 16.0 ksi
φcPn = 0.90(16.0)(4.30) = 61.9 kips < 66.1 kips — NO GOOD
Step 3 — Second trial: HSS5×5×⁵⁄₁₆. Ag = 5.26 in², r = 1.88 in.
Lc/r = 238/1.88 = 126.6; Fe = π²(29,000)/(126.6)² = 17.9 ksi
Fy/Fe = 46/17.9 = 2.57 > 2.25 → Fcr = 0.877(17.9) = 15.7 ksi
Spec. Eq. E3-1φcPn = 0.90(15.7)(5.26) = 74.3 kips ≥ 66.1 kips o.k. (D/C = 0.89)
Wall slenderness: b/t = (5 − 3(0.291))/0.291 = 14.2 ≤ λr = 1.40√(E/Fy) = 35.2 → nonslender ✓
Step 4 — Tension check on the opposite brace (§D2). Yielding: φtPn = 0.90FyAg = 0.90(46)(5.26) = 218 kips ≥ 66.1 kips o.k.
Rupture at the slotted gusset connection: with U = 0.85 and one slot each side removing 2(0.291)(0.5) = 0.29 in²,
Ae = U(Ag − Aslot) = 0.85(5.26 − 0.29) = 4.22 in²; φtPn = 0.75(62)(4.22) = 196 kips o.k.
Step 5 — The chevron beam unbalanced-load check. This is the check that makes a chevron different from a diagonal brace. After the compression brace buckles, it sheds load while the tension brace keeps pulling — leaving a net unbalanced vertical force at the beam midpoint.
Punbal = (Texpected − Cpost-buckling)·sin θ ≈ (RyFyAg − 0.3Pn)·sin θ (AISC 341 §F2.3 for SCBF)
For this R = 3 system, AISC 341 does not apply, but good practice still checks the beam for a plausible unbalance. Using T = φtPn-level force and C = 0.3Pn:
Punbal ≈ (66.1 − 0.3(82.6))(sin 40.9°) = (66.1 − 24.8)(0.655) = 27.1 kips applied downward at midspan.
The chevron beam must carry this plus its gravity load. Size it accordingly — this typically doubles the beam over a comparable gravity-only member.
Result: Use HSS5×5×⁵⁄₁₆ braces (φcPn = 74.3 kips ≥ 66.1 kips) and design the chevron beam for gravity plus a 27-kip unbalanced point load at midspan.

Engineering interpretation: notice the brace is in the elastic-buckling range (Lc/r = 127) — it is a slender strut, and the design is driven entirely by stiffness, not strength. The tension capacity (218 k) is more than triple the compression capacity (74 k). That asymmetry is the whole story of chevron bracing and the reason the unbalanced-load check exists: the two braces of a chevron are not equal partners. If the unbalanced beam becomes uneconomical, switch to a two-story X or a single diagonal, which self-balance.

Common mistakes:
  • Using Pbrace = V/2 and forgetting the 1/cos θ amplification.
  • Using the horizontal projection (15 ft) as the brace unbraced length instead of the true diagonal (19.85 ft).
  • Sizing the brace for tension and assuming compression follows. Compression governs by 3×.
  • Using Eq. E3-2 (inelastic) when Fy/Fe > 2.25. Slender braces need Eq. E3-3.
  • Using Fy = 50 ksi for A500 Gr. B square HSS (it is 46 ksi). Gr. C is 50 ksi.
  • Skipping the chevron-beam unbalanced-load check — the classic cause of a floor sagging at the brace point after a seismic event.

Worked Example 19.5 — Collector (Drag Strut) Force with the Overstrength Factor

Problem statement. The roof diaphragm force in the E-W direction is Fpx = 95 kips, distributed uniformly along the 90 ft depth of the building. It must be delivered into a 30 ft wide braced bay located at the middle of a 90 ft long collector line. Determine the maximum collector axial force, apply the seismic overstrength factor Ω0, and verify the existing W21×44 floor beam (Ag = 13.0 in², A992) can serve as the collector.

Collector: the diaphragm shear that lands outside the braced bay must be dragged into it uniform diaphragm shear q = 95 k / 90 ft = 1.06 k/ft braced bay, 30 ft collector, 30 ft collector, 30 ft Axial force diagram in the collector line 31.8 k (unamplified) 31.8 k With Ω₀ = 2.0: P_u = 63.6 kips Peak collector force occurs at the EDGE of the braced bay, not at the building end and not at midspan.
Figure 19.5a — Diaphragm shear accumulates along the collector and peaks where the collector meets the braced bay
Step 1 — Uniform diaphragm shear along the collector line.
q = Fpx / Lbuilding = 95 kips / 90 ft = 1.056 kip/ft
Step 2 — Accumulate to the edge of the braced bay. The braced bay is 30 ft wide and centred, so 30 ft of collector runs on each side.
Pcollector = q · Lcollector = 1.056(30) = 31.7 kips
This force is compression on one side of the bay and tension on the other, and it reverses when the seismic load reverses. Design the collector and its connections for both.
Step 3 — Apply the overstrength factor (ASCE/SEI 7 §12.10.2.1). Collectors and their connections in SDC C and above must be designed for the load effect including Ω0, because a collector failure would strand the diaphragm from the lateral system.
ASCE 7 Eq. 12.4-5Emh = Ω0 · QE, Ω0 = 2.0 for braced-frame systems (Table 12.2-1)
Pu,collector = 2.0(31.7) = 63.4 kips
Step 4 — Verify the W21×44 as a collector in tension (§D2). Ag = 13.0 in², A992 (Fy = 50 ksi, Fu = 65 ksi)
Spec. Eq. D2-1φtPn = 0.90FyAg = 0.90(50)(13.0) = 585 kips ≫ 63.4 kips o.k. (D/C = 0.11)
Step 5 — Verify in compression, braced by the slab (§E3). The composite slab braces the top flange continuously; take Lc = the beam spacing, 10 ft = 120 in., about the weak axis, ry = 1.26 in.
Lc/ry = 120/1.26 = 95.2; Fe = π²(29,000)/(95.2)² = 31.6 ksi
Fy/Fe = 50/31.6 = 1.58 ≤ 2.25 → Fcr = (0.658)1.58(50) = 0.520(50) = 26.0 ksi
φcPn = 0.90(26.0)(13.0) = 304 kips ≫ 63.4 kips o.k.
Step 6 — Combined axial + flexure (§H1.1) with gravity present. Under the seismic combination the beam still carries gravity. Say Mu = 85 kip-ft and φbMn = 258 kip-ft for the W21×44 (fully braced).
Pr/Pc = 63.4/304 = 0.209 < 0.2? No — 0.209 ≥ 0.2, so use Eq. H1-1a:
Spec. Eq. H1-1aPr/Pc + (8/9)(Mrx/Mcx) = 0.209 + (8/9)(85/258) = 0.209 + 0.293 = 0.502 ≤ 1.0 o.k.
CheckReferenceRequiredAvailableD/C
Collector tensionEq. D2-163.4 k585 k0.11
Collector compressionEq. E3-163.4 k304 k0.21
Combined axial + flexureEq. H1-1a——0.50
Connection to the brace gusset§J3, §J463.4 k (both directions)design required—
Result: The existing W21×44 works as the collector with a combined interaction of 0.50. Design its end connections and the shear-stud transfer for the full amplified force of 63.4 kips in tension and in compression.

Engineering interpretation: collectors are the most commonly overlooked element in a lateral system, and the failure mode is not the member — the member is almost always a gravity beam with enormous spare axial capacity. The failure mode is the connection. A standard shear tab designed for 30 kips of vertical reaction has essentially no axial capacity, and that is exactly what the collector needs. When you designate a beam as a collector, you must go back and redesign its end connections, its gusset interface, and the shear studs that transfer the diaphragm force into it. That is why ASCE 7 puts Ω0 on collectors: it forces the load path to stay elastic all the way from the slab to the brace.

Common mistakes:
  • Omitting Ω0 on the collector and its connections (ASCE 7 §12.10.2.1).
  • Taking the peak collector force at the building end. It peaks at the edge of the braced bay.
  • Designing for tension only. The force reverses with load direction — compression buckling must be checked.
  • Leaving the standard shear-tab connection in place. A collector connection must transfer axial force, not just the vertical reaction.
  • Forgetting the shear studs. The diaphragm force has to physically get from the slab into the steel collector.
  • Using Eq. H1-1b when Pr/Pc ≥ 0.2 (or H1-1a when it is < 0.2).
Chapter 19 takeaway. Framing-system design is load-path bookkeeping. Follow one pound of load from the slab it lands on, through the beam, into the girder, down the column, or sideways through the diaphragm → collector → brace → foundation. Every handoff in that chain is a design: a connection, a member, a limit state. A perfect member with an unexamined handoff is not a design — it is a gap.

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
Bay 30×30 ft, infill beams at 10 ft o.c.; service floor load 100 psf.
AISC Reference
ASCE 7 §4.7
Step-by-step solution
  1. Trib width
    10 ft
  2. w (service)
    100 psf × 10 ft = 1.0 k/ft
Answer Infill beam carries 1.0 k/ft service load.
Tributary area for a beam
Problem statement image
w (kip/ft)LSimply supported beam, UDL w
DIMDimensions from the problem statement
No numeric parameters detected in the given statement — refer to the figure above for geometry.
Beam schematic — uniform load w over span L
  • Left support A (roller/pin) at x = 0
  • Right support B at x = L
  • Uniform load w along the full span
  • R_A = R_B = wL/2 · M_max = wL²/8 at midspan

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.

C19-01AISC 341-22
1. Braced frame types include:
Braced frame · X-brace bay V
C19-02AISC 341-22 Ch. E
2. Moment frame types (seismic):
C19-03AISC 341-22 §E3
3. SMF requires beam-column connection qualified for interstory drift of:
Braced frame · X-brace bay V
C19-04ASCE 7-22 §12.10
4. Composite slab diaphragm chord force resisted by:
C19-05AISC 341-22 §E3.4a
5. Capacity design (seismic) requires:
Braced frame · X-brace bay V
C19-06ASCE 7-22 §12.8
6. Base shear V for seismic (equivalent lateral force):
C19-07ASCE 7-22 Table 12.2-1
7. R-factor for SMF:
C19-08ASCE 7-22 §12.10
8. Diaphragm design forces Fpx (ASCE 7 §12.10):
C19-09AISC 341-22 §F3
9. EBF link is designed to:
Braced frame · X-brace bay V
C19-10AISC 341-22 §F4
10. BRBF brace strain range (design):
C19-11AISC 341-22 §E3.6g
11. Column splices in SMF located:
C19-12ASCE 7-22 Comm. 12.10
12. Diaphragm chord: analog to a beam:
C19-13ASCE 7-22 Table 12.2-1
13. When both braced + moment frames used (dual system) R factor:
Braced frame · X-brace bay V

Upload your worked solution (PDF)

Attach your handwritten or typed step-by-step solution for this chapter's graded quiz. The instructor can download every submission. PDF only, up to 25 MB.

How your upload will be graded

Your file — PDF, Word document, scanned handwriting or a photo — is read page by page like an experienced structural engineering instructor would. The scan is validated first, then your reasoning, structural model, calculations, diagrams, code basis and final answers are graded on process, not just the final number. Design work is additionally reviewed against AISC 360-22 and ACI 318-19. Partial credit applies, and one early mistake carried correctly forward is only penalized once.

Before you attach the file
  • Include every page, in order and right way up — a missing page cannot earn credit.
  • Keep margins in frame: nothing cropped at the edges, especially boxed final answers.
  • Scan or photograph in good, even light — no shadows, glare or blur; 300 dpi or a steady phone scan.
  • Write in dark pen; faint pencil is the most common 'UNREADABLE — INSTRUCTOR REVIEW REQUIRED' flag.
  • Include all diagrams, FBDs, shear/moment diagrams and section sketches — label them.
  • Number each question the same way the assignment does, and note anything you skipped.
  • Show units on every line and box your final answers.
  • Combine everything into ONE file (PDF preferred; Word, JPG or PNG accepted) under 20 MB.
Sign in to upload your worked solution.
§11Section 11

Chapter summary

Formula sheet
  • Design strength
    φ Rn ≥ Ru
    AISC 360-22 B3.1
Engineering checklist
  • Module 19: Steel Framing Systems
  • 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

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.