Engineering story

A century of steel — from concept to skyline

Engineer reviewing blueprints against a steel-frame construction site at sunrise

A beam does not fail at Mp — it fails where Mp cannot be reached.

Compactness and bracing decide whether the plastic moment is available.

Chapter 6 covers laterally supported and laterally unsupported flexural members under AISC 360-22 Chapter F: compact I-shape flexure (F2) with Mp = Fy*Zx, the three lateral-torsional buckling zones (plastic, inelastic, elastic) delimited by Lp and Lr, non-compact flange penalties (F3), and the deflection serviceability check L/360.

Iconic steel structures built on engineering excellence

Navier — elastic beam theory
1826
Plastic-hinge theory (Baker et al.)
1948
AISC LRFD — Mp basis for compact shapes
1986
AISC 360-22 Chapter F (current)
2022

Load pathFloor loadDeckFiller beamGirderColumnFoundation

02

Learning objectives

What you will be able to do after finishing Chapter 1 — and why each objective matters in practice

Objective 01

Compute Mp = Fy*Zx for compact I-shapes

For a compact, laterally braced W-shape, phi_b*Mn = 0.90*Fy*Zx.

Why it matters: Mp is the ceiling — every longer/less-braced case lowers Mn below it.
Where it is used: Every gravity beam, girder, and header.
Connects to: AISC F2.1; Manual Tables 3-2, 3-10.

Objective 02

Find Lp and Lr for LTB zones

Lp = 1.76*ry*sqrt(E/Fy); Lr from F2-6. Lb<=Lp plastic; Lp<Lb<=Lr inelastic; Lb>Lr elastic Fcr.

Why it matters: The wrong zone gives the wrong Mn by 20-60%.
Where it is used: Every unbraced-length check on a beam or girder.
Connects to: AISC F2.2; Manual Table 3-10.

Objective 03

Apply Cb — moment-gradient factor

Cb = 12.5*Mmax / (2.5*Mmax + 3*MA + 4*MB + 3*MC) <= 3.0.

Why it matters: Cb rewards varying-moment segments (typical gravity beams) and is often 1.14-1.32.
Where it is used: Every inelastic/elastic LTB check.
Connects to: AISC F1(3), Eq. F1-1.

Objective 04

Read Manual Tables 3-2 and 3-10

Table 3-2 gives phi_b*Mp, phi_b*Mr, Lp and Lr for every W. Table 3-10 plots Mn vs Lb.

Why it matters: One table read replaces four Chapter F equations for standard sections.
Where it is used: Every design office; every FE flexure question.
Connects to: AISC Manual Part 3.

Objective 05

Check deflection L/360 (LL) and L/240 (Total)

For a UDL: Delta_max = 5*w*L^4 / (384*E*Ix).

Why it matters: Serviceability often governs long-span filler beams, not strength.
Where it is used: Every floor beam serviceability check.
Connects to: AISC L; ASCE 7 App. C.

Objective 06

Classify flange & web (Table B4.1b)

lambda vs lambda_p and lambda_r. Non-compact flange triggers F3 with a linear penalty on Mp.

Why it matters: Non-compact flanges cap Mn below Mp; ignoring this over-predicts strength.
Where it is used: Heavy built-up girders, some rolled W's at high Fy.
Connects to: AISC Table B4.1b; F3.

03

Engineering motivation

What each part of a steel-frame building actually does — and why it exists

Before you design any single member, you have to see the whole system. A steel-frame building is not a collection of independent shapes bolted together — it is a deliberate load path, engineered so that every kilonewton of gravity, wind, or seismic demand has a continuous route from where it starts to the ground where the earth can resist it.

The photograph below shows a typical steel framing detail. Drag each labelled chip onto the structural element it names — the drop is only accepted when it lands inside the correct element's outlined region. A correct answer locks in with a green outline and a short explanation; a wrong answer flashes the region red, tells you what you actually hit, and returns the chip so you can try again. Press Reveal expected placements to see the reference solution (that attempt is then marked as assisted).

Structural steel framing — identify each element by dragging the labels — Fig. 1.3 · Drop a chip inside the outlined element it names. Green = correct and locked; red flash = wrong element, chip returns.

Palette — drag each element onto the member you think it is

Primary beam
The long light-gray horizontal steel member that spans between columns; carries floor load in flexure and delivers its reaction to the connection.
Steel column
The vertical wide-flange column shaft; transfers axial force (and any moment from the connection) between floors down to the foundation.
Beam-to-column connection
The bolted joint where the beam frames into the column flange — plates, bolts, and the joint region that transfers beam shear (and possibly moment) into the column.
Concrete floor slab
The cast concrete floor above; acts as a diaphragm distributing lateral load and provides continuous lateral bracing to the top (compression) flange.
Secondary beam
The darker horizontal member framing in behind the primary beam; a filler / secondary beam that delivers a concentrated reaction into the primary beam or girder.

Identified: 0 / 5

04

Failure mechanisms

Why we design the way we do — six ways steel structures have failed, and what each disaster taught the profession

Every provision in AISC 360 is a scar. Behind each equation, load factor, and detailing rule is a bridge, a walkway, or a tower whose failure cost lives and rewrote the profession. The six case studies below trace the mechanisms that motivate the code you are about to learn.

Read each one as an engineer, not a spectator: identify the load, the limit state, the missing check, and the specific clause that exists today because that check was missed. When you meet those clauses again in Chapters 5–17, they will read as answers, not rules.

Failure mechanism

Lateral-torsional buckling

The compression flange bows sideways and the section twists before the plastic moment can develop.

Root cause

Lb > Lp — the unbraced length exceeds the plastic-limit length.

Lesson learned

Brace the compression flange at Lb <= Lp, or accept a reduced Mn per F2.2.

§AISC 360-22 F2.2

Failure mechanism

Flange local buckling (FLB)

A slender or non-compact compression flange crinkles locally at high moment.

Root cause

bf/2tf > lambda_p (non-compact) or > lambda_r (slender).

Lesson learned

Table B4.1b classifies flanges; F3 penalises non-compact flanges linearly toward Mr.

§AISC 360-22 F3, Table B4.1b

Failure mechanism

Web local / shear buckling

A slender web can buckle in shear (G) or in flexural compression, reducing capacity below Mp.

Root cause

h/tw > lambda_r for web compact/non-compact limits.

Lesson learned

Rolled W-shapes almost always have compact webs; built-up girders may not.

§AISC 360-22 F4/F5, G

Failure mechanism

Hyatt Regency walkway (1981)

Suspended walkway beams and hanger rods failed at the box-beam/rod connection. Standard cautionary case for flexural-member connection detailing.

Root cause

A shop-change doubled the load on the upper box-beam/rod connection; the detail could not develop the demand.

Historical case

Kansas City, 17 July 1981 — 114 killed.

Lesson learned

Flexural design is not just Mp; the connections that deliver load into the beam must be checked separately.

§AISC 360-22 J

How failure propagates

The five-stage failure progression

1Applied load

2Elastic bending

3First yield

4Lateral bow / twist

5Collapse

Design codes intervene at the transition from yield to instability. Everything before yield is elastic and reversible; everything after instability is a race to collapse. LRFD keeps the demand well below the first transition.

05

Lecture notes

The full textbook chapter — figures, equations, and engineering narrative

Reflection · Think before you read

Why can a compact, fully-braced beam reach Mp while an identical beam laterally unbraced over the same span may fail at less than half that moment? What physically changes between those two cases?

Chapter 6 — Flexural Members / Beams (AISC 360-22 Chapter F)

Chapter focus. Beams bend. If they are laterally supported and compact, they reach the full plastic moment Mp = Fy·Zx (φ = 0.90). If they are unbraced, the compression flange twists sideways — lateral-torsional buckling — and capacity drops with unbraced length Lb. This chapter walks the three flexural zones (plastic, inelastic LTB, elastic LTB), the Cb modifier for non-uniform moment, and how to enter Manual Table 3-10.

1. Behavior of a Beam under Bending

A beam resists transverse loads by developing internal bending moment M and shear V. For a compact, fully braced doubly-symmetric section, the section can reach its plastic moment M_p = F_y · Z_x. If the compression flange is not braced, the beam may fail by lateral-torsional buckling (LTB) — a coupled lateral deflection and twist.

Representative Structures

Where flexural members appear in real engineering practice. Each highlighted element carries transverse load primarily by bending — its capacity is set by M_p = F_y·Z_x when fully braced, or by lateral-torsional buckling when the compression flange is unsupported over a long L_b.

Highway bridge plate girders — Highway Bridge — Steel Plate Girders
Parallel steel plate girders carry the deck slab and live load in pure flexure between piers. The composite deck braces the top flange continuously (L_b ≈ 0 in service), so capacity approaches M_p. During construction — before the slab cures — the bare girder must still be checked against LTB.

Millennium footbridge London — Pedestrian Bridge — Deck Beams
Longitudinal edge beams span between suspension cables and take pedestrian live load in bending plus a fatigue-critical serviceability check. Pedestrian bridges are routinely governed by deflection (L/360) rather than strength.

Steel truss bridge deck beams — Truss Bridge — Transverse Floor Beams
Floor beams span between the truss panel points and carry the deck load in flexure into the truss chords. They are typically compact rolled W-shapes with short L_b, so M_n = M_p controls.

Steel building floor framing — Steel Building — Floor Beams & Girders
A W-shape floor system: filler beams frame into larger girders which frame into columns. Once the composite slab is placed, the top flange is fully braced — the classic Manual Table 3-10 design region where M_n = M_p.

Parking garage flat plate beams — Parking Garage — Long-Span Beams
Long clear spans (25–35 ft column-free bays) with heavy live load push these beams into deflection-controlled design. Precambering is common — a purely serviceability, not strength, decision.

Cellular steel roof beams — Cellular Roof Beams
Cellular (castellated) beams are the roof-girder version of a deep W-shape — the web openings let mechanical ductwork pass through without increasing floor-to-floor height. Bending governs at the tee-section between openings; web-post buckling adds a new local limit state.

2. Section Classification (Table B4.1b)

Compact: λ ≤ λ_p — reaches M_p.
Non-compact: λ_p < λ ≤ λ_r — reduced capacity between M_p and 0.7 F_y S_x.
Slender: λ > λ_r — elastic local buckling controls.

(1)

F l an g e : λ_{p} = 0.38 (E / F_{y})

λ_{r} = 1.0 (E / F_{y})

(2)

W e b : λ_{p} = 3.76 (E / F_{y})

λ_{r} = 5.70 (E / F_{y})

3. Unbraced Length L_b — the Distance that Sets LTB Capacity

Lateral-torsional buckling (LTB) is a stability failure of the compression flange. A composite floor slab attached to the top flange braces it continuously (L_b ≈ 0). A bare beam with kicker braces at the ends and one at midspan has L_b = L/2. A cantilever tip is unbraced by definition unless a stub or tie is provided.

L_b = distance between points that prevent lateral movement and twist of the compression flange.
A framing beam / joist attached to the top flange counts as a brace only if it can carry the small stabilizing force (~2 % of the flange force) per AISC Appendix 6.
Composite-slab beams: assume L_b = 0 during service, but check the beam bare during construction (before slab hardens).

4. Lateral-Torsional Buckling (§F2) — Three Zones

Nominal-moment curve for compact I-shapes

The nominal moment M_n depends on where L_b lands on the curve above:

F2-5 Limiting lengths

(3)

L_{p} = 1.76 r_{y} (E / F_{y})

F2-6 Limiting length Lr

(4)

\derive L_{r} = 1.95 r_{t s} \frac{E}{0.7 F _{y}} \frac{J c}{S _{x} h _{o}} + (\frac{J c}{S _{x} h _{o}})^{2} + 6.76 (\frac{0.7 F _{y}}{E})^{2}

F2-1 Plastic (Lb ≤ Lp)

(5)

M_{n} = M_{p} = F_{y} Z_{x}

F2-2 Inelastic LTB (Lp < Lb ≤ Lr)

(6)

M_{n} = C_{b} [M_{p} - (M_{p} - 0.7 F_{y} S_{x}) (L_{b} - L_{p}) / (L_{r} - L_{p})] \leq M_{p}

F2-3 Elastic LTB (Lb > Lr)

(7)

M_{n} = F_{cr} S_{x} \leq M_{p}, w i t h F_{cr} f r o m F 2 - 4

Design

(8)

ϕ_{b} M_{n} \geq M_{u}, ϕ_{b} = 0.90

5. Moment-Modification Factor C_b (§F1-1)

(9)

C_{b} = 12.5 M_{ma x} / (2.5 M_{ma x} + 3 M_{A} + 4 M_{B} + 3 M_{C})

Uniform moment: C_b = 1.0. Simple beam + point load at mid-span: C_b ≈ 1.32. Cantilever tip: C_b = 1.0. Never take C_b above 3.0.

6. Serviceability — Deflection and Vibration

Floor beams: Δ_L ≤ L/360; Δ_D+L ≤ L/240 (IBC Table 1604.3).
Roof beams supporting plaster: Δ_L ≤ L/360.
Vibration — see AISC Design Guide 11 (f_n ≥ 8 Hz typical office).

Always use unfactored (service) loads when computing deflections. Strength design uses factored loads; serviceability does not.

7. Two Solution Methods — Companion Tables vs. AISC Specification Formulas

Flexural design in AISC 360-22 Chapter F can be executed by either hand-calculating with the F2 equations or reading straight from the Manual companion tables and charts. Master both: the Spec route explains the mechanics; the Manual route is faster and is what you use on the FE / PE exam.

Method A — AISC Specification Formulas (Chapter F, "the long way")

Classify the section (Table B4.1b) — compact / non-compact / slender.
Compute the limiting lengths L_p (Eq. F2-5) and L_r (Eq. F2-6) from r_y, r_ts, J, S_x, h_o.
Determine the region from L_b:
- L_b ≤ L_p → plastic: M_n = M_p = F_yZ_x (Eq. F2-1).
- L_p < L_b ≤ L_r → inelastic LTB (Eq. F2-2), with C_b from F1-1.
- L_b > L_r → elastic LTB (Eq. F2-3, F2-4).
φ_bM_n = 0.90 · M_n.

Use when: non-A992 steel, built-up plate girders, singly-symmetric sections, or when a report or exam requires every substitution shown. See the C_b derivation in Worked Example 6.2.

Method B — AISC Manual Companion Tables & Charts (Part 3, "the fast way")

Table 3-2 (Z_x-sorted): tabulates φ_bM_p, φ_bM_r (= 0.7F_yS_x), φ_bBF, L_p, and L_r for every W-shape. Enter with the required Z_x (from M_u/(0.9F_y)) to pick the lightest section.
In the inelastic-LTB region, use the "BF" shortcut for Eq. F2-2:
(10)
$ϕ_{b} M_{n} = C_{b} \cdot [ϕ_{b} M_{p} - ϕ_{b} B F \cdot (L_{b} - L_{p})] \leq ϕ_{b} M_{p}$
where φ_bBF is read directly from Table 3-2 — no need to compute (M_p−0.7F_yS_x)/(L_r−L_p) by hand.
Table 3-10 (Available Moment vs. Unbraced Length): plots φ_bM_n vs. L_b for C_b=1. Enter with L_b and read φ_bM_n directly; the plateau (L_b ≤ L_p), the sloped inelastic segment, and the elastic tail are all embedded.
Table 1-1 (Dimensions and Properties) — enter with a required I_x for deflection-controlled design; pick the lightest W with I_x ≥ I_req.
Table 3-23 (Shears, Moments & Deflections): closed-form M_max and δ_max for common load cases (uniform, point at mid-span, two symmetric point loads, cantilevers, etc.).

Use when: rolled W-shape in A992, standard load cases, or trial-section selection. See Worked Example 6.1 (Table 3-2 look-up of L_p, L_r, φ_bM_p).

Which method for which problem?

Trial-section selection from M_u: Method B, Table 3-2 sorted by Z_x.
Deflection-controlled beam: Method B, Table 1-1 sorted by I_x (then confirm strength).
Given L_b, find φ_bM_n: classify the region (L_b vs. L_p, L_r) with Table 3-2, then either read Table 3-10 (Method B) or plug into Eq. F2-1/F2-2/F2-3 (Method A). The BF shortcut bridges the two.
Compute M_u for a given loading: Method B, Table 3-23 for closed-form M_max; or Method A by statics.
C_b for non-uniform moment: Method A only — always compute with Eq. F1-1 from the bending-moment diagram.

Numerical agreement between the two methods is within round-off (~1 %). Every worked example in this chapter states which method it uses and cross-checks against the other.

8. Design Procedure (put it all together)

Compute M_u from ASCE 7 combinations.
Assume compact and fully braced; required Z_x ≥ M_u/(0.9 F_y). Enter Manual Table 3-2 to pick the lightest W-shape (Method B).
Determine L_b from the bracing layout; compare to L_p and L_r from Table 3-2. Enter Table 3-10 (or Eq. F2-1/F2-2/F2-3) for φ_bM_n.
Compute C_b from the moment diagram (§F1-1) if the moment is non-uniform between brace points; multiply through and cap at φ_bM_p.
Verify shear V_u ≤ φ_v V_n (Ch. 7).
Check deflection with unfactored service loads.

⚠ Common mistakes

Using S_x instead of Z_x for plastic capacity.
Forgetting C_b when non-uniform moment increases capacity.
Checking deflection with factored loads.
Ignoring section classification when using F_y > 50 ksi.

Worked Example 6.1 — Interior Floor Beam, W-shape (A992)

Method: AISC Manual Companion Table 3-2 for section pick (Z_x, L_p, L_r, φM_p) | Formula check: Eq. F2-1 (L_b ≤ L_p plateau) and Table 1-1 for I_x in the deflection check

Given: Simply-supported interior floor beam, span L = 30 ft, tributary width 10 ft. Slab dead D = 75 psf + SDL 15 psf; live L = 80 psf (office corridor). Compression flange continuously braced by composite metal deck (L_b ≈ 0). Steel A992, F_y = 50 ksi. Deflection limit L/360 (live).

Beam geometry & loading

Step 1 — Distributed loads

(1)

w = (D \cdot tw) + (L \cdot tw)

w_D = (0.075 + 0.015) · 10 = 0.90 k/ft (add beam self-wt ≈ 0.06 → 0.96 k/ft)
w_L = 0.080 · 10 = 0.80 k/ft

Step 2 — Factored moment

(2)

w_{u} = 1.2 D + 1.6 L

M_{u} = w_{u} L^{2} /8

w_u = 1.2(0.96) + 1.6(0.80) = 1.15 + 1.28 = 2.43 k/ft… but corridor live can be higher; recompute with 100 psf → w_L = 1.00, w_u = 1.15 + 1.60 = 2.75. Add pattern → adopt w_u = 3.06 k/ft including partitions.
M_u = 3.06 · 30² / 8 = 344 k·ft

Step 3 — Required plastic modulus

(3)

Z_{x, r e q} = M_{u} / (ϕ_{b} F_{y}) = M_{u} \cdot 12/ (0.9 \cdot 50)

Z_x,req = 344·12 / 45 = 91.7 in³

Step 4 — Trial section (Manual Table 3-2)

Try W18×50: Z_x = 101 in³, S_x = 88.9 in³, I_x = 800 in⁴, r_y = 1.65 in, L_p = 5.83 ft, L_r = 17.0 ft, φM_p = 379 k·ft. Compact for A992.

Step 5 — Flexural strength (F2-1)

Compression flange continuously braced → L_b = 0 < L_p, so LTB not applicable.

(4)

ϕ_{b} M_{n} = ϕ_{b} M_{p} = 0.9 \cdot F_{y} \cdot Z_{x}

φ_b M_n = 0.9 · 50 · 101 / 12 = 379 k·ft ≥ 344 ✓

Step 6 — Live-load deflection

(5)

Δ_{L} = 5 w_{L} L^{4} / (384 E I_{x})

w_L = 1.0 k/ft = 0.0833 k/in; L = 360 in
Δ_L = 5 · 0.0833 · 360⁴ / (384 · 29000 · 800)
= 5 · 0.0833 · 1.680×10¹⁰ / 8.928×10⁹
= 6.997×10⁹ / 8.928×10⁹ = 0.78 in
Limit: L/360 = 360/360 = 1.00 in ✓

Step 7 — Shear (preview of Ch. 7)

V_u = w_u L / 2 = 3.06·30/2 = 45.9 k; φV_n (W18×50) = 192 k ≫ ✓

DESIGN: W18×50, A992. φ_bM_n = 379 k·ft ≥ M_u = 344 k·ft ✓; Δ_L = 0.78 in ≤ L/360 ✓.

Worked Example 6.2 — Computing C_b from the Bending Moment Diagram (Four-Point Load Test Beam)

Method: AISC Specification only — Eq. F1-1 from the BMD (no companion table gives C_b). Support tables: Table 3-23 Case 9 for M_max; Table 3-2 for φ_bBF when C_b is later plugged into F2-2.

Given: A simply-supported W-shape spans L = 30 ft and carries two equal concentrated loads P = 20 kip at the third points (a = 10 ft from each support). The compression flange is unbraced along the full span (L_b = 30 ft). Compute the AISC moment-modification factor C_b (Eq. F1-1) so it can be used in the F2-2 inelastic-LTB equation.

Loading, dimensions, and bending moment diagram (BMD) with the four C_b ordinate stations

Step 1 — Support reactions

(6)

Σ M_{@ R_{2}} = 0 \Rightarrow R_{1} \cdot L - P \cdot (L - a) - P \cdot a = 0 \Rightarrow R_{1} = P (b y sy mm e t r y R_{2} = P)

R₁ = R₂ = P = 20 kip

Step 2 — Piecewise moment equation M(x)

Cut the beam at a station x measured from the left support and sum moments of forces to the left of the cut.

(7)

S e g m e n t ①0 \leq x \leq a : M (x) = R_{1} \cdot x

(8)

S e g m e n t ② a \leq x \leq L - a : M (x) = R_{1} \cdot x - P \cdot (x - a) = P \cdot a (co n s t an tpl a t e a u)

(9)

S e g m e n t ③ L - a \leq x \leq L : M (x) = R_{2} \cdot (L - x)

Step 3 — Draw the BMD (values needed to sketch it)

M(0) = 0 • M(a) = P·a = 20·10 = 200 k·ft • M(L−a) = 200 k·ft • M(L) = 0
Shape: two straight ramps (0 → 200 k·ft over 0–10 ft), a flat plateau at 200 k·ft (10–20 ft), then a mirror ramp back to zero. Peak moment M_max = 200 k·ft.

Step 4 — Read the four C_b ordinates from the BMD

AISC Eq. F1-1 evaluates the moment at four fixed stations of the unbraced segment: the two ends (only their maximum matters), the quarter-point, midspan, and three-quarter point. Here L_b = L, so:

(10)

M_{A} = ∣ M (L /4) ∣, M_{B} = ∣ M (L /2) ∣, M_{C} = ∣ M (3 L /4) ∣

x = L/4 = 7.5 ft → Segment ① ⇒ M_A = R₁·x = 20·7.5 = 150 k·ft
x = L/2 = 15 ft → Segment ② (plateau) ⇒ M_B = P·a = 200 k·ft
x = 3L/4 = 22.5 ft → Segment ③ ⇒ M_C = R₂·(L−x) = 20·(30 − 22.5) = 150 k·ft
M_max = 200 k·ft (anywhere on the plateau)

Step 5 — Apply AISC Eq. F1-1

(11)

C_{b} = 12.5 \cdot M_{ma x} / (2.5 \cdot M_{ma x} + 3 \cdot M_{A} + 4 \cdot M_{B} + 3 \cdot M_{C}) \leq 3.0

Numerator = 12.5 · 200 = 2500
Denominator = 2.5·200 + 3·150 + 4·200 + 3·150 = 500 + 450 + 800 + 450 = 2200
C_b = 2500 / 2200 = 1.136 (≤ 3.0 ✓)

Step 6 — Interpretation & use in F2-2

C_b = 1.14 > 1.0 → the non-uniform moment shape is less critical than uniform moment; the beam's LTB capacity may be scaled up by ~14 %.
If L_p < L_b ≤ L_r, plug this C_b into F2-2: M_n = C_b·[M_p − (M_p − 0.7F_yS_x)(L_b−L_p)/(L_r−L_p)] ≤ M_p.
If lateral braces were added at the two load points (three 10-ft segments), the center segment has uniform moment → C_b = 1.0, while the end segments have a triangular BMD → C_b ≈ 1.67. Always compute C_b segment-by-segment and use the lowest φM_n that results.

⚠ Common mistakes with C_b

Reading M_A, M_B, M_C at the load points instead of at L/4, L/2, 3L/4 of the unbraced segment.
Using signed moments — Eq. F1-1 uses absolute values.
Applying C_b when L_b ≤ L_p — C_b only helps in the LTB range and is still capped at M_p.
Forgetting to recompute C_b for each unbraced segment when lateral braces divide the span.

ANSWER: C_b = 12.5(200) / [2.5(200) + 3(150) + 4(200) + 3(150)] = 2500/2200 = 1.14.

Engineering tips

The plastic moment Mp. Zx > Sx always. Zx/Sx ~= 1.10-1.15 for a W-shape — the 'shape factor'.
The three LTB zones. Plot Mn vs Lb: horizontal at Mp, linear drop to 0.7*Fy*Sx at Lr, then a 1/Lb^2 decay.
Moment-gradient factor Cb. Cb = 1.0 is always conservative; use the full equation for economy.
Serviceability deflection. Long-span filler beams: deflection often governs, not strength.

Common mistakes

The plastic moment Mp. Using Sx (elastic) instead of Zx (plastic) in F2.
The three LTB zones. Using the elastic Fcr formula when Lb is still below Lr.
Moment-gradient factor Cb. Applying Cb to the plastic zone — Cb caps at Mp regardless.
Serviceability deflection. Using unfactored loads for strength or factored loads for deflection.

Real-world use

The plastic moment Mp. Every compact laterally braced W-shape.
The three LTB zones. Every unbraced-length check.
Moment-gradient factor Cb. Continuous beams, cantilevers, girders with concentrated loads.
Serviceability deflection. Every floor beam.

06

Professional practice, safety & ethics

Flexural member practice

Professional practice

Specify Lb on the drawings, or specify the bracing (joists, deck, kickers) that creates it — LTB capacity is meaningless without it.
Note whether the deck is attached to the beam and how; only a positively attached deck counts as continuous bracing.
Camber must be called out with a tolerance and never used to compensate for a strength deficiency.

Safety in design & construction

During deck placement, the top flange may be unbraced — check the construction-stage Lb, not the final one.
Cb > 1.0 is a real benefit but must reflect the actual moment diagram, including construction loading.
Web openings for MEP cut after fabrication can destroy flexural and shear capacity; require EOR review of every penetration.

Engineering ethics

Do not use Cb to rescue a beam if the moment diagram assumption is not documented.
Serviceability complaints (bouncy floors) are an ethical as well as a technical issue — disclose vibration limitations up front.
Refuse to certify a beam whose bracing depends on work by others not shown in the contract documents.

Ironworkers bolting a steel beam connection while tied off at height — Erection safety: OSHA Subpart R fall protection and stable temporary bracing.

Engineers reviewing sealed structural drawings across a conference table — Design review: documenting assumptions before the drawings are sealed.

ABET / licensure link. These points map to ABET Student Outcomes 2 and 4 — engineering design within realistic constraints, and recognition of ethical and professional responsibilities. Expect NCEES FE and PE exam questions on the NSPE Code of Ethics, OSHA construction requirements, and the engineer's standard of care.

07

Cost analysis

Beam economy and depth

Approach

Deeper is cheaper: for the same Mu, a deeper beam weighs less. Depth is limited by floor-to-floor height, not by strength.
Camber costs ~$60–100 per beam; sometimes a heavier uncambered beam is cheaper than a cambered light one.
Bracing (joists/deck) increases capacity for free — make sure you take credit for real bracing before upsizing.

Worked cost example — W18 vs W21 for the same moment

Basis: Mu = 320 k-ft, Lb fully braced, 30 ft span

Line item	Qty	Rate	Cost
W18×55 — 55 lb/ft × 30 ft	0.825 ton installed	$2,700	$2,228
W21×50 — 50 lb/ft × 30 ft	0.75 ton installed	$2,700	$2,025
Camber (W21 option)	1 ea	$80	$80
Estimated total			$4,333

Takeaway. Going one nominal depth deeper saves ~$120 per beam — multiply by 60 beams per floor before dismissing it.

Unit rates are representative US averages for teaching purposes. On a real project, price with current local rates (RSMeans, fabricator quotes, or contractor pricing) and state the estimate date.

08

Animated concepts

Key mechanics visualised — watch the strain profile, stress block, or buckled shape evolve

Lateral-torsional buckling

Between braces, the compression flange deflects laterally and the section twists. Lb vs Lp / Lr sets the flexural regime (F2-1 / F2-2 / F2-3).

09

Engineering figures

Full-page reference diagrams — the visual vocabulary you will use for the rest of the course

§6.6.1

Beam in service

Fig. 6.1Steel filler beam supporting composite deck
Filler beam under composite deck — the deck provides continuous lateral bracing to the compression flange.

§6.6.2

Beam-to-girder connection

Fig. 6.2Bolted double-angle shear tab
Shear connections transfer beam reaction into the girder without moment.

§6.6.3

Lateral-torsional buckling

Fig. 6.3LTB test specimen
The compression flange bows out laterally and the section twists — the F2.2 mode.

§6.6.4

Flange local buckling

Fig. 6.4Non-compact flange after test
Local flange buckling — F3. Preceded by bf/2tf > lambda_p.

10

Worked examples

Full textbook solutions — problem, theory, step-by-step, verification, interpretation

Example 6.1

W18x50 floor beam — phi_b*Mn per F2 (fully braced)

A W18x50 (A992) filler beam spans 30 ft simply supported and carries a composite deck that provides continuous lateral bracing. Compute phi_b*Mn per F2 and check L/360 live-load deflection.

Problem statement

A W18x50 (A992) filler beam spans 30 ft, simply supported, with the compression flange continuously braced by a composite deck. Compute phi_b*Mn per AISC 360-22 F2 and verify live-load deflection under w_LL = 0.80 klf against L/360.

Wide-flange steel floor beam with composite deck above — FIG. 6.1 — Deck-braced W18x50 filler beam typical of this example.

Floor plan

3-D isometric

Element detail

DIMPlan locates the filler beam, isometric shows the floor framing, detail carries the factored UDL — φbMn per §F2/F3.

Given

W18x50: Zx = 101 in^3, Sx = 88.9 in^3, Ix = 800 in^4, ry = 1.65 in
A992 steel: Fy = 50 ksi, E = 29,000 ksi
L = 30 ft = 360 in, simply supported
Lb ~ 0 (deck brace) => plastic zone
w_LL = 0.80 klf (service)

Find

phi_b*Mn per F2
Mid-span Delta_LL vs L/360

Assumptions

Compact section (verify vs Table B4.1b)
Full lateral bracing (Lb = 0)
Simple supports; no continuity

Code references

AISC 360-22 F1 — phi_b = 0.90
AISC 360-22 F2 — compact I-shape flexure
AISC Manual Table 3-2
ASCE 7 Table C.C.1.1

Theory & approach

For a compact, laterally braced I-shape bent about the strong axis, the entire section can reach Fy. F2.1 gives Mn = Mp = Fy*Zx. With Lb <= Lp the plastic zone applies. Deflection is a serviceability check independent of the strength calculation.

Step-by-step solution

1
Section & regime
FormulaAISC F2
Compact + L_b <= L_p → M_n = M_p
Deck brace: L_b ~ 0 → plastic zone.
2
Plastic moment Mp
FormulaF2.1
M_p = F_y * Z_x
M_p = 50 * 101 = 5,050 kip*in = 421 kip*ft
3
phi_b*Mn
FormulaF1
phi_b*M_n = 0.90 * M_n
phi_b*M_n = 0.90 * 421 = 379 kip*ft
Manual Table 3-2: phi_b*M_p (W18x50) = 379 kip*ft OK
4
Live-load deflection
Formula
Delta = 5*w*L^4 / (384*E*I)
w = 0.80/12 = 0.0667 kip/in; L = 360 in
Delta = 5*0.0667*(360)^4 / (384*29,000*800) = 0.62 in
5
Deflection check
Formula
L/360 = 360/360 = 1.00 in
0.62 in < 1.00 in → OK

Verification

phi_b*Mn = 379 kip*ft matches Manual Table 3-2 exactly. Delta_LL = 0.62 in is 62% of the L/360 limit.

Final answer

phi_b*Mn = 379 kip*ft and Delta_LL = 0.62 in < 1.00 in.

Design interpretation

The deck brace lets the beam reach Mp. Remove the deck brace (Lb = 30 ft, beyond Lr ~ 17 ft for W18x50) and Mn drops into elastic LTB — roughly 40-50% of Mp. Bracing is often free.

Common mistakes

Using Sx (elastic) in place of Zx.
Applying phi = 0.75 (rupture) instead of 0.90 (flexure).
Running deflection with factored loads — deflection uses service loads.

Engineering insight

For 25-35 ft filler beams under office LL, deflection usually governs before strength. Screen deflection first with Delta = 5*w*L^4/(384*E*I).

References

· AISC 360-22 Chapter F
· AISC Manual 16th ed., Tables 3-2 and 3-10
· ASCE 7-22 App. C

11

Guided practice

Compute the governing variables — hints unlock as you need them

W16x40 (A992) has Zx=73.0, Sx=64.7, Lp=5.55 ft, Lr=15.9 ft. Cb=1.0. Estimate phi_b*Mn at Lb=10 ft via F2.2 linear interpolation.

Your turn

M_p =kip*ftM_r =kip*ftphi_{b*M_n} =kip*ft

Hints

1.Mp = Fy*Zx; Mr = 0.7*Fy*Sx.

12

Independent practice

Solve the chapter's design task — compute each governing variable

Design task

A W16×26 (A992) simply-supported beam is fully braced along the compression flange (Lb ≤ Lp). Verify it is compact, then compute the plastic moment Mp and the design flexural strength φb Mn.

Given

Fy = 50 ksi
W16×26: Zx = 44.2 in³, Sx = 38.4 in³
Lb ≤ Lp → plastic-yielding limit state (F2.1)
Compact section (flange & web OK)

Approach

For a compact, fully-braced I-shape: Mn = Mp = Fy·Zx.
Cap: Mp ≤ 1.5·My (My = Fy·Sx). Check the cap is not active.
Design: φb Mn = 0.90·Mn. Report in k-ft (÷12 from k-in).

Submit your answer

M_p =k-ftphi_{b*M_n} =k-ft

13

Mini design challenge

Select the option that satisfies every code and serviceability requirement in the brief

Brief

Select the lightest W18 A992 filler beam for a 30 ft simply-supported floor. Factored wu = 1.6 klf, service w_LL = 0.80 klf. Deck bracing.

Requirements

Mu = wu*L^2/8 = 180 kip*ft
phi_b*Mn >= 180 kip*ft (strength)
Delta_LL <= L/360 = 1.00 in (service)
A992, W18 family

Section

Wt (lb/ft)

Δ (in)

Ru/Rn

Cost

Pick

14

Chapter summary

A mind map of how every concept connects

Graded Chapter Quiz(25 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.

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.

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16

FE exam preparation

NCEES-style practice with timer, equation sheet, and mastery tracking

Exam mode

30:00 Calculator

Question 1 / 10

A 32-ft simply supported W18×35 roof beam with a 12-ft tributary width supports service loads of $D = 28 psf$ , $L_{r} = 20 psf$ , $S = 35 psf$ , and $W = \pm 18 psf$ . Determine the governing factored uniform load $w_{u}$ using ASCE 7-22 LRFD combinations.

◆◆ MediumASCE 7-22 §2.3.1

Fig. W18×35 roof beam — 12 ft tributary width, 32 ft span.

A. 0.829 klf
B. 1.075 klf
C. 1.183 klf
D. 1.291 klf

Flexural Members / Beams

Compute Mp = Fy*Zx for compact I-shapes

Find Lp and Lr for LTB zones

Apply Cb — moment-gradient factor

Read Manual Tables 3-2 and 3-10

Check deflection L/360 (LL) and L/240 (Total)

Classify flange & web (Table B4.1b)

Lateral-torsional buckling

Flange local buckling (FLB)

Web local / shear buckling

Hyatt Regency walkway (1981)

Chapter 6 — Flexural Members / Beams (AISC 360-22 Chapter F)

1. Behavior of a Beam under Bending

Representative Structures

2. Section Classification (Table B4.1b)

3. Unbraced Length Lb — the Distance that Sets LTB Capacity

4. Lateral-Torsional Buckling (§F2) — Three Zones

5. Moment-Modification Factor Cb (§F1-1)

6. Serviceability — Deflection and Vibration

7. Two Solution Methods — Companion Tables vs. AISC Specification Formulas

Method A — AISC Specification Formulas (Chapter F, "the long way")

Method B — AISC Manual Companion Tables & Charts (Part 3, "the fast way")

Which method for which problem?

8. Design Procedure (put it all together)

⚠ Common mistakes

Worked Example 6.1 — Interior Floor Beam, W-shape (A992)

Step 1 — Distributed loads

Step 2 — Factored moment

Step 3 — Required plastic modulus

Step 4 — Trial section (Manual Table 3-2)

Step 5 — Flexural strength (F2-1)

Step 6 — Live-load deflection

Step 7 — Shear (preview of Ch. 7)

Worked Example 6.2 — Computing Cb from the Bending Moment Diagram (Four-Point Load Test Beam)

Step 1 — Support reactions

Step 2 — Piecewise moment equation M(x)

Step 3 — Draw the BMD (values needed to sketch it)

Step 4 — Read the four Cb ordinates from the BMD

Step 5 — Apply AISC Eq. F1-1

Step 6 — Interpretation & use in F2-2

⚠ Common mistakes with Cb

Beam in service

Beam-to-girder connection

Lateral-torsional buckling

Flange local buckling

W18x50 floor beam — phi_b*Mn per F2 (fully braced)

Section & regime

Plastic moment Mp

phi_b*Mn

Live-load deflection

Deflection check

Graded Chapter Quiz(25 FE-style questions · AISC Manual required)

Upload your worked solution (PDF)

3. Unbraced Length L_b — the Distance that Sets LTB Capacity

5. Moment-Modification Factor C_b (§F1-1)

Worked Example 6.2 — Computing C_b from the Bending Moment Diagram (Four-Point Load Test Beam)

Step 4 — Read the four C_b ordinates from the BMD

⚠ Common mistakes with C_b