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LecturesModule 6

Flexural Members / Beams

Plastic moment, lateral-torsional buckling, compact / noncompact / slender sections.

AISC Reference Box
  • AISC 360-22 Chapter F — Design of Members for Flexure
  • AISC §F2 — Doubly Symmetric Compact I-Shapes Bent About Major Axis

Lecture Notes

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

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

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

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

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

Chapter 6 — Flexure (filler beam & girder)
Typical 30 ft floor filler beam, Lb = 10 ft (slab restraint)
Demand carried forward
From Chapter 2: wu = 1.70 klf Mu = wu L² / 8 = 1.70 · 30² / 8 = 191 k-ft Vu = 1.70 · 30 / 2 = 25.5 k.
This chapter contributes
Selects from AISC Manual Table 3-2 with Mu and Lb. Confirms Mp not reduced by LTB (Lb < Lp because the deck braces the top flange).
Feeds into next chapter
Section becomes the basis for Chapter 7 (shear), Chapter 11 (composite action) and Chapter 18 (deflection).
Unbraced length LbMnLpLrMp0.7FySxMp = Fy·ZxElastic LTB
Mn vs Lb: plastic plateau Mp for Lb ≤ Lp, linear inelastic LTB to Lr, then elastic LTB.

Formula Sheet

NameEquationAISC Ref
Plastic momentMp = Fy · ZxAISC §F2.1
Lp (LTB lower limit)Lp = 1.76 ry √(E/Fy)AISC §F2.5
Lr (LTB upper limit)Lr = 1.95 rts (E/0.7Fy) √(Jc/(Sx ho) + √((Jc/(Sx ho))² + 6.76(0.7Fy/E)²))AISC §F2.6
LTB inelastic (Lp < Lb ≤ Lr)Mn = Cb [Mp − (Mp − 0.7 Fy Sx)·((Lb − Lp)/(Lr − Lp))] ≤ MpAISC §F2.2
Design strengthφb Mn = 0.90 · MnAISC §F1

Worked Example

Flexural Members / Beams

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

FE-Style Worked Examples (7)

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
W18×50, A992, Zx=101 in³.
AISC Reference
AISC §F2.1
Step-by-step solution
  1. Mp
    Fy·Zx = 50(101) = 5050 k-in = 421 k-ft
  2. φMn
    0.90(421) = 379 k-ft
Answer φMn = 379 k-ft (full Lb ≤ Lp).
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Course Materials — Lecture & Worked Examples

Lecture content and worked examples below come from the instructor's uploaded 'Design of Steel Beams in Flexure' notes. The figures are the original sketches of beam loading, the three failure modes, and the moment–unbraced-length curve.

Lecture highlights (from instructor notes)
  • A beam is any element loaded transverse to its longitudinal axis; it carries shear V(x) and bending M(x). Elastic stresses (doubly symmetric section): fc = ft = Mc/I, fv = V/(d·tw).
  • Three flexural failure modes (AISC §F): (1) plastic-hinge yielding of the section; (2) lateral-torsional buckling (LTB); (3) local buckling of the flange or web (when the section is noncompact / slender).
  • If Lb ≤ Lp the beam reaches its full plastic moment Mp = Fy·Zx. For Lp < Lb ≤ Lr the section is in the inelastic-LTB regime, and Mn linearly interpolates between Mp and 0.7 Fy Sx. For Lb > Lr buckling is elastic and Mn = Fcr·Sx.
  • Design strength: φb·Mn ≥ Mu, with φb = 0.90. Tabulated φb·Mp for every W-shape is in AISC Manual Table 3-2 (use Cb = 1.0 to be conservative).
  • Always add the section's self-weight to wu after picking a trial section, then verify the design demand again before locking in the size.
Given
L = 20 ft simply supported; wD = 1.0 k/ft (incl. self-wt), wL = 3.0 k/ft. Continuously braced. Fy = 50 ksi.
AISC Reference
AISC §F2.1; AISC Manual Table 3-2
Step-by-step solution
  1. Factored load
    wu = 1.2(1.0) + 1.6(3.0) = 6.0 k/ft (note: source used 1.25/1.75 → wu = 6.5 k/ft)
  2. Mu
    Mu = wu·L²/8 = 6.5(20)²/8 = 325 k-ft
  3. Select
    From Table 3-2 (p. 3-25): W10×77, φb·Mp = 366 k-ft ≥ 325 ✓
Answer Use W10×77 (A992).
Instructor note: Source notes used the older 1.25D + 1.75L factors. Modern ASCE 7 uses 1.2D + 1.6L → Mu ≈ 300 k-ft, W10×68 (φMp = 320) is enough.
20-ft simply-supported beam, wD = 1.0 k/ft, wL = 3.0 k/ft.
20-ft simply-supported beam, wD = 1.0 k/ft, wL = 3.0 k/ft.

Textbook — Aghayere & Vigil (2009)(5 worked examples with figures + numerical answers)

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

Chapter summary

Chapter 6 develops flexural strength. For compact, doubly symmetric I-shapes bent about the strong axis, four LTB regimes apply: Mp (Lb ≤ Lp), linear inelastic between Lp and Lr, and elastic LTB for Lb > Lr. Cb modifies the inelastic branch; serviceability deflection limits are checked at service loads.

  • Mp = Fy·Zx ; φMp = 0.90·Mp (compact, Lb ≤ Lp).
  • Inelastic LTB: Mn = Cb [Mp − (Mp − 0.7 Fy Sx)·(Lb−Lp)/(Lr−Lp)] ≤ Mp.
  • Cb = 12.5 Mmax / (2.5 Mmax + 3 MA + 4 MB + 3 MC) — quarter-point moments.
  • Beam selection plot in Manual Part 3 — pick by Lb vs φbMn.
  • Shear: φvVn = 1.0·0.6·Fy·Aw for rolled I-shapes (h/tw ≤ 2.24√(E/Fy)).
  • Deflection: L/360 LL, L/240 D+LL — serviceability uses unfactored loads.
Setup
W21×50, A992, fully braced compact shape. Zx = 110 in³. Find φMp.
AISC Reference
AISC §F2.1
Numerical practice
φMp (k·ft)?
Textbook page 225 — Plastic moment capacity
Aghayere & Vigil (2009), p. 225 — full worked solution & sketch.

FE Practice Bank (10)

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

P4B. RC Beam Flexure CheckEasyFEExam Structural Design – Problem 4
RC beam: Mu = 260 k·ft, Mn = 310 k·ft, εt = 0.0055. Adequate?
P5B. RC Nominal Moment MnEasyFEExam Structural Design – Problem 5
Singly-reinforced RC beam: b = 12, d = 22 in, As = 4.00 in², f′c = 5, fy = 60 ksi. Mn (k·ft)?
P6B. Net Tensile StrainEasyFEExam Structural Design – Problem 6
Same section but As = 5.10 in², f′c = 5 ksi (β1 = 0.80). εt at nominal strength?
P7B. Design Moment φMnEasyFEExam Structural Design – Problem 7
RC beam b = 14, d = 25 in, 4 #10 (As = 5.06 in²), f′c = 5, fy = 60 ksi. φMn (k·ft)?
P27E. φbMn W18×65EasyFEExam Structural Design – Problem 27
W18×65 Grade 50, Lb = 13 ft, Cb = 1.0. φbMn? (Zx = 133, Sx = 117 in³, Lp = 5.97 ft, Lr ≈ 17.1 ft)
P28E. Cb Moment-GradientEasyFEExam Structural Design – Problem 28
Two 25-ft unbraced segments, Mu1 = 797 k·ft. Cb1 and controlling Mu/Cb for span 1?
P29E. Select Lightest W (Zx)EasyFEExam Structural Design – Problem 29
Fully braced, 30 ft, wD = 0.5, wL = 1.0 k/ft, Fy = 50. Lightest W via Zx?
P30E. Lightest W (Table 3-10)EasyFEExam Structural Design – Problem 30
Simply supported, Lb = 20 ft (Cb = 1), Mu = 310 k·ft. Lightest W?
P38C. RC Beam Min WidthEasyFEExam Structural Design – Problem 38
Mu = 648 k·ft, f′c = 4, fy = 60, 8 #8 (As = 6.32), h = 30. Min b?
P39E. Max wu on W21×57EasyFEExam Structural Design – Problem 39
W21×57, 30 ft span, 2 intermediate braces (Lb = 10 ft), Cb = 1, Fy = 50. Max wu?

Interactive Calculator

Beam Flexure (Compact, Lb ≤ Lp)

AISC §F2.1
Mp = Fy·Zx326.7 kip-ft
φb Mn = 0.90 Mp294.0 kip-ftOK

Assumes a compact section with full lateral support (Lb ≤ Lp). For LTB regions use AISC §F2.2.

Practice Problems

  1. [E] State Mp = Fy·Zx and φb = 0.90.
  2. [E] Define Lb, Lp, Lr.
  3. [E] State Lp = 1.76·ry·√(E/Fy) and its meaning.
  4. [E] State COMPACT flange limit λp = 0.38√(E/Fy) and web limit 3.76√(E/Fy).
  5. [E] Give φb·Mp for W18x35, A992.
  6. [M] W21x62 (A992), 30 ft span, Lb = 30 ft (no bracing). Compute φMn.
  7. [M] W14x38 (A992), Lb = 8 ft. Confirm Lb ≤ Lp; compute φMn.
  8. [M] W16x26 carries wu = 1.2 k/ft on 24 ft span. Pick lightest W (A992), Lb = L/3.
  9. [M] Compute Cb for a simply supported beam with concentrated factored load P at midspan.
  10. [M] W24x62 roof girder, Lb = 30 ft, only end bracing. Check φMn vs Mu = 320 k-ft.
  11. [H] Design W-shape for Mu = 410 k-ft, Lb = 14 ft, Cb = 1.0, A992 (Table 3-10).
  12. [H] Plot φMn vs Lb for W16x40 (A992) from Lb = 0 to 30 ft; identify Lp and Lr.
  13. [H] 28 ft floor beam, wu = 2.6 k/ft + P = 12 k at midspan. Pick W-shape, Lb = 7 ft, Cb = 1.30.
  14. [H] Non-compact flange W12x65: compute FLB-controlled Mn per §F3.2.
  15. [H] Weak-axis bending W18x35: compute φMny per §F6.
Structured Clues
  • Three zones on the Mn–Lb curve: plastic (Lb ≤ Lp), inelastic LTB (Lp < Lb ≤ Lr), elastic LTB (Lb > Lr).
  • Cb = 1.0 is conservative; use Eq. F1-1 when end moments are unequal.
  • For COMPACT W-shapes, FLB and WLB are not checks.
Code References
  • AISC 360-22 §F2, F3, F6
  • AISC Manual Tables 3-2, 3-10

Quiz (10 FE-bank + 2 concept)

1. [P4] RC beam: Mu = 260 k·ft, Mn = 310 k·ft, εt = 0.0055. Adequate?
2. [P5] Singly-reinforced RC beam: b = 12, d = 22 in, As = 4.00 in², f′c = 5, fy = 60 ksi. Mn (k·ft)?
3. [P6] Same section but As = 5.10 in², f′c = 5 ksi (β1 = 0.80). εt at nominal strength?
4. [P7] RC beam b = 14, d = 25 in, 4 #10 (As = 5.06 in²), f′c = 5, fy = 60 ksi. φMn (k·ft)?
5. [P27] W18×65 Grade 50, Lb = 13 ft, Cb = 1.0. φbMn? (Zx = 133, Sx = 117 in³, Lp = 5.97 ft, Lr ≈ 17.1 ft)
6. [P28] Two 25-ft unbraced segments, Mu1 = 797 k·ft. Cb1 and controlling Mu/Cb for span 1?
7. [P29] Fully braced, 30 ft, wD = 0.5, wL = 1.0 k/ft, Fy = 50. Lightest W via Zx?
8. [P30] Simply supported, Lb = 20 ft (Cb = 1), Mu = 310 k·ft. Lightest W?
9. [P38] Mu = 648 k·ft, f′c = 4, fy = 60, 8 #8 (As = 6.32), h = 30. Min b?
10. [P39] W21×57, 30 ft span, 2 intermediate braces (Lb = 10 ft), Cb = 1, Fy = 50. Max wu?
11. Which AISC 360-22 chapter primarily governs flexural members / beams?
12. In LRFD, the basic design inequality is:

Common Student Mistakes

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

"Professor Explains" Script

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