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LecturesModule 7

Beam Shear Design

Web shear yielding and shear buckling per AISC Chapter G.

AISC Reference Box
  • AISC 360-22 Chapter G — Design of Members for Shear

Lecture Notes

This module introduces beam shear design. 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 7 — Beam shear
Same 30 ft filler beam — web shear at the support
Demand carried forward
From Chapter 6: Vu = 25.5 k.
This chapter contributes
Checks φv Vn = 1.00 · 0.6 · Fy · Aw for the selected section (Chapter G web yielding). Confirms web buckling is not triggered (h/tw of a typical W18 is well below 2.24√(E/Fy)).
Feeds into next chapter
Vu is the bolt-group demand at the shear-tab connection in Chapter 12 and Chapter 14.
Flange (low shear)τ ≈ V/(d·tw)Aw = d · tw | Vn = 0.6·Fy·Aw·Cv1Web carries ~95% of shear in a W-shape
Beam web carries ~95% of shear: Aw = d·tw, Vn = 0.6·Fy·Aw·Cv1 (AISC §G2.1).

Formula Sheet

NameEquationAISC Ref
Nominal shearVn = 0.6 · Fy · Aw · Cv1AISC §G2.1
Design shearφv Vn = 1.00 · Vn (for most rolled I-shapes)AISC §G1

Worked Example

Beam Shear Design

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 (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
W18×50: d=17.99, tw=0.355, h/tw=45.2, Fy=50.
AISC Reference
AISC §G2.1
Step-by-step solution
  1. Cv1
    h/tw=45.2 ≤ 2.24√(E/Fy)=53.9 → Cv1=1.0 φv=1.00
  2. Aw
    d·tw = 17.99(0.355) = 6.39 in²
  3. φVn
    1.00 × 0.6(50)(6.39)(1.0) = 192 k
Answer φVn = 192 k.
VMMmax

Textbook — Aghayere & Vigil (2009)(2 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 §6.5 covers beam shear. For nearly all rolled W-shapes h/tw is below 2.24√(E/Fy), so φv = 1.00 and Cv1 = 1.0 → φvVn = 0.6·Fy·Aw. Slender webs need stiffeners (Chapter G2 & G3).

  • Vn = 0.6·Fy·Aw·Cv1 (AISC §G2.1).
  • Rolled I-shapes with h/tw ≤ 2.24√(E/Fy): φv = 1.00, Cv1 = 1.0.
  • Beyond that limit: φv = 0.90, Cv1 < 1.0.
  • Aw = d·tw for I-shapes.
  • Concentrated load → check web yielding (J10.2), web crippling (J10.3), sidesway buckling (J10.4).
Setup
W21×50, A992 (Fy = 50 ksi), d = 20.8 in, tw = 0.380 in.
AISC Reference
AISC §G2.1
Numerical practice
φvVn (kips)?
Textbook page 240 — Shear capacity of a W-shape
Aghayere & Vigil (2009), p. 240 — full worked solution & sketch.

FE Practice Bank (9)

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.

P8C. Concrete Shear VcEasyFEExam Structural Design – Problem 8
bw = 12, d = 20 in, f′c = 4 ksi. Vc (kips)?
P9C. Stirrup VsEasyFEExam Structural Design – Problem 9
Av = 0.40 in², fyt = 60 ksi, d = 22, s = 10 in. Vs?
P10C. Vn with Vs CapEasyFEExam Structural Design – Problem 10
bw = 12, d = 20, f′c = 4 ksi, Av = 0.62 in², fy = 60, s = 4 in. Vn considering Vs ≤ 8√f′c·bw·d?
P11C. φVn CheckEasyFEExam Structural Design – Problem 11
Same beam, Vu = 110 k. φVn and adequacy?
P12C. Stirrup RequirementEasyFEExam Structural Design – Problem 12
bw = 12, d = 20, f′c = 4, Vu = 20 k. Stirrup requirement?
P13C. Min Stirrup SpacingEasyFEExam Structural Design – Problem 13
bw = 12, d = 24, f′c = 4, Av = 0.40, fy = 60, Vu = 18 k. Required s (in)?
P14C. Spacing Vu > φVcEasyFEExam Structural Design – Problem 14
Same section, Vu = 60 k. Required s (in)?
P15C. d/4 Spacing LimitEasyFEExam Structural Design – Problem 15
Same section, Vu = 150 k. Required s (in)?
P31E. Steel Beam ShearEasyFEExam Structural Design – Problem 31
Simply supported W, L = 24 ft, wD = 0.9, wL = 1.2 k/ft, A992, d = 18, tw = 0.35 in. Vu and φvVn?

Interactive Calculator

Beam Shear

AISC §G2.1
Aw = d·tw5.638 in²
φv Vn = 1.00·0.6·Fy·Aw (Cv1=1)169.2 kipsOK

Practice Problems

  1. [E] State Vn = 0.6·Fy·Aw·Cv1 (§G2.1).
  2. [E] Define Aw for a rolled W-shape (Aw = d·tw).
  3. [E] State φv = 1.00 for most rolled I-shapes.
  4. [E] When is Cv1 = 1.0 for a rolled W (h/tw ≤ 2.24√(E/Fy))?
  5. [E] State the AISC chapter for beam shear (Chapter G).
  6. [M] W21x62 (A992) carries Vu = 110 k. Compute φVn and check.
  7. [M] W16x26 with h/tw = 56.8. Compute Cv1 and φVn (A992).
  8. [M] Simply supported beam, 30 ft, wu = 3 k/ft. Find Vu; pick lightest W12.
  9. [M] HSS 10x6x1/4 beam carrying Vu = 35 k. Compute φVn.
  10. [M] Coped W18x35 (2 in. top flange cope). Compute effective d and check shear at the cope.
  11. [H] Plate-girder shear with stiffeners a/h = 1.5. Compute Cv2 (Table G2.2) and Vn.
  12. [H] Tension-field action: A36 plate-girder web, a/h = 1.0, h/tw = 200. Compute Vn with and without TFA.
  13. [H] W21x50 with 50 k point load 4 ft from support. Compute Vu at d/2 and check φVn.
  14. [H] Channel C12x30 beam, Vu = 30 k. Compute φVn including shear-center torsion warning.
  15. [H] Bearing stiffener at a 60 k reaction on W21x62 web: check WLY (§J10.2) and crippling (§J10.3).
Structured Clues
  • Aw = d·tw for rolled W; Cv1 = 1.0 if h/tw ≤ 2.24√(E/Fy).
  • φv = 1.00 for most rolled I-shapes (not 0.90).
  • Check shear at d from face of support.
Code References
  • AISC 360-22 §G2
  • AISC Manual Table 3-6

Quiz (9 FE-bank + 2 concept)

1. [P8] bw = 12, d = 20 in, f′c = 4 ksi. Vc (kips)?
2. [P9] Av = 0.40 in², fyt = 60 ksi, d = 22, s = 10 in. Vs?
3. [P10] bw = 12, d = 20, f′c = 4 ksi, Av = 0.62 in², fy = 60, s = 4 in. Vn considering Vs ≤ 8√f′c·bw·d?
4. [P11] Same beam, Vu = 110 k. φVn and adequacy?
5. [P12] bw = 12, d = 20, f′c = 4, Vu = 20 k. Stirrup requirement?
6. [P13] bw = 12, d = 24, f′c = 4, Av = 0.40, fy = 60, Vu = 18 k. Required s (in)?
7. [P14] Same section, Vu = 60 k. Required s (in)?
8. [P15] Same section, Vu = 150 k. Required s (in)?
9. [P31] Simply supported W, L = 24 ft, wD = 0.9, wL = 1.2 k/ft, A992, d = 18, tw = 0.35 in. Vu and φvVn?
10. Which AISC 360-22 chapter primarily governs beam shear design?
11. 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 beam shear design. 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.