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LecturesModule 23

RC — Shear Design of Beams

Concrete Vc, vertical stirrups Vs, spacing limits, and minimum shear reinforcement per ACI Chapter 22.

AISC Reference Box
  • ACI 318-19 Ch. 22.5 — One-Way Shear Strength
  • ACI §9.5.3 — Beam Shear Strength Requirement
  • ACI §9.6.3 — Minimum Shear Reinforcement
  • ACI §9.7.6.2 — Stirrup Spacing Limits
  • ACI Table 21.2.1 — φ for Shear = 0.75

Lecture Notes

Shear design of RC beams uses Vu ≤ φVn, where Vn = Vc + Vs (ACI §22.5.1.1). Vc is the concrete contribution and Vs is the contribution of transverse reinforcement (stirrups). φ = 0.75 for shear (ACI Table 21.2.1).

Concrete shear strength (ACI Table 22.5.5.1, simplified): Vc = 2·λ·√f'c·bw·d (psi units) for non-prestressed members without axial load when Av ≥ Av,min. ACI 318-19 also provides a detailed equation that includes the ratio of longitudinal reinforcement and axial load. For routine building beams the 2λ√f'c form is acceptable.

Stirrup contribution: Vs = Av·fyt·d / s where Av is the total cross-sectional area of stirrup legs crossing the crack (2·Ab for a single U-stirrup), fyt is the stirrup yield strength (60 ksi typical), and s is the spacing. Required stirrup spacing s = Av·fyt·d / (Vu/φ − Vc).

Maximum spacing (ACI §9.7.6.2.2): smax = min(d/2, 24 in.) when Vs ≤ 4·√f'c·bw·d; smax = min(d/4, 12 in.) when Vs > 4·√f'c·bw·d. Vs must not exceed 8·√f'c·bw·d — beyond that, the cross-section must be enlarged or f'c increased.

Minimum shear reinforcement (ACI §9.6.3) is required when Vu > 0.5·φVc, with Av,min/s = max(0.75·√f'c·bw/fyt, 50·bw/fyt). Critical section is located at distance d from the face of support (ACI §9.4.3.2) for typical beams supported by compression on bottom.

Project case study — Cardinal Square RC — 4-story cast-in-place concrete 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 @ 30 ft × 30 ft. Stories: 4 @ 13 ft (52 ft roof). Floor: 7 in one-way slab on RC beams (b = 14 in, h = 24 in) framing to RC columns (16 in × 16 in tied) on grid. Roof: same beam grid + 6 in slab. Materials: Normal-weight concrete f'c = 4,000 psi (λ = 1.0). Reinforcement Grade 60 (fy = 60,000 psi); ties/stirrups Grade 60 #3 / #4. Clear cover: 1.5 in interior beams/slabs, 2 in columns, 3 in footings (ACI §20.5.1).

Chapter 23 — Shear at the beam supports
Same 30 ft floor beam — shear at face of column
Demand carried forward
From Chapter 22: Vu = wu L/2 = 2.42 · 30/2 = 36.3 k Check at d from face: Vu,d ≈ 33.9 k.
This chapter contributes
Computes φVc = 0.75·2·1.0·√4,000·14·21.5/1000 ≈ 28.5 k. Since Vu,d > φVc, designs #3 U-stirrups: Vs req = Vu/φ − Vc = 45.3 − 38.0 = 7.3 k → s ≤ Av·fy·d/Vs = 0.22·60·21.5/7.3 ≈ 39 in, but limit s ≤ d/2 = 10.75 in governs. Use #3 @ 10 in.
Feeds into next chapter
Stirrup detailing here is anchored using the hook development length checked in Chapter 24.
crack ~45°Vu ≤ φ(Vc + Vs), φ = 0.75Vc = 2λ√f'c·bw·d ; Vs = Av·fyt·d/s ; smax = d/2 (or d/4)s
Stirrups cross the 45° shear crack: Vn = Vc + Vs; smax = d/2 (or d/4 if Vs > 4√f'c·bw·d).

Formula Sheet

NameEquationAISC Ref
Design inequalityVu ≤ φ Vn = φ (Vc + Vs)ACI §22.5.1.1
Concrete shear (simplified)Vc = 2·λ·√f'c·bw·d (psi)ACI Table 22.5.5.1
Stirrup contributionVs = Av·fyt·d / sACI §22.5.8.5
Required spacings = Av·fyt·d / (Vu/φ − Vc)ACI §22.5.8.5
Max spacing (Vs ≤ 4√f'c bw d)smax = min(d/2, 24 in.)ACI §9.7.6.2.2
Max spacing (Vs > 4√f'c bw d)smax = min(d/4, 12 in.)ACI §9.7.6.2.2
Av,min /sAv,min/s = max( 0.75·√f'c·bw / fyt , 50·bw / fyt )ACI §9.6.3.3
Upper limitVs ≤ 8·√f'c·bw·dACI §22.5.1.2

Worked Example

Stirrup spacing design for a rectangular beam

Given
Rectangular beam bw = 12 in., d = 20 in., f'c = 4000 psi (λ = 1.0), fyt = 60,000 psi. Factored shear at distance d from support Vu = 45 kips. Use #3 U-stirrups (Av = 2·0.11 = 0.22 in²).
Load combination
Vu = 45 kips at the critical section x = d (already factored).
Required strength
Provide stirrups so that φ(Vc + Vs) ≥ Vu = 45 kips.
Limit states
  • φVc adequacy
  • Stirrup spacing
  • Maximum spacing
  • Minimum reinforcement
AISC reference
ACI 318-19 §22.5 and §9.7.6.2.2
Solution steps
  1. 1. Given
    bw = 12 in. d = 20 in. f'c = 4000 psi, λ = 1.0. fyt = 60,000 psi. Av = 0.22 in² (#3 U). Vu = 45 kips.
  2. 2. Find
    Stirrup spacing s satisfying ACI §22.5.
  3. 3. ACI reference
    Vn = Vc + Vs (ACI §22.5.1.1) with φ = 0.75 (ACI Table 21.2.1).
  4. 4. Concrete shear Vc
    Vc = 2·λ·√f'c·bw·d. Vc = 2·1.0·√4000·12·20. Vc = 2·63.25·12·20. Vc = 30,360 lb. Vc = 30.36 kips.
  5. 5. φVc check
    φVc = 0.75·30.36. φVc = 22.77 kips. 0.5·φVc = 11.4 kips. Vu = 45 > φVc → stirrups required.
  6. 6. Required Vs
    Vs,req = Vu/φ − Vc. Vs,req = 45/0.75 − 30.36. Vs,req = 60.00 − 30.36. Vs,req = 29.64 kips.
  7. 7. Required spacing
    s = Av·fyt·d / Vs,req. s = 0.22·60,000·20 / 29,640. s = 264,000 / 29,640. s = 8.91 in.
  8. 8. Max spacing
    4·√f'c·bw·d = 4·63.25·12·20 = 60,720 lb = 60.72 kips. Vs,req = 29.64 kips < 60.72 → use smax = min(d/2, 24). smax = min(20/2, 24). smax = 10 in.
  9. 9. Minimum reinforcement check
    (Av/s)min = max(0.75·√4000·12/60,000, 50·12/60,000). (Av/s)min = max(0.0095, 0.0100). (Av/s)min = 0.0100 in²/in. Provided Av/s = 0.22/8.5 = 0.0259 ≥ 0.0100 ✓
  10. 10. Design check
    Use s = 8.5 in. ≤ min(8.91, 10) → governs. Vs = 0.22·60,000·20/8.5 = 31,059 lb = 31.1 k. φVn = 0.75·(30.36 + 31.1) = 46.1 k ≥ 45 k ✓
Final design decision
Provide #3 U-stirrups at 8.5 in. on center starting d/2 from the face of support. φVn = 46.1 k ≥ Vu = 45 k.
Common mistakes in this example
  • Forgetting that Av for a U-stirrup is TWO legs (2·Ab).
  • Using fy of longitudinal bars in place of fyt of stirrups.
  • Skipping the maximum spacing limit d/2 (or d/4).
  • Mixing psi and ksi inside √f'c — the 2λ√f'c form is in psi.

FE-Style Worked Examples (5)

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
bw = 14 in., d = 22 in.; f'c = 4 ksi; NWC, λ = 1.0.
AISC Reference
ACI 318-19 §22.5.5.1
Step-by-step solution
  1. 1. Problem Statement
    Compute the simplified concrete shear capacity Vc.
  2. 2. Variables & Values
    bw = 14 d = 22 f'c = 4,000 psi λ = 1.0.
  3. 3. Approach
    Use Vc = 2·λ·√f'c·bw·d.
  4. 4. Formula Setup
    Vc = 2·λ·√f'c·bw·d (lb)
  5. 5. Substitution
    Vc = 2·1.0·√4,000·14·22 = 2·63.25·14·22 = 38,948 lb
  6. 6. Solution Steps
    Vc ≈ 38.9 kips.
Answer Vc ≈ 38.9 k; φVc = 0.75·38.9 = 29.2 k.
w (kip/ft)LSimply supported beam, UDL w

Practice Problems

  1. [E] State Vu ≤ φVn = φ(Vc + Vs); φ = 0.75.
  2. [E] State Vc = 2λ√f'c·bw·d (simplified).
  3. [E] State Vs = Av·fyt·d/s.
  4. [E] State smax = min(d/2, 24 in.) when Vs ≤ 4√f'c·bw·d.
  5. [E] State Vs upper limit 8√f'c·bw·d.
  6. [M] bw = 12, d = 20, f'c = 4 ksi, Vu = 45 k. Design #3 U-stirrups.
  7. [M] Determine if a beam bw = 10, d = 18, Vu = 14 k, f'c = 4 needs stirrups.
  8. [M] bw = 14, d = 22, Vu = 60 k, f'c = 5 ksi — size #3 stirrups.
  9. [M] Compute Av,min/s for bw = 12, f'c = 4, fyt = 60.
  10. [M] When Vs > 4√f'c·bw·d, restate smax.
  11. [H] Vu = 90 k, d = 24, bw = 14, f'c = 4 — choose #3 vs #4 stirrups; report spacing for each.
  12. [H] Design stirrups along full length of simply supported 24 ft beam, wu = 4 k/ft, bw = 14, d = 22, f'c = 4.
  13. [H] Compare simplified vs detailed Vc (with ρw, Nu) for a sample beam.
  14. [H] Critical-section at d from face per §9.4.3.2 — design the max-shear stirrup region.
  15. [H] Deep beam (ℓn/d < 4): why §22.5 does NOT apply; outline strut-and-tie per Ch. 23.
Structured Clues
  • Compute Vc in psi (the 2·λ·√f'c form expects psi).
  • Av for a U-stirrup is TWO legs (2·Ab).
  • Check 0.5·φVc to decide if stirrups are required at all.
Code References
  • ACI 318-19 §22.5, 9.5.3, 9.6.3, 9.7.6.2

Quiz

1. The strength reduction factor for shear is:
2. For a non-prestressed beam, the simplified concrete shear strength (psi) is:
3. When Vs ≤ 4√f'c·bw·d, the maximum stirrup spacing is:
4. The upper-bound shear that any cross-section may carry (Vs limit) is:
5. Av of a single #3 U-stirrup (two legs) is:
6. Compute Vc (kips) for bw = 14, d = 22, f'c = 4000 psi, λ = 1.0.
7. For Vu = 70 k at d, the required Vs equals (φ = 0.75) using Vc above:
8. Spacing s (in.) using #3 U-stirrups (Av = 0.22 in²), fyt = 60 ksi, d = 22, Vs = 54.4 k is:
9. Maximum stirrup spacing when Vs ≤ 4√f'c·bw·d, d = 22 in., is:
10. Upper bound on Vs for the section above (kips):
11. Av of one #4 U-stirrup (two legs) is:
12. Stirrups are required at all when Vu exceeds:
13. If Vs > 4√f'c·bw·d the max spacing reduces to:
14. Critical section for shear in a normally supported beam is at:
15. (Av/s)min for bw = 12, fyt = 60 ksi, f'c = 4000 psi using 50·bw/fyt 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 rc — shear design of 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.