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LecturesModule 22

RC — Flexural Design of Beams

Singly reinforced, doubly reinforced, and T-beams per ACI 318 Chapter 9 with the Whitney stress block.

AISC Reference Box
  • ACI 318-19 Ch. 9 — Beams — Strength Design
  • ACI §22.2 — Design Assumptions for Strength
  • ACI §22.2.2.4 — Equivalent Rectangular (Whitney) Stress Block
  • ACI §9.6.1 — Minimum Flexural Reinforcement
  • ACI §21.2.2 — Strength Reduction Factor φ

Lecture Notes

Flexural design of RC beams uses the Whitney equivalent rectangular stress block (ACI §22.2.2.4): the actual non-linear concrete compressive stress distribution is replaced by a uniform stress of 0.85·f'c acting over a depth a = β1·c, where c is the neutral-axis depth and β1 = 0.85 for f'c ≤ 4000 psi, decreasing 0.05 per 1000 psi above 4000 psi, with a floor of 0.65.

Singly reinforced rectangular beams: equilibrium ΣF = 0 gives a = As·fy / (0.85·f'c·b). Nominal moment Mn = As·fy·(d − a/2), and design moment φMn = 0.90·Mn for tension-controlled sections (εt ≥ 0.005). ACI requires As ≥ As,min = max(3·√f'c·bw·d / fy, 200·bw·d / fy) per §9.6.1.2.

Tension-controlled vs compression-controlled: check the net tensile strain εt = 0.003·(dt − c)/c. If εt ≥ 0.005 the section is tension-controlled (φ = 0.90); if εt ≤ fy/Es = εty the section is compression-controlled (φ = 0.65); in between is the transition region with linear-interpolated φ (ACI §21.2.2).

Doubly reinforced beams add compression steel A's when the singly reinforced design cannot fit the required Mu in the allowed depth. The compression steel contribution is (A's·f's)·(d − d'). When the strain compatibility shows compression steel yielding, f's = fy. Otherwise compute f's = Es·ε's < fy.

T-beams (ACI §6.3.2): the effective flange width bf is the least of (1) L/4, (2) bw + 16·hf, (3) center-to-center spacing of beams. If the compressive block depth a ≤ hf, the beam behaves as a rectangular beam with width bf. If a > hf, split into flange-overhang force and web force.

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 22 — Flexure (typical interior floor beam)
30 ft simply-supported floor beam, b = 14 in, h = 24 in, d ≈ 21.5 in
Demand carried forward
Tributary 10 ft of slab: wD = 0.110 ksf · 10 + self-weight ≈ 1.35 klf wL = 0.050 · 10 = 0.50 klf. wu = 1.2(1.35) + 1.6(0.50) = 1.62 + 0.80 = 2.42 klf Mu = wu L²/8 = 2.42 · 30²/8 = 272 k-ft.
This chapter contributes
Sizes longitudinal tension steel with the Whitney block: required As ≈ Mu/(φ·fy·(d − a/2)) — for Mu = 272 k-ft, As ≈ 3.1 in² → try 4 #8 (As = 3.16 in²). Verifies εt ≥ 0.005 (tension-controlled, φ = 0.90) and As ≥ As,min = max(3√f'c·bw·d/fy, 200·bw·d/fy).
Feeds into next chapter
d and As selected here drive Chapter 23 (Vc depends on d) and Chapter 24 (ℓd of the bottom bars at supports).
bdAsεcu = 0.003εt ≥ 0.005c0.85 f'ca = β1·cC = 0.85f'c·b·aT = As·fy
Whitney block: 0.85·f'c over depth a = β1·c; C = T → As·fy = 0.85·f'c·b·a; Mn = As·fy·(d − a/2).

Formula Sheet

NameEquationAISC Ref
Equivalent stress block deptha = (As·fy) / (0.85·f'c·b)ACI §22.2.2.4
Stress block factorβ1 = 0.85 − 0.05·(f'c − 4000)/1000 (0.65 ≤ β1 ≤ 0.85)ACI §22.2.2.4.3
Neutral axis depthc = a / β1ACI §22.2.2.4
Nominal moment (singly)Mn = As·fy·(d − a/2)ACI §22.3
Design momentφ Mn ≥ Mu (φ = 0.90 if εt ≥ 0.005)ACI §9.5.1
Net tensile strainεt = 0.003·(dt − c)/cACI §22.2.1
Minimum steelAs,min = max( 3√f'c·bw·d / fy , 200·bw·d / fy )ACI §9.6.1.2
Doubly reinforcedMn = (As − A's)·fy·(d − a/2) + A's·fy·(d − d')ACI §22.3

Worked Example

Singly reinforced rectangular beam — design moment check

Given
Rectangular beam b = 12 in., h = 22 in., d = 19.5 in. Reinforcement: 3 #8 bars (As = 3·0.79 = 2.37 in²). Materials: f'c = 4000 psi (β1 = 0.85), fy = 60,000 psi. Factored moment Mu = 165 k·ft.
Load combination
Mu = 165 k-ft (assumed already factored per ASCE 7).
Required strength
Verify φMn ≥ Mu = 165 k-ft.
Limit states
  • Flexural strength φMn
  • Minimum steel
  • Tension-controlled check (φ = 0.90)
AISC reference
ACI 318-19 §9.5 and §22.2
Solution steps
  1. 1. Given
    b = 12 in. d = 19.5 in. As = 2.37 in². f'c = 4000 psi. fy = 60,000 psi. β1 = 0.85.
  2. 2. Find
    Verify φMn ≥ Mu = 165 k-ft (1980 k-in).
  3. 3. ACI reference
    ACI 318-19 §22.2.2.4 (Whitney block) and §9.5.1.1 (φMn ≥ Mu).
  4. 4. Assumptions
    Tension steel yields (verify εt later). Whitney stress block valid. No compression steel.
  5. 5. Stress-block depth a
    a = (As·fy)/(0.85·f'c·b). a = (2.37·60,000)/(0.85·4000·12). a = 142,200 / 40,800. a = 3.485 in.
  6. 6. Neutral axis c and εt
    c = a/β1. c = 3.485/0.85. c = 4.100 in. εt = 0.003·(d − c)/c. εt = 0.003·(19.5 − 4.10)/4.10. εt = 0.003·3.756. εt = 0.01127 > 0.005 → tension-controlled. φ = 0.90.
  7. 7. Nominal moment Mn
    Mn = As·fy·(d − a/2). Mn = 2.37·60,000·(19.5 − 3.485/2). Mn = 142,200·(19.5 − 1.7425). Mn = 142,200·17.7575. Mn = 2,525,308 lb·in. Mn = 2,525 k·in. Mn = 210.5 k·ft.
  8. 8. Design moment φMn
    φMn = 0.90·210.5. φMn = 189.4 k·ft.
  9. 9. Minimum steel check
    As,min = max(3·√4000·12·19.5/60,000, 200·12·19.5/60,000). As,min = max(3·63.25·12·19.5/60,000, 0.780). As,min = max(0.740, 0.780). As,min = 0.780 in² < 2.37 in² ✓
  10. 10. Design check
    φMn = 189.4 k·ft ≥ Mu = 165 k·ft ✓ Ratio = 189.4/165 = 1.15 (about 15% reserve).
Final design decision
Use the trial section b×h = 12×22 in. with d = 19.5 in. and 3 #8 bottom bars. Section is tension-controlled (φ = 0.90) and exceeds As,min.
Common mistakes in this example
  • Forgetting to convert Mu to k-in (or Mn to k-ft) before comparing.
  • Using 0.85 for β1 when f'c > 4000 psi without applying the reduction.
  • Assuming φ = 0.90 without checking εt.
  • Skipping the As,min check — required by ACI §9.6.1.

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
b = 12 in., d = 20 in., As = 3·#8 = 2.37 in²; f'c = 4 ksi; fy = 60 ksi.
AISC Reference
ACI 318-19 §22.2
Step-by-step solution
  1. 1. Problem Statement
    Compute φMn for a tension-controlled rectangular beam.
  2. 2. Variables & Values
    b = 12 d = 20 As = 2.37 in² f'c = 4,000 psi fy = 60,000 psi β1 = 0.85.
  3. 3. Approach
    Find a (stress-block depth) → check εt → compute Mn → apply φ.
  4. 4. Formula Setup
    a = As·fy/(0.85·f'c·b) c = a/β1 εt = 0.003·(d−c)/c Mn = As·fy·(d − a/2) φMn = φ·Mn.
  5. 5. Substitution
    a = 2.37·60/(0.85·4·12) = 142.2/40.8 = 3.49 in. c = 3.49/0.85 = 4.10 in. εt = 0.003·(20−4.10)/4.10 = 0.01164 > 0.005 → tension-controlled, φ = 0.90 Mn = 2.37·60·(20 − 1.74) = 142.2·18.26 = 2,597 k·in.
  6. 6. Solution Steps
    φMn = 0.90·2,597 = 2,337 k·in = 195 k·ft.
Answer φMn ≈ 195 k·ft (tension-controlled, εt = 0.012).
w (kip/ft)LSimply supported beam, UDL w

Practice Problems

  1. [E] State a = As·fy/(0.85·f'c·b).
  2. [E] State Mn = As·fy·(d − a/2).
  3. [E] State β1 for f'c = 3000, 4000, 5000, 6000 psi.
  4. [E] State tension-controlled limit εt ≥ 0.005.
  5. [E] State As,min per §9.6.1.2.
  6. [M] b = 14 in., d = 21 in., As = 4 #8 (3.16 in²), f'c = 4 ksi, fy = 60 ksi. Compute φMn.
  7. [M] Max Mu for a 12 x 20 in. (d = 17.5) beam with 3 #7 bars.
  8. [M] Verify tension-controlled: As = 4 #9 in b = 12 in., d = 21 in., f'c = 4 ksi.
  9. [M] T-beam bf = 36, bw = 12, hf = 5, d = 22 in., As = 6 #8 — compute φMn.
  10. [M] 12 x 24 beam (d = 21) carrying Mu = 240 k-ft — design singly reinforced As.
  11. [H] Doubly reinforced: b = 12, d = 22, d' = 2.5, As = 5 #9, A's = 2 #6, f'c = 4 ksi. Verify A's yields; compute φMn.
  12. [H] Continuous beam negative-moment region: As for Mu− = 320 k-ft, b = 14, d = 23, f'c = 5 ksi.
  13. [H] T-beam with a > hf: bf = 40, bw = 14, hf = 4, d = 22, As = 8 #9. Compute φMn (flange + web split).
  14. [H] Design singly reinforced beam for Mu = 350 k-ft, choosing b and d for ρ ≈ 0.5·ρmax.
  15. [H] Plot Mn vs εt for a 12 x 20 in. beam from As = 2.0 to 6.0 in²; identify the tension-controlled boundary.
Structured Clues
  • Always check εt to confirm tension-controlled (φ = 0.90).
  • β1 decreases 0.05 per 1000 psi over 4000 psi; floor at 0.65.
  • Don't forget As,min = max(3√f'c·bw·d/fy, 200·bw·d/fy).
Code References
  • ACI 318-19 §22.2, 9.5, 9.6.1, 21.2.2

Quiz

1. The Whitney stress block uses a uniform concrete stress of:
2. For f'c = 5000 psi, β1 equals:
3. A flexural section is tension-controlled (φ = 0.90) when:
4. Compute a (in.) for As = 2.0 in², fy = 60 ksi, f'c = 4 ksi, b = 12 in.:
5. Minimum flexural steel per ACI §9.6.1.2 is the LARGER of 3√f'c·bw·d/fy and:
6. Compute a (in.) for As = 3.16 in², fy = 60 ksi, f'c = 4 ksi, b = 14 in.
7. For a = 3.98 in. and d = 21 in., As = 3.16 in², fy = 60 ksi, compute Mn (k-ft).
8. If c = 4.20 in. and dt = 19.5 in., net tensile strain εt equals:
9. As,min (in²) per 200·bw·d/fy for bw = 12, d = 20, fy = 60 ksi is:
10. Effective T-beam flange width — for L = 30 ft, bw = 12 in., centerline spacing 10 ft, hf = 5 in. — is the SMALLEST of:
11. For the values above, bf (in.) =
12. A tension-controlled rectangular beam with φ = 0.90, b = 12, d = 20, f'c = 4, fy = 60 ksi, As = 2.0 in² has φMn (k-ft) closest to:
13. Doubly reinforced beam Mn formula requires verifying that compression steel:
14. Reinforcement ratio ρ = As/(bd) for As = 3.0 in², b = 12, d = 18 is:
15. For f'c = 5000 psi, fy = 60 ksi, the balanced ratio ρb = 0.85β1(f'c/fy)·(εcu/(εcu+εy)) gives ρb ≈

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 — flexural 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.