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LecturesModule 24

RC — Development Length and Splices

Tension and compression development length, hooked bars, and lap splices per ACI Chapter 25.

AISC Reference Box
  • ACI 318-19 §25.4.2 — Development of Deformed Bars in Tension
  • ACI §25.4.3 — Standard Hooks in Tension
  • ACI §25.4.9 — Development of Deformed Bars in Compression
  • ACI §25.5 — Splices of Reinforcement
  • ACI §25.3 — Standard Hook Geometry

Lecture Notes

Development length ℓd is the embedment required for a reinforcing bar to develop its yield strength fy by bond with the concrete (ACI §25.4). Bars terminated short of ℓd cannot reach yield and the beam fails prematurely in bond/splitting.

Tension development length (ACI §25.4.2.4 — simplified equation): ℓd / db = (3/40)·(fy / (λ·√f'c)) · (ψt·ψe·ψs·ψg / cb + Ktr/db). For typical bottom bars, uncoated, normal-weight concrete, with adequate cover: ψt = ψe = ψs = 1.0 and ψg = 1.0; the bracket simplifies and minimum ℓd = 12 in.

Modification factors: ψt = 1.3 for top bars (more than 12 in. of fresh concrete below); ψe = 1.5 for epoxy-coated bars with low cover; ψs = 0.8 for #6 and smaller; ψg = 1.0 for Grade 60, 1.15 for Grade 80, 1.3 for Grade 100; λ = 0.75 for lightweight concrete.

Standard hooks (ACI §25.4.3) — when straight ℓd cannot fit, use a 90° or 180° hook. Hooked development length ℓdh = (fy·ψe·ψr·ψo·ψc/(55·λ·√f'c))·db^1.5 with minimum 8db and 6 in. Hook geometry: 90° hook has a tail extension of 12db; 180° hook has 4db (≥ 2.5 in.) past the bend.

Compression development length ℓdc (ACI §25.4.9): ℓdc = max( 0.0003·fy·ψr·db, (fy·ψr / (50·λ·√f'c))·db, 8 in. ). Compression development is roughly half of tension because bond is enhanced by the bar bearing on the concrete.

Lap splices: tension lap splice Class A = 1.0·ℓd, Class B = 1.3·ℓd (ACI §25.5.2.1). Class B applies when more than 50 % of bars are spliced at the same location OR when As,provided/As,required < 2. Compression lap = 0.0005·fy·db (fy ≤ 60 ksi) or (0.0009·fy − 24)·db (fy > 60 ksi), minimum 12 in. (ACI §25.5.5).

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 24 — Development length & splices
Bottom #8 bars from Chapter 22 anchored into the support
Demand carried forward
Bars must develop fy = 60 ksi at the face of the interior column Bottom bars (not top), uncoated, NWC: ψt = ψe = 1.0 #8 → ψs = 1.0 ψg = 1.0 λ = 1.0.
This chapter contributes
Tension ℓd (simplified ACI Eq. 25.4.2.4) ≈ (fy·ψt·ψe·ψs / (20·λ·√f'c))·db = (60,000·1·1·1)/(20·1·√4,000) · 1.0 ≈ 47 in. Provides standard 90° hook (ℓdh ≈ 15 in) where straight ℓd doesn't fit, or Class B lap = 1.3·ℓd at midspan splices.
Feeds into next chapter
Confirms the column-to-beam joint detail used in Chapter 25 (column ties pass through the joint).
ℓdStraight tension barℓdh90° hookHooked bar (ACI §25.4.3)ℓd = (fy·Πψ)/(20 or 25·λ√f'c)·db ≥ 12 in. • ℓdh ~ db^1.5
Tension ℓd = (fy·Πψ)/(20 or 25·λ√f'c)·db ≥ 12 in.; hooked ℓdh ∝ db^1.5 (ACI §25.4).

Formula Sheet

NameEquationAISC Ref
Tension ℓd (simplified, normal cases)ℓd = (fy·ψt·ψe·ψs·ψg) / (20·λ·√f'c) · db (#6 and smaller)ACI Table 25.4.2.3
Tension ℓd (#7 and larger)ℓd = (fy·ψt·ψe·ψs·ψg) / (25·λ·√f'c) · dbACI Table 25.4.2.3
Hooked barℓdh = (fy·ψe·ψr·ψo·ψc) / (55·λ·√f'c) · db^1.5 ≥ max(8db, 6 in.)ACI §25.4.3.1
Compression developmentℓdc = max( 0.0003·fy·ψr·db , (fy·ψr/(50·λ·√f'c))·db , 8 in. )ACI §25.4.9.2
Tension lap (Class A / B)ℓst = 1.0·ℓd (Class A) or 1.3·ℓd (Class B) ≥ 12 in.ACI §25.5.2.1
Compression lap (fy ≤ 60 ksi)ℓsc = 0.0005·fy·db ≥ 12 in.ACI §25.5.5.1

Worked Example

Tension development length of a #8 bottom bar

Given
Bottom #8 bar (db = 1.00 in.) in a beam. f'c = 4000 psi NWC (λ = 1.0), fy = 60,000 psi (Grade 60). Uncoated, adequate cover and spacing.
Load combination
Detailing check — no load combination.
Required strength
Compute ℓd and Class B lap splice length.
Limit states
  • Tension development length ℓd
  • Class B lap splice ℓst
AISC reference
ACI 318-19 §25.4.2 and §25.5.2
Solution steps
  1. 1. Given
    db = 1.00 in. (#8). fy = 60,000 psi (Grade 60). f'c = 4000 psi. λ = 1.0 (normal weight).
  2. 2. Find
    ℓd (tension) and Class B lap length.
  3. 3. ACI reference
    ACI Table 25.4.2.3 (#7 and larger) and §25.5.2.1 (Class B = 1.3·ℓd).
  4. 4. Assumptions / factors
    ψt = 1.0 (bottom bar). ψe = 1.0 (uncoated). ψs = 1.0 (#8, not small bar). ψg = 1.0 (Grade 60).
  5. 5. Tension ℓd
    ℓd = (fy·ψt·ψe·ψs·ψg)/(25·λ·√f'c) · db. ℓd = (60,000·1·1·1·1)/(25·1.0·√4000) · 1.00. ℓd = 60,000 / (25·63.25) · 1.00. ℓd = 60,000 / 1581.1. ℓd = 37.95 in. ℓd ≈ 38 in.
  6. 6. Minimum check
    ℓd,min = 12 in. 38 in. > 12 in. ✓
  7. 7. Class B lap splice
    ℓst = 1.3·ℓd. ℓst = 1.3·38. ℓst = 49.4 in. Use ℓst = 50 in.
  8. 8. Design check
    Provide straight embedment ≥ 38 in. past the point where the bar is fully stressed. If splicing > 50% of bars at one location: use 50 in. lap (Class B).
Final design decision
ℓd = 38 in. and Class B tension lap = 50 in. for a Grade 60 #8 bottom bar in 4000 psi NWC.
Common mistakes in this example
  • Using the (20·λ·√f'c) denominator for #7 or larger bars (must use 25·λ·√f'c).
  • Forgetting ψt = 1.3 for top bars when more than 12 in. of fresh concrete is below.
  • Using 0.8 ψs for #7 and larger (it is only for #6 and smaller).
  • Forgetting the 12 in. minimum on ℓd and lap splices.

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
#8 bar (db = 1.0 in.); bottom bar; uncoated; NWC; f'c = 4 ksi; fy = 60 ksi; (cb+Ktr)/db = 2.5 (use max).
AISC Reference
ACI 318-19 §25.4.2.4
Step-by-step solution
  1. 1. Problem Statement
    Compute the tension development length using the general expression.
  2. 2. Variables & Values
    db = 1.0 ψt = 1.0 (bottom) ψe = 1.0 (uncoated) ψs = 1.0 (#7+) λ = 1.0 f'c = 4,000 fy = 60,000 (cb+Ktr)/db capped at 2.5.
  3. 3. Approach
    Use ld = [3/40·fy/(λ·√f'c)·(ψt·ψe·ψs)/((cb+Ktr)/db)]·db.
  4. 4. Formula Setup
    ld = (3·fy·ψt·ψe·ψs)/(40·λ·√f'c·((cb+Ktr)/db))·db (in.)
  5. 5. Substitution
    ld = (3·60,000·1·1·1)/(40·1·√4,000·2.5)·1.0 = 180,000/(40·63.25·2.5) = 180,000/6,325 = 28.5 in.
  6. 6. Solution Steps
    Round up to ld = 29 in (≥ 12 in. minimum, OK).
Answer ld ≈ 29 in.
w (kip/ft)LSimply supported beam, UDL w

Practice Problems

  1. [E] Minimum ℓd floor = 12 in.
  2. [E] ψt = 1.3 for top bars (>12 in. fresh concrete below).
  3. [E] ψe = 1.5 for epoxy with low cover.
  4. [E] ψs = 0.8 for #6 and smaller; 1.0 for #7 and larger.
  5. [E] Class A = 1.0·ℓd; Class B = 1.3·ℓd.
  6. [M] ℓd for a #6 BOTTOM bar, uncoated, f'c = 4 ksi NWC, fy = 60 ksi.
  7. [M] ℓd for a #5 TOP bar, uncoated, f'c = 5 ksi NWC, fy = 60 ksi.
  8. [M] ℓdh for 90° hook on #6 bar, f'c = 4, fy = 60, side cover ≥ 2.5 in.
  9. [M] Compression ℓdc for a #9 column bar, fy = 60, f'c = 4.
  10. [M] Class B tension lap for a #8 bar, fy = 60, f'c = 4.
  11. [H] At a beam-column joint a #8 negative-moment bar terminates inside column — required embedment or hook.
  12. [H] All bars spliced at one location in a beam: identify splice class B; compute ℓst for a #9.
  13. [H] Grade 100 rebar (#7), f'c = 6 ksi NWC, ψg = 1.3 — compute ℓd.
  14. [H] Bar cutoff diagram for a continuous beam with two #9 bent up — verify ld and 1/3 development per §9.7.3.
  15. [H] Column compression lap: 4 #10 verticals, all spliced at floor, fy = 60, f'c = 5. Verify §10.7.5.
Structured Clues
  • #7 and larger bars use the 25·λ·√f'c denominator, not 20.
  • Class B lap = 1.3·ℓd whenever > 50% of bars splice at one location.
  • Hook length depends on db^1.5 — small bar size penalty disappears quickly.
Code References
  • ACI 318-19 §25.4, §25.5

Quiz

1. Tension development length for #7 and larger uses the denominator:
2. Top-bar factor ψt is taken as:
3. Class B tension lap splice length is:
4. Minimum compression lap splice length is:
5. The hooked-bar development length ℓdh depends on db raised to which power?
6. Compute ℓd (in.) for a #6 BOTTOM bar, f'c = 4000 psi, fy = 60 ksi (small-bar formula).
7. For a TOP #6 bar (ψt = 1.3) instead of bottom, ℓd would be approximately:
8. Class B tension lap of a Grade 60 #8 bar with f'c = 4000 psi, bottom, normal cover (use ℓd = 38 in.):
9. Compute ℓdh (in.) for #6 90° hook, ψ-product = 1.0, f'c = 4000 psi.
10. Compression development length ℓdc lower bound is:
11. For a #9 bar (fy = 60 ksi), compression lap length 0.0005·fy·db is:
12. Lightweight-concrete factor λ in development equations equals:
13. Top-bar factor ψt is applied when:
14. If 100% of bars are spliced at one location, the splice class is:
15. For #5 bar (db = 0.625), fy = 60 ksi, f'c = 5000 psi, ψ-product = 1.0, bottom, ℓd (small-bar) ≈

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 — development length and splices. 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.