Free Web Tool

Development & Splice Length

Compute tension and compression development lengths, hook anchorage, and Class A / Class B lap splices per ACI 318-19 §25.4.

Inputs

References — ACI 318-19 §25.4
  • Tension: ld = (fy·ψt·ψe·ψs) / (1.1·λ·√f'c · ((cb+Ktr)/db)) · db. Simplified form used here with (cb+Ktr)/db = 1.5.
  • Compression: ldc = max( 0.24·fy·db/(λ·√f'c) , 0.043·fy·db ).
  • Hooked bar (90°/180°): ldh = 0.24·ψe·ψc·ψr·fy·db / (λ·√f'c) ≥ 8db, 150 mm.
  • Splice: Class A = 1.0·ld, Class B = 1.3·ld; minimum 300 mm.

Results

Tension Development ld
Compression Development ldc
90° Hook ldh
Tension Splice (Class A)
Tension Splice (Class B)
Compression Splice

About this calculator

This tool computes ACI 318-style tension and compression development lengths, standard hook anchorage length, and Class A / Class B tension lap-splice lengths (plus a compression lap length) for a single deformed reinforcing bar. It's meant for quick bar-anchorage and splice checks by engineers and detailers working out rebar layouts.

Method & formulas

Tension development length follows the ACI 318 confinement-based expression, using a fixed, conservative confinement term in place of a case-by-case (cb+Ktr)/db calculation:

ld = [fy·ψt·ψe·ψs / (1.1·λ·√f'c · (cb+Ktr)/db)] · db , with (cb+Ktr)/db = 1.5 assumed

Compression development and standard hook anchorage:

ldc = max(0.24·fy·db/(λ√f'c), 0.043·fy·db), not less than 200 mm ldh = max(0.24·ψe·fy·db/(λ√f'c), 8db, 150 mm)

Lap splice lengths, and the modifier factors used above:

Tension splice: Class A = 1.0·ld ; Class B = 1.3·ld ; minimum 300 mm Compression splice ≈ ldc , minimum 300 mm ψt = top-bar factor (1.3 when >300 mm fresh concrete is cast below the bar) ψe = coating factor (1.0 uncoated, up to 1.5 epoxy-coated with tight cover/spacing) ψs = bar-size factor (0.8 for db ≤ 20 mm, 1.0 otherwise) ; λ = 0.75 lightweight, 1.0 normal-weight

Worked example

Tool defaults: db=16 mm, f'c=28 MPa, fy=420 MPa, ψtes=λ=1.0.

√f'c = 5.29 ld = (420×1×1×1) / (1.1×1×5.29×1.5) × 16 ≈ 770 mm ldc = max(0.24×420×16/5.29, 0.043×420×16, 200) = max(305, 289, 200) ≈ 305 mm ldh = max(0.24×420×16/5.29, 128, 150) ≈ 305 mm Class A splice = max(770, 300) ≈ 770 mm Class B splice = 1.3 × 770 ≈ 1001 mm Compression splice = max(305, 300) ≈ 305 mm

Assumptions & limitations

FAQ

Q: When is Class A allowed instead of Class B?

A: Class A applies only when the steel area provided is at least twice what analysis requires over the full splice length, and no more than half the bars are spliced at that location; otherwise Class B (1.3·ld) governs.

Q: Why doesn't the compression splice use a class factor?

A: Compression lap splices aren't classified A/B like tension splices — the length is based directly on the compression development length ldc, subject to a 300 mm absolute minimum.

Q: Does this account for confining transverse reinforcement (Ktr)?

A: No — it uses a fixed conservative value for (cb+Ktr)/db rather than computing Ktr from your actual stirrup/tie layout, so results correspond to the simplified-table equivalent rather than the full expression.