Interactive lap length calculator
Simplified EC2 method: straight bars in tension, full steel utilisation (σsd = fyd), α1=α2=α3=α5=1.0.
—design lap length l0 (rounded up to 5 mm)
For the full check — hooks, bends, compression laps, cover and spacing effects, custom fck/fyk — use SteelCut’s built-in EC2 anchorage & lap calculator in the app.
The formula (EN 1992-1-1 §8.7.3)
A lap transfers force from one bar to the next through bond with the concrete. EC2 computes the design lap length in five steps:
- Design tensile strength of concrete: fctd = fctk,0.05 / γc, with fctk,0.05 = 0.7 · fctm and fctm = 0.3 · fck2/3 (classes up to C50/60). γc = 1.5.
- Design bond stress: fbd = 2.25 · η1 · η2 · fctd, where η1 = 1.0 for good bond (0.7 for poor) and η2 = 1.0 for Φ ≤ 32 mm, otherwise (132 − Φ)/100.
- Basic required anchorage length: lb,rqd = (Φ/4) · (σsd / fbd), with σsd = fyd = fyk/1.15 at full utilisation.
- Design lap length: l0 = α1α2α3α5α6 · lb,rqd. For straight bars the first four factors are 1.0, so l0 = α6 · lb,rqd, with α6 = √(ρl/25), limited to 1.0–1.5 (ρl = % of bars lapped at the section).
- Minimum: l0 ≥ l0,min = max(0.3·α6·lb,rqd, 15Φ, 200 mm).
Worked example
Φ16, C25/30, good bond, B500, 100% of bars lapped at the section, tension:
- fctd = 0.7 · 0.3 · 252/3 / 1.5 = 1.20 MPa
- fbd = 2.25 · 1.0 · 1.0 · 1.20 = 2.69 MPa
- lb,rqd = (16/4) · (434.8/2.69) = 646 mm (40.4Φ)
- α6 = 1.5 (100% lapped) → l0 = 1.5 · 646 = 969 mm
- l0,min = max(291, 240, 200) = 291 mm < 969 mm — does not govern
Result: design lap length 970 mm (rounded up to 5 mm). On site this is usually specified as 970 mm or 1000 mm.
Quick reference table
Lap lengths in mm — tension, B500, good bond, 100% lapped at the section (α6 = 1.5), straight bars, rounded up to 5 mm:
| Φ (mm) | C20/25 | C25/30 | C30/37 |
|---|---|---|---|
| 8 | 565 | 485 | 430 |
| 10 | 705 | 610 | 540 |
| 12 | 845 | 730 | 645 |
| 16 | 1125 | 970 | 860 |
| 20 | 1405 | 1215 | 1075 |
| 25 | 1760 | 1515 | 1345 |
| 32 | 2250 | 1940 | 1720 |
Common mistakes
- Forgetting α6. Lapping every bar at one section without the 1.5 factor under-designs the lap by a third. Stagger laps where you can.
- Assuming good bond everywhere. Bars in the top zone of deep pours (and sections over 600 mm) are “poor” bond: η1 = 0.7, which makes laps ~43% longer.
- Ignoring l0,min. For small diameters the minimum can govern — always check it.
- Mixing up anchorage and lap. The lap (l0) is longer than the anchorage (lbd) because α6 ≥ 1.0. They are not interchangeable.
Assumptions & limits: this page uses the simplified method — straight bars, tension, full steel utilisation, α1=α2=α3=α5=1.0, γc=1.5, γs=1.15. National annexes can modify values — always check yours. For hooks, bends, compression laps and cover/spacing effects, use SteelCut’s full EC2 calculator in the app or consult your structural engineer.
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