Tension and compression lap lengths for reinforcing bars per IS 456:2000 Cl 26.2.1 & 26.2.5 — M20–M60 concrete, Fe415–Fe600 steel, deformed or plain bars.
Laps not permitted above 36 mm — weld instead (Cl 26.2.5.1(a))
Result
mm
lap length ≈ Ø
Development length (IS 456 Cl 26.2.1): Lₗ = Ø·σₖ / (4·τₓₖ) with σₖ = 0.87·fₔ.
Design bond stress τₓₖ for plain bars in tension (Cl 26.2.1.1): M20 1.2 · M25 1.4 · M30 1.5 · M35 1.7 · M40 and above 1.9 N/mm². For deformed bars conforming to IS 1786, multiply by 1.60; for bars in compression, multiply by 1.25.
Lap length (Cl 26.2.5.1): flexural tension max(Lₗ, 30Ø) · direct tension max(2·Lₗ, 30Ø) · compression max(Lₗ in compression, 24Ø). The straight portion of a lap must be at least max(15Ø, 200 mm).
All values verified against IS 456:2000 (AMD 5) — see our verified constants table.
Given: M25 beam, Fe500D bars, Ø16, flexural tension, deformed.
τₓₖ = 1.4 × 1.6 = 2.24 N/mm² → Lₗ = 16 × 435 / (4 × 2.24) = 777 mm (48.5Ø)
Lap = max(777, 30×16 = 480) = 777 mm. Stagger successive laps ≥ 1.3 × 777 = 1010 mm apart.
No. Development length Lₗ anchors a bar's stress into surrounding concrete (Cl 26.2.1). A lap transfers force from one bar to another through the concrete between them — the lap is built on Lₗ but has its own minimums (30Ø tension, 24Ø compression) and staggering rules (Cl 26.2.5).
IS 456 Cl 26.2.5 says not more than half the bars should be spliced at a section. Where unavoidable — or where splices fall at points of maximum stress — take special precautions such as longer laps or closely-spaced stirrups/spirals around the splice.
Bond stress in compression is 25% higher than in tension (Cl 26.2.1.1), so Lₗ in compression is shorter, and Cl 26.2.5.1(d) permits the compression lap to equal that shorter Lₗ (with a 24Ø floor) instead of the 30Ø tension minimum.
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