Anchorage length Ld and lap-splice length per IS 456:2000.
Bar type: deformed HYSD (IS 1786) — τ_bd × 1.6
Reference only. Base bond τbd per IS 456 Cl. 26.2.1.1 (×1.6 for deformed bars, +25% in compression); lap splices per Cl. 26.2.5.1 (flexural tension max(Ld, 30φ), direct tension max(2Ld, 30φ), compression max(Ld, 24φ); no laps for bars > 36 mm). Bundled bars per Cl. 26.2.1.2; with excess steel σs is the stress at design load, 0.87fy·As,req/As,prov(Cl. 26.2.1, SP 34). Hooks per Cl. 26.2.2.1 count only in tension. Real detailing also accounts for splice location (stagger laps, Cl. 26.2.5.1) and confinement. Read the clause in the Code Clause Compass.
How the Development & Lap Length Calculator works
A bar can only carry its design stress if it is embedded far enough into the concrete for bond to transfer that force. This length is the development length Ld. The calculator works it out from IS 456:2000 Cl 26.2.1, Ld = φσs / (4τbd), with the bar stress taken as σs = 0.87 fy.
The design bond stress τbd starts from the IS 456 Cl 26.2.1.1 values for plain bars in tension: 1.2, 1.4, 1.5, 1.7 and 1.9 MPa for M20 to M40. Deformed bars get 1.6 times these values and bars in compression a further 25 per cent. The bar type follows the steel grade: Fe 250 is treated as plain mild steel and Fe 415, Fe 500 and Fe 550 as HYSD deformed bars.
The lap length then follows IS 456 Cl 26.2.5.1. In flexural tension it is the greater of Ld and 30φ, in direct tension the greater of 2Ld and 30φ, and in compression the greater of the compression Ld and 24φ. For bars larger than 36 mm the calculator gives no lap length, because Cl 26.2.5.1(a) does not permit lapping them; they must be welded or joined with a mechanical coupler.
Results are shown in millimetres and as a multiple of the bar diameter, with Ld rounded up to the next 10 mm for site use.
- Development length
- Ld = φ × σs / (4 × τbd)
- Bar stress
- σs = 0.87 × fy
- Design bond stress
- τbd = τbd,plain × 1.6 (deformed) × 1.25 (compression)
- Lap, flexural tension
- Lap = max(Ld, 30φ)
- Lap, direct tension
- Lap = max(2Ld, 30φ)
- Lap, compression
- Lap = max(Ld, 24φ)
Follows: IS 456:2000 Cl 26.2.1 · IS 456:2000 Cl 26.2.1.1 · IS 456:2000 Cl 26.2.5.1
Worked example: development and lap length of a 16 mm Fe 500 bar in M25 concrete
Given
- Bar diameter φ = 16 mm
- Steel Fe 500 (HYSD deformed bar)
- Concrete M25
- Condition: flexural tension
Steps
- Bar stress: σs = 0.87 × 500 = 435.0 MPa.
- Bond stress: plain-bar value for M25 is 1.4 MPa; for a deformed bar τbd = 1.4 × 1.6 = 2.24 MPa.
- Development length: Ld = 16 × 435 / (4 × 2.24) = 6960 / 8.96 = 776.8 mm, shown as 777 mm.
- In bar diameters: 776.8 / 16 = 48.5φ. Rounded up for site: 780 mm.
- Lap in flexural tension: max(777, 30 × 16) = max(777, 480) = 777 mm.
Result: Ld = 777 mm (48.5φ), lap splice 777 mm.
Enter the same values in the calculator above to check each step.
Frequently asked questions
What is the development length of Fe 500 bars in M25 concrete?
For a deformed Fe 500 bar in tension in M25 concrete, τbd = 1.4 × 1.6 = 2.24 MPa and Ld = 0.87 × 500 × φ / (4 × 2.24), which is about 48.5φ. A 16 mm bar therefore needs 777 mm. In compression τbd rises to 2.80 MPa and Ld drops to 38.8φ, or 621 mm.
How is lap length calculated as per IS 456?
IS 456 Cl 26.2.5.1 sets the lap in flexural tension as the greater of Ld and 30φ, in direct tension as the greater of 2Ld and 30φ, and in compression as the greater of the compression development length and 24φ. For a 16 mm Fe 500 bar in M25 that gives 777 mm in flexural tension and 1554 mm in direct tension.
What is the design bond stress τbd in IS 456?
For plain bars in tension IS 456 Cl 26.2.1.1 gives 1.2 MPa for M20, 1.4 for M25, 1.5 for M30, 1.7 for M35 and 1.9 for M40. Deformed bars use 1.6 times these values, and bars in compression a further 25 per cent more.
Can bars larger than 36 mm be lapped?
No. IS 456 Cl 26.2.5.1(a) does not allow lap splices for bars larger than 36 mm; they should be welded or joined with mechanical couplers. The calculator still gives Ld for a 40 mm bar (1942 mm for Fe 500 in M25, tension) but shows a warning in place of the lap length.
What does this development length calculator not include?
It gives the straight Ld and lap length only. It does not credit hooks or bends as anchorage, check the location or staggering of splices, or allow for confinement or bundled bars. Concrete grades above M40 and steel grades other than Fe 250, 415, 500 and 550 are not offered.
Step-by-step guides
- Mat Foundation Design (IS 456 & ACI 318) — GuideMat foundation design with drop panels for punching shear: step-by-step calculation per IS 456 Clause 31.6 and ACI 318-19 Section 22.6, critical perimeter, drop panel sizing, and a worked example.
- BBS of Slab: Bar Bending Schedule for a Slab with Worked ExampleHow to prepare the bar bending schedule of an RCC slab: bar spacing, number of bars, cutting lengths, bend deductions and steel weight, with a 3 m × 4 m slab.
- Steel Weight Formula D²/162: Rebar Weight per Metre with TableHow to calculate the weight of a steel bar with D²/162: where 162 comes from, a kg/m table for 6 to 40 mm bars, and worked examples for 12 mm bars.
- Development & Lap Length (IS 456) — Formula & ExampleHow to calculate development length (Ld) and lap length to IS 456:2000 Cl 26.2, with a ready Ld/φ table for Fe 415, Fe 500 and Fe 550 and a worked example for a 16 mm bar.