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Home›IS Codes›IS 456:2000›Clauses›Cl. 26.2.1
IS 456:2000 — Plain and Reinforced Concrete — Code of Practice
IS 456:2000 — Clause 26.2.1

Development Length of Bars

Clause 26.2.1 specifies the development length (Ld) required to develop the design stress in reinforcement through bond with concrete. The bar must extend at least Ld beyond the point where it is fully stressed. Inadequate development length is a major cause of reinforcement pull-out failure.

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Key Requirements

  • •Development length Ld = (φ × σs) / (4 × τbd)
  • •σs = 0.87 fy for limit state design
  • •Design bond stress τbd from Table 26.2.1.1 — increased by 60% for deformed bars
  • •Bars must extend at least Ld beyond the point of maximum stress
  • •At simple supports, positive moment reinforcement must extend Ld from the face of support or satisfy the check: M1/V + Lo ≥ Ld
  • •For bundled bars, development length of each bar is increased by 10% for 2-bar bundles and 20% for 3-bar bundles

Reference Tables

Design Bond Stress τbd (N/mm²) — Table 26.2.1.1 (Plain bars)
GradeM20M25M30M35M40+
τbd (plain bars)1.21.41.51.71.9
τbd (deformed bars)1.922.242.42.723.04
For deformed bars (TMT), multiply plain bar τbd by 1.6 (60% increase per Clause 26.2.1.1).
Development Length Ld (mm) — Fe500, Deformed Bars (σs = 0.87 × 500 = 435 MPa)
Bar φ (mm)M20M25M30M35M40
8181155145128114
10227194181160143
12272233217192171
16362310290256228
20453388362320286
25566485453399357
28634543507447400
32724621580511457
36815698652575514

Formulas

Ld = (φ × σs) / (4 × τbd)
Development length for reinforcement bars
Ld = Development length in mmφ = Nominal diameter of bar in mmσs = Stress in bar = 0.87 × fy for limit state methodτbd = Design bond stress from Table 26.2.1.1 (use 1.6× for deformed bars)

Practical Notes

✓For Fe500 in M25, Ld ≈ 47φ. This is the most common combination on Indian sites. For 16 mm bars: Ld = 47 × 16 = 752 mm ≈ 760 mm (rounded up).
✓At beam-column junctions, development length often governs over lap length. If column width is insufficient (e.g., 230 mm column with 20 mm beam bar requiring 388 mm Ld), use 90° bends.
✓IS 13920 (seismic detailing) requires Ld to be calculated with full fy (not 0.87fy) in plastic hinge regions. This increases Ld by ~15%.

Common Mistakes

⚠Forgetting the 1.6 multiplier for deformed bars — using plain bar τbd gives an Ld that's 60% too long.
⚠Not providing Ld from the point of maximum stress — Ld is from the critical section, not from the face of support.
⚠Ignoring the M1/V + Lo ≥ Ld check at simple supports (Clause 26.2.3.3) — this is commonly violated in simply supported beams.

Frequently Asked Questions

Related Resources

Cl. 26.3Cl. 26.5Cl. 38.1Development Length & Lap LengthSplice & Lap Length TablesBbsRcc DesignLap Length & Development Length Explained
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Cl. 23.2
Effective Span of Beams and Slabs
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Cl. 26.2.5
Lap Length & Splicing of Reinforcement
View all 15 clauses of IS 456:2000 →