Development length
A bar can only work at full stress if it is anchored far enough into concrete on each side of the point where that stress is needed. The bond between bar and concrete transfers force gradually, so a certain length is needed to build up the full bar force. That length is the development length. If it is not available, the bar pulls out before it yields and the member fails early.
Development length is found by equating the bar force to the bond force: the bar force is its area times design stress, and the bond force is the perimeter times length times design bond stress. This gives Ld = φ σs / (4 τbd). Deformed (ribbed) bars grip much better than plain bars, and bars in compression need less length because the bar end bears on concrete. Codes give design bond stresses that rise with concrete grade. Bundled bars need extra length.
Where space is short, hooks and bends add anchorage, each counted as an equivalent length. Laps, where one bar ends and another starts alongside it, transfer force through the surrounding concrete and are sized from development length, with extra rules for tension laps and for laps in highly stressed zones. At simple supports, the positive moment bars must satisfy Ld ≤ M1/V + L0, and the M1/V term may be increased by 30 percent where the bar ends are confined by a compressive reaction; this limits bar size at supports. Good detailing staggers laps and keeps them out of plastic hinge regions in seismic design.
Ld = φ σs / (4 τbd), with σs = 0.87 fyDeformed bars: design bond stress increased (60 percent over plain bars in IS 456)Compression: bond stress increased by 25 percentAt simple supports: Ld ≤ 1.3 M1 / V + L0 where bar ends are confined by a compressive reaction; otherwise Ld ≤ M1 / V + L0
Problem. Find the tension development length of a 16 mm Fe 500 deformed bar in M25 concrete. IS 456 gives a design bond stress of 1.4 MPa for plain bars in M25, increased by 60 percent for deformed bars.
- τbd = 1.4 × 1.6 = 2.24 MPa.
- σs = 0.87 × 500 = 435 MPa.
- Ld = 16 × 435 / (4 × 2.24) = 6 960 / 8.96 = 777 mm.
- That is about 48.6 φ; adopt 780 mm.
Answer. About 780 mm (≈ 49 bar diameters)
Quick quiz
1What does development length ensure?
That the bar can reach its design stress before pulling out.
2Why do deformed bars need less length than plain bars?
Ribs give much better bond.
3Where should laps be placed?
Away from points of maximum stress, and staggered.
- IS 456 (India): Plain and reinforced concrete: general rules for materials, durability, design and construction.
- IS 1786 (India): High-strength deformed steel bars for concrete reinforcement.
Clauses and documents to read
- IS 456:2000 cl. 26.2.1 (development length)
- IS 456:2000 cl. 26.2.2 (anchoring bars)
- IS 456:2000 cl. 26.2.3.3 (positive moment bars at supports)
- IS 456:2000 cl. 26.2.5 (laps)
Cited by number and clause only. We do not reproduce clause text, tables or figures (BIS Act 2016, s.10(5) and s.11). Check the current edition and amendments before use.
References and further reading
Standard texts, listed in no order of merit. We do not copy text, figures, tables or solved problems from any book; our explanations and examples are original.
- S. Unnikrishna Pillai, Devdas Menon, Reinforced Concrete Design, McGraw Hill Education (India), latest edition.
- J. K. Wight, J. G. MacGregor, Reinforced Concrete: Mechanics and Design, Pearson, latest edition.
- A. H. Nilson, D. Darwin, C. W. Dolan, Design of Concrete Structures, McGraw Hill, latest edition.
- R. Park, T. Paulay, Reinforced Concrete Structures, Wiley, 1975.
Free and official sources
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