Ductility
Ductility is the ability to deform well beyond yield without losing much strength. A ductile member bends, cracks and yields visibly but keeps carrying load; a brittle one fails suddenly. It is measured as a ratio, the displacement ductility, maximum displacement divided by yield displacement, or as curvature or rotation ductility at a section.
Earthquake design depends on it. Elastic design for the full earthquake would make buildings very heavy. Instead, codes divide the elastic force by a response reduction factor and rely on the structure to yield and dissipate energy through hysteresis. Two classic rules link the force reduction to ductility. For long-period structures, the equal-displacement rule says the inelastic displacement is about the same as the elastic one, so the reduction factor equals the ductility. For short-period structures, the equal-energy rule gives a smaller reduction, the square root of two times ductility minus one.
Ductility in concrete comes from under-reinforced sections, confinement by closely spaced hoops that let the core strain far beyond the crushing strain of plain concrete, and reinforcement with adequate elongation and a suitable ratio of tensile to yield strength. In steel, compact sections and capacity-designed connections give ductility. Brittle modes, shear failure, bond splitting, buckling of bars and weld fracture, must be suppressed by giving them more strength than the ductile mode.
Ductility demand is not uniform: it concentrates in the hinges that form. That is why detailing rules focus on beam ends, column ends, joints and wall boundaries.
Ductility needs both detailing and material: check bar elongation and the ultimate-to-yield strength ratio too.
Displacement ductility μ = Δ_u / Δ_yEqual displacement (long period): R ≈ μEqual energy (short period): R ≈ √(2 μ − 1)
Assuming a high R value for a system that is not detailed for ductility.
Problem. A structure has a yield displacement of 25 mm and can reach 100 mm before significant strength loss. Find its ductility and the force reduction factor by both rules.
- μ = 100 / 25 = 4.0.
- Equal displacement: R ≈ 4.0.
- Equal energy: R ≈ √(2 × 4 − 1) = √7 = 2.65.
- A short-period building of the same ductility therefore gets a smaller reduction, so its design force must be higher.
Answer. μ = 4; R ≈ 4.0 (long period) or 2.65 (short period)
Quick quiz
1Define displacement ductility.
Maximum displacement divided by yield displacement.
2Which rule applies to short-period structures?
Equal energy, R = √(2μ − 1).
3Name two brittle modes to suppress.
Shear failure and bar buckling (also bond splitting, weld fracture).
- IS 13920 (India): Ductile design and detailing of reinforced concrete structures for seismic forces.
- IS 1893 (Part 1) (India): Criteria for earthquake-resistant design: seismic zones and how to work out earthquake forces.
- IS 1786 (India): High-strength deformed steel bars for concrete reinforcement.
- EN 1998 (Eurocode 8) (Europe): Design of structures for earthquake resistance.
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.
- A. K. Chopra, Dynamics of Structures, Pearson, latest edition.
- T. Paulay, M. J. N. Priestley, Seismic Design of Reinforced Concrete and Masonry Buildings, Wiley, 1992.
- P. Agarwal, M. Shrikhande, Earthquake Resistant Design of Structures, PHI Learning, latest edition.
Free and official sources
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