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Limit state of collapse in flexure

Intermediate Reinforced Concrete Design Reviewed 24 Oct 2026 2 min read

Limit state design checks that a member is safe against collapse under factored loads and serviceable under working loads. For flexure at collapse, IS 456 (Annex G and Cl 38) models the concrete compression zone with an idealised stress block and assumes the steel stress reaches its design yield value in an under-reinforced section.

The resulting expressions for a singly reinforced rectangular beam are widely taught. The depth of the neutral axis comes from equating the compression in concrete to the tension in steel. A limit on that depth, depending on steel grade, keeps the section under-reinforced so that steel yields first and the beam gives visible warning by cracking and deflecting before it fails.

compressiontension steel
Diagram: Indiacon original, not to scale.

If the moment exceeds the limiting moment of the section, the designer increases the depth, raises the concrete grade, or adds compression steel to make a doubly reinforced section.

The partial safety factors are 1.5 on concrete strength (built into the 0.36 and 0.446 coefficients) and 1.15 on steel, so 0.87 fy is the design steel stress. Other codes use similar ideas with different block shapes; we do not rank them.

Tip

Always compare x_u with x_u,max before using the moment formula.

Key formulas
  • x_u = 0.87 f_y A_st / (0.36 f_ck b)
  • M_u = 0.87 f_y A_st (d − 0.42 x_u)
  • Limiting x_u,max / d: 0.53 (Fe 250), 0.48 (Fe 415), 0.46 (Fe 500)
  • M_u,lim = 0.36 f_ck b x_u,max (d − 0.42 x_u,max); for Fe 415 this is about 0.138 f_ck b d²
Common mistake

Using total depth instead of effective depth d.

Solved example

Problem. Beam 230 mm × 450 mm effective depth, M20 concrete, Fe 415, A_st = 3 bars of 16 mm. Find M_u.

  1. A_st = 3 × 201 = 603 mm².
  2. x_u = 0.87 × 415 × 603 / (0.36 × 20 × 230) = 217 711 / 1656 = 131.5 mm.
  3. x_u,max = 0.48 × 450 = 216 mm, so the section is under-reinforced.
  4. M_u = 0.87 × 415 × 603 × (450 − 0.42 × 131.5) = 217 711 × 394.8 = 85.9 × 10⁶ Nmm.

Answer. M_u ≈ 86 kNm

Quick quiz

1What is the design stress in steel at collapse?

0.87 f_y.

2Why limit x_u / d?

So steel yields first and the beam fails in a ductile, warned manner.

3Name one fix when M exceeds M_u,lim.

Increase depth, raise grade, or add compression steel.

Code note
  • IS 456 (India): Plain and reinforced concrete: general rules for materials, durability, design and construction.

Clauses and documents to read

  • IS 456:2000, cl. 38.1: assumptions for flexure and limiting depth of neutral axis
  • IS 456:2000, Annex G: moment of resistance of rectangular and flanged sections

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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