InfraLens
HomeIS CodesIRCHandbookDesign RulesPMCQA/QCBIMGATE PrepArticlesToolsAbout Join Channel
Join
HomeIS CodesIRCHandbookDesign RulesPMCQA/QCBIMGATE PrepArticlesToolsAbout Join WhatsApp Channel
Home›IS Codes›IS 456:2000›Clauses›Cl. 38.1
IS 456:2000 — Plain and Reinforced Concrete — Code of Practice
IS 456:2000 — Clause 38.1

Limit State of Collapse — Flexure

Clause 38.1 gives the theory and design procedure for flexural members (beams and slabs) under the limit state of collapse. It uses a rectangular stress block with a parabolic-rectangular stress-strain curve for concrete and bilinear curve for steel. The key design parameter is the neutral axis depth xu which determines whether the section is under-reinforced (desirable) or over-reinforced.

Quick Calculator

Key Requirements

  • •Maximum strain in concrete in bending: 0.0035 at the extreme compression fibre
  • •Maximum strain in steel at failure: 0.87fy/Es + 0.002
  • •Limiting neutral axis depth xu,max/d: 0.4791 (Fe415), 0.4560 (Fe500)
  • •Parabolic-rectangular stress block: average stress = 0.36 fck, depth of stress block = 0.42 xu
  • •Design shall ensure xu ≤ xu,max (under-reinforced section) — over-reinforced design is not permitted

Reference Tables

Limiting Neutral Axis Depth xu,max/d (Clause 38.1, Table E)
Steel Gradefy (MPa)xu,max / d
Fe2502500.531
Fe4154150.4791
Fe5005000.456
Fe5505500.438
Limiting Moment of Resistance Factor Mu,lim / (fck × b × d²)
Steel GradeMu,lim / (fck·b·d²)
Fe2500.149
Fe4150.1389
Fe5000.1338
Fe5500.1292

Formulas

Mu = 0.36 fck · b · xu (d − 0.42 xu)
Moment of resistance from concrete compression (singly reinforced)
Mu = Ultimate moment of resistance (N·mm)fck = Characteristic compressive strength (MPa)b = Width of section (mm)xu = Actual neutral axis depth (mm)d = Effective depth (mm)
Mu,lim = 0.36 fck · b · xu,max (d − 0.42 xu,max)
Limiting moment of resistance (balanced section)
Mu,lim = Maximum moment capacity as singly reinforced (N·mm)xu,max = Limiting neutral axis depth = xu,max/d × d
Ast = (0.5 fck / fy) [1 − √(1 − 4.6 Mu / (fck·b·d²))] × b × d
Area of tension steel required for a given Mu
Ast = Area of tension reinforcement (mm²)Mu = Factored bending moment (N·mm)

Practical Notes

✓For Fe500 in M25: Mu,lim = 0.1338 × 25 × b × d² = 3.345 b·d² (N·mm). For a 230 × 450 beam (d ≈ 410 mm): Mu,lim = 3.345 × 230 × 410² = 129.3 kN·m.
✓If Mu > Mu,lim, the section needs compression steel (doubly reinforced). Consider increasing depth first — it's more economical than adding compression steel.
✓SP:16 tables give direct Ast for given Mu/(b·d²) — faster than manual calculation. Most design offices use SP:16 or software.

Common Mistakes

⚠Using xu,max/d = 0.48 for Fe500 — the correct value is 0.456. Using 0.48 (Fe415 value) overestimates Mu,lim by ~5%.
⚠Not checking minimum steel: Ast,min = 0.85 × b × d / fy (Clause 26.5.1.1). For Fe500, 150 mm slab: Ast,min = 0.85 × 1000 × 125/500 = 212.5 mm²/m.
⚠Confusing factored moment with working moment — IS 456 load factors are 1.5 DL + 1.5 LL (for combination 1).

Frequently Asked Questions

Related Resources

Cl. 23.2Cl. 26.2.1Cl. 40.1Cl. 43.1Two-Way Slab Bending Moment CoefficientsSlab & Beam Span/Depth RatiosBeam & Slab Deflection LimitsRcc Design
← Previous
Cl. 26.5
Spacing of Reinforcement
Next →
Cl. 39.3
Limit State of Collapse — Short Column Under Axial Load
View all 15 clauses of IS 456:2000 →