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Home›IS Codes›IS 456:2000›Clauses›Cl. 43.1
IS 456:2000 — Plain and Reinforced Concrete — Code of Practice
IS 456:2000 — Clause 43.1

Limit State of Serviceability — Deflection

Clause 43.1 (read with Clause 23.2.1) provides the deemed-to-satisfy span-to-depth ratio method for controlling deflection. Instead of computing actual deflection, the code allows checking that the actual span/depth ratio does not exceed the permissible value (basic ratio × modification factors). This is the most commonly used deflection check in Indian practice.

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Key Requirements

  • •Basic L/d ratios (Clause 23.2.1): Cantilever = 7, Simply supported = 20, Continuous = 26
  • •Modify basic ratio by kt (tension reinforcement factor, Fig. 4) and kc (compression reinforcement factor, Fig. 5)
  • •For spans > 10 m: multiply basic ratio by 10/span
  • •For flanged beams: further reduce by factor (0.8 to 1.0) depending on bw/bf ratio
  • •Steel stress fs = 0.58 fy × (Ast,required / Ast,provided) — used to enter Fig. 4

Reference Tables

Basic Span-to-Effective-Depth Ratios (Clause 23.2.1)
Support ConditionBasic L/d
Cantilever7
Simply Supported20
Continuous26
Modification Factor kt for Tension Reinforcement (Fig. 4 — approximate)
pt%fs=145 MPa (Fe250)fs=240 MPa (Fe415)fs=290 MPa (Fe500)
0.2221.68
0.41.951.61.35
0.61.751.351.16
0.81.561.181.05
1.01.41.060.98
1.21.280.980.92
1.51.140.890.85
2.010.80.76
3.00.90.720.68
fs = 0.58 fy × (Ast,req/Ast,prov). Interpolate for intermediate values.

Formulas

L/d (allowable) = Basic ratio × kt × kc
Permissible span-to-depth ratio
Basic ratio = 7 (cantilever), 20 (simply supported), 26 (continuous)kt = Modification factor for tension reinforcement (Fig. 4) — depends on pt% and fskc = Modification factor for compression reinforcement (Fig. 5) — depends on pc%
fs = 0.58 fy (Ast,req / Ast,prov)
Service stress in tension steel for Fig. 4 entry
fs = Estimated service stress in steel (MPa)fy = Characteristic yield strength (MPa)Ast,req = Area of steel required by calculation (mm²)Ast,prov = Area of steel actually provided (mm²)

Practical Notes

✓For a simply supported slab with Fe500 and pt ≈ 0.4%: kt ≈ 1.35. Allowable L/d = 20 × 1.35 = 27. For a 4 m span: minimum d = 4000/27 = 148 mm → use 150 mm slab (overall 175 mm with 25 mm cover).
✓Providing more steel than required (Ast,prov > Ast,req) reduces fs and increases kt — this is a practical way to pass a tight deflection check without increasing depth.
✓For continuous slabs, the L/d ratio of 26 applies to the shorter span. Don't use 26 for end spans — use 20 (simply supported) for the end bay.

Common Mistakes

⚠Using basic L/d = 20 for a continuous slab — it's 26 for intermediate spans, but only 20 for end spans (simply supported edge).
⚠Not computing fs correctly — fs depends on ratio of Ast required to Ast provided, not just on fy.
⚠Ignoring the 10/span reduction for spans exceeding 10 m — a 12 m continuous beam: basic L/d = 26 × 10/12 = 21.7.

Frequently Asked Questions

Related Resources

Cl. 23.2Cl. 38.1Slab & Beam Span/Depth RatiosBeam & Slab Deflection LimitsRcc Design
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Cl. 40.4
Design of Shear Reinforcement
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Cl. 46.2
Permissible Stresses — Working Stress Method
View all 15 clauses of IS 456:2000 →