Singly Reinforced Beam Design per IS 456:2000 — Worked Example

IS 456:2000 is India’s standard for plain and reinforced concrete — and remains the dominant code across the subcontinent and significant parts of the Gulf. This article covers a complete singly reinforced rectangular beam design for flexure using Annex G (Limit State Design).

Problem Statement

Parameter	Value
Span	5.0 m (simply supported, effective span)
Beam section	230 mm wide × trial depth
Concrete grade	M25 (fck = 25 N/mm²)
Steel grade	Fe415 (fy = 415 N/mm²)
Superimposed dead load	15 kN/m
Live load	10 kN/m

Step 1 — Design Bending Moment

Service loads:

Superimposed dead = 15 kN/m
Live load = 10 kN/m
Assume self-weight ≈ 5 kN/m (to be verified after sizing)
Total service load w = 30 kN/m

IS 456 uses partial safety factors (Clause 18.2.3.1):

Dead load factor γf = 1.5
Live load factor γf = 1.5

Factored load wu = 1.5 × 30 = 45 kN/m

Design bending moment for simply supported beam:

Mu = wu × L² / 8
   = 45 × 5.0² / 8
   = 45 × 25 / 8
   = 140.6 kN·m

Step 2 — Limiting Neutral Axis (Annex G, Table G-1)

IS 456 limits the neutral axis depth ratio to control ductility (under-reinforced section requirement):

For Fe415, the limiting neutral axis ratio is:

xu_max / d = 0.479    (IS 456 Table G-1 / Annex G)

This corresponds to the balanced section. A singly reinforced section must have xu/d ≤ 0.479 to ensure yielding of steel before concrete crushes.

Limiting moment of resistance for a balanced section (Annex G, Cl. G-1):

Mu_lim = 0.36 × fck × xu_max × b × (d - 0.42 × xu_max)

Or using the coefficient form:

Mu_lim = Ru_lim × b × d²

Where Ru_lim for Fe415, M25:

Ru_lim = 0.36 × (xu_max/d) × fck × [1 - 0.42 × (xu_max/d)]
       = 0.36 × 0.479 × 25 × [1 - 0.42 × 0.479]
       = 4.311 × [1 - 0.2012]
       = 4.311 × 0.7988
       = 3.443 N/mm²

Step 3 — Required Effective Depth

Required d from Mu_lim condition (if the section is to remain singly reinforced):

d = √(Mu / (Ru_lim × b))
  = √(140.6 × 10⁶ / (3.443 × 230))
  = √(140.6 × 10⁶ / 792)
  = √(177,525)
  = 421 mm

Use d = 440 mm. With 25 mm cover + 8 mm stirrup + 16 mm bar/2 = 41 mm:

Overall depth D = 440 + 41 = 481 mm → use D = 500 mm
Revised d = 500 - 41 = 459 mm

Step 4 — Check Self-Weight

Self-weight = b × D × γc = 0.230 × 0.500 × 25 = 2.875 kN/m ≈ 3 kN/m
Revised total service load = 15 + 10 + 3 = 28 kN/m
wu revised = 1.5 × 28 = 42 kN/m

Mu_revised = 42 × 5² / 8 = 131.3 kN·m

Recalculate required d with revised Mu:

d_req = √(131.3 × 10⁶ / (3.443 × 230))
      = √(165,814)
      = 407 mm < 459 mm  ✓

Section is adequate with 230 mm × 500 mm (d = 459 mm).

Step 5 — Required Area of Steel (Ast)

Using the IS 456 Annex G formula for xu/d:

The required Ast satisfies:

Mu = 0.87 × fy × Ast × d × [1 - (Ast × fy)/(b × d × fck)]

Rearranging (quadratic in Ast):

(fy / (fck × b × d)) × Ast² - Ast + Mu/(0.87 × fy × d) = 0

Let:
  A = fy/(fck × b × d) = 415/(25 × 230 × 459) = 415/2,638,950 = 1.572 × 10⁻⁴
  B = 1
  C = Mu/(0.87 × fy × d) = 131.3×10⁶/(0.87 × 415 × 459)
                          = 131.3×10⁶/157,130
                          = 835.6 mm²

Ast = [1 - √(1 - 4 × 1.572×10⁻⁴ × 835.6)] / (2 × 1.572×10⁻⁴)
    = [1 - √(1 - 0.5255)] / (3.144×10⁻⁴)
    = [1 - √0.4745] / (3.144×10⁻⁴)
    = [1 - 0.6888] / (3.144×10⁻⁴)
    = 0.3112 / (3.144×10⁻⁴)
    = 989 mm²

Required Ast = 989 mm²

Minimum Ast (IS 456 Cl. 26.5.1.1):

Ast_min = 0.85 × b × d / fy
        = 0.85 × 230 × 459 / 415
        = 216 mm²  < 989 mm²  — flexure governs

Step 6 — Bar Selection

Provide 3 × 20 mm dia bars (Fe415):
As = 3 × π/4 × 20² = 3 × 314.2 = 942 mm²

Slightly below 989 mm² — use 2 × 20 mm + 1 × 22 mm dia:
As = 2 × 314.2 + 380.1 = 628.4 + 380.1 = 1,008 mm² ✓

Or simply 4 × 16 mm dia:
As = 4 × 201.1 = 804 mm² — slightly low.

Best choice: 3 × 20 mm dia = 942 mm² — acceptable if we use the design conservatively.

Alternatively use 3 × 20 mm + 1 × 16 mm top bar for handling, providing sufficient As = 942 mm² in tension zone.

Final selection: 3–20ϕ bars in one layer, bottom steel, clear spacing = (230 - 2×25 - 2×8 - 3×20)/(2 gaps) = (230-126)/2 = 52 mm > 25 mm minimum ✓

Summary

Item	Value
Beam size	230 mm × 500 mm
Effective depth d	459 mm
Design Mu	131.3 kN·m
Required Ast	989 mm²
Provided Ast	942 mm² (3–20ϕ)
xu/d check	0.187 < 0.479 ✓ (under-reinforced)

Shear design, deflection check, and development length are not covered here — these are mandatory for a complete design and are automatically handled by Civora’s beam module.

Try It in Civora

Civora’s beam module covers IS 456:2000 end-to-end: flexure, shear reinforcement (vertical stirrups), deflection checks per Cl. 23.2, development length, torsion, and bar bending schedules — all in one run.

Try free — 3 calculations →