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IS 456:2000 · Beams

RCC Beam Design as per IS 456 — Complete Guide with 3 Worked Examples

⏱ 22 min read📅 June 2026✅ IS 456:2000🎓 GATE relevant

The beam is the horizontal load-carrying member that transfers slab loads to columns. Beam design involves three critical checks: flexure (will it break by bending?), shear (will it crack diagonally?), and deflection (will it sag too much?). IS 456:2000 provides the limit state method for all three. This guide covers singly and doubly reinforced beams, shear reinforcement design, and deflection checks — with three worked examples and 10 GATE MCQs.

1. Introduction

In a typical Indian building, beams span 3–6m between columns and carry uniformly distributed loads from slabs, walls, and self-weight. The most common beam sizes are 230×450mm, 230×600mm, and 300×600mm. A well-designed beam has enough tension steel to resist bending, enough stirrups to resist shear, and enough depth to control deflection — all while staying within the IS 456 limits for minimum and maximum reinforcement.

2. Concept and Theory

How beams resist bending

When a simply supported beam carries a downward load, the top fibres compress and the bottom fibres stretch (go into tension). Concrete is strong in compression but weak in tension — it cracks at about 3–5 MPa. Steel reinforcement bars placed near the bottom face carry all the tension. The concrete in the compression zone and the steel in the tension zone form an internal couple that resists the external bending moment.

Singly vs doubly reinforced beams

A singly reinforced beam has steel only in the tension zone. This is sufficient when M_u ≤ M_u,lim (the balanced moment capacity). A doubly reinforced beam has steel in both tension and compression zones — the compression steel helps when the beam depth is restricted (by architectural constraints) and the moment exceeds M_u,lim. Most beams in practice are singly reinforced; doubly reinforced beams are used at supports of continuous beams or when depth is limited.

The balanced section and xu,max

At the balanced condition, the concrete reaches its crushing strain (0.0035) at the same time the steel yields. The depth of the neutral axis at this point is x_u,max. For Fe415: x_u,max = 0.48d. For Fe500: x_u,max = 0.46d. If the actual neutral axis x_u is less than x_u,max, the section is under-reinforced (desirable — steel yields first, giving warning). If x_u > x_u,max, it is over-reinforced (undesirable — concrete crushes suddenly).

3. IS Code Background

Clause	Subject	Plain English
38.1	Assumptions for flexure	Plane sections remain plane. Max concrete strain = 0.0035. Stress block is parabolic-rectangular. Tensile strength of concrete ignored.
Annex G	Flexure formula	M_u = 0.87 fy Ast (d − 0.42 x_u). Equivalent to the standard flexure equation.
40.1	Shear design	Nominal shear stress τ_v = V_u/(bd). Compare with τ_c (Table 19) and τ_c,max (Table 20).
26.5.1	Min tension steel	Ast,min = 0.85bd/fy. Ensures ductile failure, not sudden brittle fracture.
26.5.1.2	Max tension steel	Ast,max = 0.04bD (4% of gross section). Limits congestion.
26.5.1.6	Stirrup requirements	Min stirrup Asv = 0.4b×sv/0.87fy. Max spacing = 0.75d or 300mm.

4. Key Formulas

Limiting Moment Capacity (Singly Reinforced)

M_u,lim = 0.138 × fck × b × d² (for Fe415)
M_u,lim = 0.133 × fck × b × d² (for Fe500)
If M_u ≤ M_u,lim → singly reinforced design
If M_u > M_u,lim → doubly reinforced needed

Tension Steel (Singly Reinforced)

Ast = [0.5 fck/fy × {1 − √(1 − 4.6M_u/(fck·b·d²))} ] × b × d

Alternative: Ast = M_u / (0.87 × fy × (d − 0.42x_u))
where x_u = 0.87 × fy × Ast / (0.36 × fck × b)

Shear Design (Cl 40)

τ_v = V_u / (b × d) — nominal shear stress
If τ_v ≤ τ_c: no shear reinforcement needed (but provide minimum stirrups)
If τ_c < τ_v ≤ τ_c,max: design stirrups for V_us = V_u − τ_c·b·d
Stirrup spacing: s_v = 0.87 × fy × Asv × d / V_us

Minimum Tension Reinforcement

Ast,min = 0.85 × b × d / fy
For Fe415: Ast,min = 0.00205 × b × d
For Fe500: Ast,min = 0.0017 × b × d

5. Important Tables

Limiting Moment Coefficients

Steel Grade	x_u,max/d	M_u,lim/(fck·b·d²)
Fe250	0.53	0.149
Fe415	0.48	0.138
Fe500	0.46	0.133

Permissible Shear Stress τ_c (IS 456 Table 19, extract)

100Ast/(bd)	M20	M25	M30
0.25	0.36	0.36	0.37
0.50	0.48	0.49	0.50
0.75	0.56	0.57	0.59
1.00	0.62	0.64	0.66
1.50	0.72	0.74	0.76
2.00	0.79	0.82	0.84

6. Step-by-Step Design Procedure

Determine factored load w_u on the beam (from slab loads, wall loads, self-weight).
Calculate M_u = w_uL²/8 (SS) or from moment coefficients (continuous).
Calculate V_u = w_uL/2 (SS) at supports.
Assume beam size (b × D). Common: 230×450, 230×600, 300×600.
Check M_u vs M_u,lim. If M_u ≤ M_u,lim → singly reinforced.
Calculate Ast using flexure formula. Check ≥ Ast,min and ≤ 0.04bD.
Check shear: τ_v = V_u/(bd). Compare with τ_c. Design stirrups.
Check deflection: L/d ≤ basic ratio × modification factor (IS 456 Fig. 4).
Select bar sizes and detailing.

7. Worked Examples

Example 1 — Singly Reinforced Beam (Beginner)

Simply supported beam, span 4m, w_u = 25 kN/m. Size 230×450mm (d = 415mm). M20, Fe415.

Step 1 — BM & Shear

M_u = 25 × 4² / 8 = 50 kN·m
V_u = 25 × 4 / 2 = 50 kN

Step 2 — Check M_u,lim

M_u,lim = 0.138 × 20 × 230 × 415² / 10⁶ = 109.3 kN·m
50 < 109.3 → Singly reinforced OK

Step 3 — Tension Steel

Ast = [0.5×20/415 × {1−√(1−4.6×50×10⁶/(20×230×415²))}] × 230 × 415
= 367 mm²
Min = 0.85×230×415/415 = 230 mm² → 367 > 230 ✅
Provide 2 nos. 16mm (Ast = 402 mm²)

Step 4 — Shear Check

τ_v = 50000/(230×415) = 0.524 N/mm²
pt = 402/(230×415) = 0.42% → τ_c ≈ 0.43 N/mm² (M20)
τ_v > τ_c → design stirrups
V_us = 50000 − 0.43×230×415 = 50000 − 41055 = 8945 N
Using 8mm 2-legged stirrups (Asv = 100.6 mm²):
s_v = 0.87×415×100.6×415/8945 = 1688mm → but max = min(0.75×415, 300) = 300mm
Provide 8mm 2L stirrups @ 150mm c/c near supports, 300mm at midspan

Example 2 — Beam with Higher Load (Intermediate)

Beam 300×600mm (d = 550mm), span 5m, w_u = 45 kN/m. M25, Fe500.

Step 1

M_u = 45 × 25/8 = 140.6 kN·m
M_u,lim = 0.133 × 25 × 300 × 550² / 10⁶ = 301.5 kN·m → singly reinforced

Step 2 — Steel

Ast = 633 mm²
Provide 3 nos. 16mm (Ast = 603 mm²) — slightly less, so use 2-16mm + 1-20mm = 402 + 314 = 716 mm² ✅

Step 3 — Shear

V_u = 45 × 5/2 = 112.5 kN
τ_v = 112500/(300×550) = 0.682 N/mm²
τ_c ≈ 0.44 N/mm² → stirrups needed
V_us = 112500 − 0.44×300×550 = 112500 − 72600 = 39900 N
8mm 2L @ s_v = 0.87×500×100.6×550/39900 = 603mm → use 150mm c/c near supports

Example 3 — Doubly Reinforced Beam (Advanced)

Beam 230×400mm (d = 360mm), span 5m, M_u = 120 kN·m. M20, Fe415. Depth restricted by architectural constraint.

Step 1 — Check

M_u,lim = 0.138 × 20 × 230 × 360² / 10⁶ = 82.3 kN·m
120 > 82.3 → Doubly reinforced beam required!

Step 2 — Excess Moment

M_u2 = 120 − 82.3 = 37.7 kN·m

Step 3 — Tension Steel

Ast1 (for M_u,lim) = pt,lim × b × d = 0.96% × 230 × 360 = 795 mm²
Ast2 (for M_u2) = 37.7×10⁶ / (0.87×415×(360−40)) = 326 mm²
Total Ast = 795 + 326 = 1121 mm²
Provide 2-20mm + 2-16mm = 628 + 402 = 1030 mm² + add 1-12mm = 1143 mm²

Step 4 — Compression Steel

Asc = M_u2 / ((fsc − 0.45fck) × (d − d')) = 37.7×10⁶ / ((353−9)×320) = 342 mm²
Provide 2-16mm at top (402 mm²) ✅

8. GATE MCQs

Q1. For Fe415 steel, the limiting neutral axis depth x_u,max/d is:

(a) 0.53
(b) 0.48
(c) 0.46
(d) 0.44

Answer: (b)
Fe250: 0.53, Fe415: 0.48, Fe500: 0.46. These are fundamental values to memorise.

Q2. Minimum tension reinforcement in a beam (IS 456) is:

(a) 0.12% of bD
(b) 0.85bd/fy
(c) 0.8% of Ag
(d) 0.15% of bD

Answer: (b)
IS 456 Cl 26.5.1.1: Ast,min = 0.85bd/fy. Option (a) is for slabs, (c) is for columns.

Q3. M_u,lim for a 230×500mm beam (d=460mm) with M20 and Fe415 is approximately:

(a) 85 kN·m
(b) 135 kN·m
(c) 170 kN·m
(d) 200 kN·m

Answer: (b)
M_u,lim = 0.138 × 20 × 230 × 460² / 10⁶ = 0.138 × 20 × 230 × 211600 / 10⁶ = 134.5 ≈ 135 kN·m.

Q4. If Mu > Mu,lim for a given beam section, the beam must be designed as:

(a) Singly reinforced
(b) Doubly reinforced
(c) Under-reinforced
(d) Balanced

Answer: (b)
When the applied moment exceeds the limiting moment for a singly reinforced section, compression steel is needed → doubly reinforced.

Q5. Maximum spacing of stirrups in a beam is:

(a) d or 300mm
(b) 0.75d or 300mm
(c) 3d or 300mm
(d) 2d or 200mm

Answer: (b)
IS 456 Cl 26.5.1.5: max spacing = min(0.75d, 300mm).

Q6. In a singly reinforced beam, a balanced section means:

(a) Steel yields and concrete crushes simultaneously
(b) Only steel yields
(c) Only concrete crushes
(d) No failure occurs

Answer: (a)
At balanced condition, concrete reaches 0.0035 strain and steel reaches yield strain simultaneously.

Q7. The nominal shear stress in a beam is calculated as:

(a) V_u/(b×D)
(b) V_u/(b×d)
(c) V_u/(b×d²)
(d) M_u/(b×d)

Answer: (b)
τ_v = V_u/(bd) where d = effective depth. Not D (overall depth).

Q8. An under-reinforced beam fails by:

(a) Sudden concrete crushing
(b) Steel yielding first (ductile failure)
(c) Bond failure
(d) Shear failure

Answer: (b)
Under-reinforced: steel yields first, giving visible deflection and cracking as warning before failure. This is the desirable mode — IS 456 ensures this by limiting x_u ≤ x_u,max.

Q9. The maximum percentage of tension steel in a beam as per IS 456 is:

(a) 2%
(b) 4% of bD
(c) 6%
(d) 0.8%

Answer: (b)
IS 456 Cl 26.5.1.1(b): max steel = 4% of gross cross-section (bD).

Q10. The minimum stirrup reinforcement in a beam is given by:

(a) Asv ≥ 0.4b×sv/(0.87fy)
(b) Asv ≥ 0.87fy×b/sv
(c) Always 8mm @ 300mm
(d) No minimum required

Answer: (a)
IS 456 Cl 26.5.1.6: Asv/b×sv ≥ 0.4/(0.87fy). This ensures a minimum level of shear resistance even where τ_v < τ_c.

9. Common Mistakes

Mistake 1: Using D instead of d in formulas. Most beam formulas use effective depth d (to centroid of tension steel), not overall depth D. The difference (cover + stirrup + bar/2) is typically 35–50mm.

Mistake 2: Not checking minimum steel. Ast,min = 0.85bd/fy is a critical check. A beam with less steel than this can fail suddenly without warning.

Mistake 3: Mixing up beam and slab minimum steel rules. Beams: 0.85bd/fy. Slabs: 0.12% of bD. These are different clauses and different values.

Mistake 4: Ignoring shear design. Students often focus on flexure and forget shear. Shear failure is sudden and catastrophic — always design stirrups.

10. Quick Revision Summary

Memorise:

M_u,lim = 0.138 fck bd² (Fe415) | 0.133 (Fe500)
x_u,max/d: 0.53 (Fe250), 0.48 (Fe415), 0.46 (Fe500)
Min steel: Ast = 0.85bd/fy
Max steel: 4% of bD
τ_v = V_u/(bd) → compare with τ_c from Table 19
Max stirrup spacing: min(0.75d, 300mm)
Under-reinforced = desirable (steel yields first)
M_u > M_u,lim → doubly reinforced

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