RCC Beam Design as per IS 456:2000 — Step-by-Step with Worked Example

Reinforced concrete beam design is one of the most fundamental skills in structural engineering. Whether you are preparing for GATE, ESE, or designing real structures, mastering beam design as per IS 456:2000 (Limit State Method) is essential. This article walks through the complete design procedure with a fully worked numerical example.

Types of RCC Beams

• Simply supported beam — most common for floor beams
• Continuous beam — better material efficiency, used in most real buildings
• Cantilever beam — projecting slabs, sunshades, canopies
• T-beam / L-beam — monolithic with slab; effective in compression

Assumptions in LSM Beam Design (IS 456 Clause 38.1)

1. Plane sections remain plane after bending (Bernoulli's assumption)
2. Maximum strain in extreme concrete fibre = 0.0035
3. Strain in steel at design strength: ε_su = 0.87f_y/E_s + 0.002
4. Tensile strength of concrete is ignored
5. Stress-strain of concrete: parabolic-rectangular block; design stress = 0.67f_ck/γ_m = 0.446f_ck
6. Stress-strain of steel: from IS 456 Fig. 23A/23B

Limiting Depth of Neutral Axis (x_u,lim)

To ensure ductile failure (steel yields before concrete crushes):

Limiting Moment of Resistance (M_u,lim)

Mu,lim = 0.36 × (xu,lim/d) × [1 - 0.42 × (xu,lim/d)] × b × d² × fck

Complete Beam Design Procedure

Step 1: Determine Design Loads

Dead Load (DL) = self-weight of beam + floor/wall loads
Live Load (LL) per IS 875 Part 2
Factored Load w_u = 1.5(DL + LL)

Step 2: Calculate Design Moments and Shears

For simply supported beam of span L:
M_u = w_uL²/8 (UDL)
V_u = w_uL/2

Step 3: Preliminary Sizing

From IS 456 Cl. 23.2.1, for simply supported beam: d/l ≥ 1/20 (basic ratio)
Modified by f_s factor and flange effect. Rule of thumb: d ≈ L/15 for residential spans.

Step 4: Check if Singly or Doubly Reinforced

If M_u ≤ M_u,lim → singly reinforced
If M_u > M_u,lim → doubly reinforced (add compression steel)

Step 5: Calculate Tension Steel Area (Singly Reinforced)

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

Rearrange as quadratic in A_st, or use design tables from SP 16.

Step 6: Shear Design

Nominal shear stress: τ_v = V_u / (b × d)
Design shear strength of concrete: τ_c from IS 456 Table 19 (depends on %A_st and f_ck)
If τ_v > τ_c: provide shear reinforcement
V_us = V_u - τ_c·b·d
Spacing of 2-legged stirrups: s_v = 0.87·f_y·A_sv·d / V_us

Step 7: Deflection Check

IS 456 Cl. 23.2.1: l/d ≤ (basic ratio) × k_t × k_c × k_f
For simply supported, Fe 415, 100A_st/bd ≈ 0.5%: basic ratio = 26, k_t ≈ 1.5, effective l/d ≤ 39

Worked Example — Simply Supported Beam

Given: Simply supported beam, effective span = 6 m, factored UDL w_u = 50 kN/m. Materials: M 25 concrete, Fe 415 steel. Beam width b = 300 mm.

Solution:

1. Effective depth: Try d = L/15 = 6000/15 = 400 mm. Use D = 450 mm, d = 415 mm (cover 25 mm + 10 mm bar)

2. Design Moment: M_u = 50 × 6²/8 = 225 kN·m

3. M_u,lim: = 0.1389 × 300 × 415² × 25 = 179.7 kN·m < 225 kN·m → Doubly reinforced beam required

Revise depth: Try D = 550 mm, d = 515 mm

M_u,lim = 0.1389 × 300 × 515² × 25 = 276.8 kN·m > 225 kN·m → Singly reinforced ✓

4. Steel Area:
225 × 10⁶ = 0.87 × 415 × A_st × 515 × [1 - A_st×415/(300×515×25)]
Solving: A_st = 1373 mm²
Use 3 – 25φ bars: A_st provided = 3 × 491 = 1473 mm² ✓

5. Shear Design:
V_u = 50 × 6/2 = 150 kN
τ_v = 150000/(300×515) = 0.97 N/mm²
100A_st/bd = 100×1473/(300×515) = 0.95%
τ_c from Table 19 (M25, 0.95%): 0.61 N/mm²
V_us = 150000 - 0.61×300×515 = 55,585 N
Use 2-legged 8φ stirrups (A_sv = 100 mm²):
s_v = 0.87×415×100×515/55585 = 334 mm → Provide @ 200 mm c/c (also < 0.75d = 386 mm ✓)

6. Deflection Check:
l/d = 6000/515 = 11.65 << 20 (basic ratio) → Pass ✓

Detailing Requirements (IS 456 Cl. 26)

• Minimum bar diameter: 10 mm for main steel
• Maximum bar spacing: The greater of 25+d_g (where d_g = max agg size) or bar diameter
• Side cover: 25 mm or bar diameter (whichever is more)
• Anchorage of main bars: extend past supports by development length L_d = 47φ for Fe 415 in M25
• Minimum stirrup diameter: 6 mm (IS 456 Cl. 26.5.1.6)

Frequently Asked Questions