Design of Reinforcement for a Slab

Introduction

• Topic: Designing the reinforcement of a slab, including design of main bars and distribution bars.
• Objective: Check the safety of provided reinforcement, slab depth, shear, and development length of steel bars.
• Example: Design a one-way slab (6m x 3m) using M20 grade concrete and Fe415 steel.

Assumptions and Given Data

• Dimensions: 6m x 3m
• Concrete Grade: M20
• Steel Grade: Fe415
• Effective Depth: 115 mm

Main Reinforcement Design

• Formula:

  Ast = 0.5 * (Fck/Fy) * [1 - (√(1 - 4.6 * Mu / (Fck * b * d^2)))] * b * d
  • Fck: 20 N/mm² (Characteristic Compressive Strength)
  • Fy: 415 N/mm² (Yield Strength of Steel)
  • Mu: Design Factor Moment (from previous calculation)
  • b: 1m
  • d: Effective Depth 115mm

• Calculation of Area of Steel (Ast): 290 mm²

• Using 10mm diameter bars:

  • Area of one bar = π/4 * d² = 78.5 mm²
  • Spacing = (Area of one bar) / Ast * 1000 = 270 mm
  • Maximum Permissible Spacing: Lesser of 3*d or 300 mm, Minimum 200 mm.
  • Provided Spacing: 10mm steel bars at 270mm center-to-center.

Check Minimum Reinforcement

• As per IS 456:2000:
  • Mild Steel: ≥0.15% of cross-sectional area
  • HYSD Bars: ≥0.12% of cross-sectional area
• Calculation:
  • Minimum Ast: 0.12% of b*d = 160 mm²
  • Provided: 290 mm² (Safe)*

Deflection Check

• Formula for fs: fs = 0.58 * Fy * (Ast required / Ast provided)

  • fs: Calculated as 240 N/mm²

• Percentage of Steel: Ast / (b * d) * 1000 = 0.25%

  • Compared to assumed percentage: 0.4% (Safe)

• Depth required based on Deflection:

  • Span / Basic Value * Modification Factor
  • Basic Value for simply supported slab = 20
  • Deflection Depth: 100mm (less than 115mm, hence Safe)

Distribution Steel Design

• Minimum Reinforcement (as per IS 456:2000) for HYSD bars: 0.12%
• Calculation:
  • Required Ast: 0.12% of b*d = 138 mm²
  • Using 8mm diameter bars:
    • Area of one bar = π/4 * d² = 50.24 mm²
    • Spacing: 364 mm (provided as 360mm center-to-center)
    • Max spacing as per IS 456:2000: Lesser of 5*d or 450mm.
    • Provided spacing: 360mm (Safe)*

Shear Check

• Design Shear (Vu): w * lx / 2 = 16.875 kN
• Shear Stress (τbd): Vu / (b * d) = 0.146 N/mm²
• Shear Strength of Concrete (τc) as per Table 19 of IS 456:2000: τc = 0.28 N/mm² (for M20 and 0.25% of steel)
  • Corrected τc: k * τc = 1.3 * 0.28 = 0.36 N/mm²
  • τc < k * τc (Safe)

Development Length Check

• Formula: M/V + l0 ≥ Ld
  • M: Design moment = 11.7 kNm
  • V: Design shear = 16.87 kN
  • L₀: Number of hooks (zero)
  • Ld: 0.87 * Fy * φ / (4 * τbd)
    • τbd = 1.2 N/mm² for M20
  • Calculated Ld: 470.11mm (Safe, as M/V + l0 = 693.33mm)

Conclusion

• Reinforcement designed and checked for safety in terms of area, deflection, shear, and development length.
• Detailed steps align with IS 456:2000 standards.