Solution: Two-way Slab Design

Given Parameters
  • Short Span: 3.5 m
  • Long Span: 5.0 m
  • Live Load: 4.0 kN/m²
  • Floor Finish: 1.0 kN/m²
  • Grade of Concrete: M25
  • Grade of Steel: Fe500

Step-by-Step Solution

1

Preliminary Sizing

1.1 Effective Depth
  • Assume slab thickness = 180mm
  • Effective depth (d) = 180 - 20 - 6 = 154mm
1.2 Loading
  • Self weight = 0.18 × 25 = 4.5 kN/m²
  • Floor finish = 1.0 kN/m²
  • Live load = 4.0 kN/m²
  • Total load = 9.5 kN/m²
  • Factored load = 1.5 × 9.5 = 14.25 kN/m²
2

Analysis

2.1 Bending Moments (IS 456 Table 26)
  • Short span moment = αx × wu × lx²
  • αx = 0.045 (for two adjacent edges discontinuous)
  • Short span moment = 0.045 × 14.25 × 3.5² = 7.85 kN⋅m/m
  • Long span moment = αy × wu × lx²
  • αy = 0.032
  • Long span moment = 0.032 × 14.25 × 3.5² = 5.58 kN⋅m/m
3

Design

3.1 Flexural Reinforcement (Short Span)
  • Mu = 7.85 kN⋅m/m
  • d = 154mm
  • Required As = 145 mm²/m
  • Provide 10mm @ 200mm c/c (393 mm²/m)
3.2 Flexural Reinforcement (Long Span)
  • Mu = 5.58 kN⋅m/m
  • d = 154mm
  • Required As = 103 mm²/m
  • Provide 8mm @ 200mm c/c (251 mm²/m)
4

Final Design

Design Summary
  • Depth: 180mm
  • Main steel (short span): 10mm @ 200mm c/c
  • Main steel (long span): 8mm @ 200mm c/c
  • Clear cover: 20mm