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