Designing a slab is a crucial part of reinforced concrete construction. Slabs are flat horizontal structural elements that transfer loads to beams, columns, and walls. They form floors and ceilings in buildings. A proper understanding of slab design ensures the structure’s safety, serviceability, and durability.
Types of Slabs
a) One-Way Slab
- Load is carried in one direction (shorter span).
- Ratio of longer span (Ly) to shorter span (Lx) > 2.
b) Two-Way Slab
- Load is carried in both directions.
- Ratio Ly/Lx ≤ 2.
c) Flat Slab
- No beams between columns and slabs.
- Suitable for commercial buildings.
d) Cantilever Slab
- Fixed at one end and free at the other.
- Common in balconies.
Thumb Rules for Slab Design (Site Use)
✅ Thickness of Slab:
- Residential Building: 100 mm – 150 mm (generally 125 mm)
- Commercial or heavy load: 150 mm – 200 mm
✅ Minimum Steel Reinforcement:
- Main reinforcement: 0.12% of cross-sectional area for HYSD bars.
- Distribution of steel: 0.12% of the cross-sectional area.
✅ Spacing of Reinforcement Bars:
- Max spacing:
- Main bars: 3d or 300 mm (whichever is less)
- Distribution bars: 5d or 450 mm (whichever is less)
- (d = effective depth)
✅ Concrete Cover:
- Minimum 15 mm or the diameter of the bar, whichever is greater.
Load Consideration on Slabs
Dead Load:
- Self-weight = Density × Thickness
- For M25 concrete, density ≈ 25 kN/m³
- E.g., 0.125 m thick slab = 25 × 0.125 = 3.125 kN/m²
Live Load (as per IS 875 Part 2):
- Residential: 2.0 kN/m²
- Office: 2.5 kN/m²
- School/Classroom: 3.0 kN/m²
Floor Finish:
- Generally taken as 1 kN/m²
Total Load Calculation Example
For a 3m x 4m slab, 125 mm thick:
- Dead Load = 25 × 0.125 = 3.125 kN/m²
- Live Load = 2.0 kN/m²
- Floor Finish = 1.0 kN/m²
Total Load (W) = 3.125 + 2.0 + 1.0 = 6.125 kN/m²
Add Factor of Safety (FOS) = 1.5
Factored Load = 6.125 × 1.5 = 9.19 kN/m²
Design of One-Way Slab (Step-by-Step)
Assume slab thickness
Say, 125 mm → Effective depth (d) = 125 – 20 mm (cover) – 8 mm (half of bar dia) ≈ 97 mm
Calculate the bending moment
M=wL2/8
Where:
- w = factored load (9.19 kN/m² = 9190 N/m²)
- L = span (say 3 m)
M=9190×32/8
=9190×98 =10376.25 N\m =10.38 kN\m
Find the required effective depth (d)
M=0.138×fck×b×d2
Assume:
- fck=20 MPa (M20)
- b=1000 mm
Rearranged:
d= √ (M×106/0.138×fck×b)
=√ (10.38×1060.138×20×1000) ≈ 61.3 mm
Use 97 mm → OK.
Calculate Steel Area (Ast)
Ast=M×106/0.87×fy×jd
Assume:
fy=415MPa, j≈0.9
Ast= 10.38×106/0.87×415×0.9×97 ≈319.1mm2
Use 8 mm diameter bars:
Area of 1 bar=π/4×82 =50.26mm2 ⇒Spacing = (1000×50.26) /319.1 ≈ 157.5mm
Use 8 mm bars @ 150 mm c/c.
Distribution Steel (Shrinkage/Temperature)
Minimum = 0.12% of cross-section
Astmin=0.12×(1000×125)/100 =150mm2⇒Use 6 mm bars @ 200 mm c/c (Area = 28.27 mm² per bar)
✅Summary Table (Quick Reference)
| Parameter | Thumb Rule/Value |
|---|---|
| Slab Thickness | 125 mm (residential), 150 mm+ (commercial) |
| Cover to Reinforcement | 15–20 mm |
| Main Steel | 8–12 mm dia @ 100–150 mm c/c |
| Distribution Steel | 6–8 mm dia @ 150–200 mm c/c |
| Minimum Steel Area | 0.12% of cross-sectional area |
| Max Spacing of Bars | 3d (main), 5d (secondary) |
| Concrete Grade | M20 for residential, M25+ for commercial |
| Steel Grade | Fe415 or Fe500 |
Design of Two-Way Slab (Step-by-Step)
Assumptions and Given Data
Let’s assume we have a two-way supported slab with corners not held down.
- Size of slab: 4 m × 3 m
(Ly = 4 m, Lx = 3 m → Ly/Lx = 1.33 < 2 → Two-way slab) - Thickness of slab (assumed): 150 mm
- Concrete grade: M20 (fck = 20 MPa)
- Steel grade: Fe415 (fy = 415 MPa)
- Clear cover: 20 mm
- Live load: 2.0 kN/m² (residential building)
- Floor finish: 1.0 kN/m²
- Unit weight of RCC: 25 kN/m³
Calculate Loads
Dead Load (Self-weight):
=Density×Thickness=25×0.15=3.75kN/m2
Live Load:
=2.0kN/m2
Floor Finish:
=1.0kN/m2
Total Service Load (w):
=3.75+2.0+1.0=6.75kN/m2
Factored Load (wu):
=1.5×6.75=10.125kN/m2
Effective Depth (d)
Assume 12 mm bars, clear cover = 20 mm
Effective depth,
d=150−20−(12/2) =124mm
Bending Moment Calculation Using Coefficients
From IS 456:2000 Table 26, for a two-way simply supported slab with corners free (not restrained):
αx=0.049
αy=0.036
Where:
Mx=αx⋅wu⋅Lx2
My=αy⋅wu⋅Lx2
(Note: Use Lx in both for uniformity.)
M_x = Bending moment in short span
Mx=0.049×10.125×32 =0.049×10.125×9 =4.464kN/m
My = Bending moment in long span
My=0.036×10.125×32 =0.036×10.125×9 =3.285kN/m
Steel Area (Ast) Calculation
For Short Span (Lx):
Use:
Ast=(M×106) / 0.87×fy×j⋅d
Where j≈0.9
Astx = (4.464×106) / 0.87×415×0.9×124 ≈115.7mm2/m
For Long Span (Ly):
Asty= (3.285×106) / 0.87×415×0.9×124 ≈85.2mm2/m
Provide Reinforcement
Short Span (Main Reinforcement):
Use 10 mm bars
Area of 10 mm bar = (π/4)×102=78.5 mm2
Spacing:
Sx=(1000×78.5)/115.7 ≈ 678 mm Long Span (Distribution Reinforcement):
Use 8 mm bars
Area of 8 mm bar = 50.3 mm²
⇒Use 10 mm @ 150 mm c/c
Spacing:
Sy=(1000×50.3)/85.2 ≈ 590mm⇒Use 8 mm @ 150 mm c/c
(Spacing rounded to standard value for site practicality and minimum steel requirements)
Check for Minimum Reinforcement (IS 456 Clause 26.5.2.1)
Astmin=0.12%×b×D=0.0012×1000×150=180mm2/m
Our provided:
- Short span: 10 mm @ 150 mm → 522 mm²/m → OK
- Long span: 8 mm @ 150 mm → 335 mm²/m → OK
✅ Minimum reinforcement satisfied
Deflection Check (Depth Check)
From IS 456 (Clause 23.2.1), basic span/depth ratio = 20
Modified by tension steel percentage and compression reinforcement (if any). Assuming normal conditions:
L/d = 3000/124 ≈24.2<26 (modifiedlimit)→OK
✅ Deflection control OK
✅ Summary of Design
| Item | Value |
| Slab Size | 4 m × 3 m |
| Type | Two-way simply supported |
| Slab Thickness | 150 mm |
| Concrete Grade | M20 |
| Steel Grade | Fe415 |
| Main Reinforcement (Lx) | 10 mm @ 150 mm c/c |
| Distribution Reinforcement (Ly) | 8 mm @ 150 mm c/c |
| Effective Depth | 124 mm |
| Factored Load | 10.125 kN/m² |
| Bending Moments | Mx = 4.464, My = 3.285 kNm/m |
Reinforcement Detailing Guidelines (IS 456:2000)
- Provide crank bars (bent up at supports).
- Alternate bars are bent up to counter negative bending moments.
- Use corner reinforcement if the slab is restrained.
- Provide sufficient anchorage length.
Important IS Codes for Slab Design
- IS 456:2000 – Code of Practice for Plain and Reinforced Concrete
- IS 875 Part 2 – Live Loads
- IS 13920 – Ductile Detailing for Earthquake-Resistant Structures
Conclusion
Slab design is a blend of theoretical understanding and practical considerations. While thumb rules help in quick decisions on-site, proper calculations ensure safety and code compliance. Always refer to IS codes and consult with a structural engineer for critical structures.
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