Concrete - DiyCivil-Calc | Cement, Sand & Aggregate Estimator
📖 Beam Concrete Example
- Length = 10 ft
- Width = 9 in ÷ 12 = 0.75 ft
- Depth = 18 in ÷ 12 = 1.5 ft
- Cement: 1440 kg/m³
- Sand: 1600 kg/m³
- Aggregate: 1500 kg/m³
- Volume = (1 / 5.5) × 17.33 = 3.15 cu.ft
- Volume in m³ = 0.0892 m³
- Weight = 0.0892 × 1440 = 128.45 kg
- 50-kg bags = 3 bags
- Volume = 4.72 cu.ft
- Volume in m³ = 0.1336 m³
- Volume = 9.45 cu.ft
- Volume in m³ = 0.267 m³
- Wet Concrete: 11.25 cu.ft (0.318 m³)
- Dry Concrete: 17.33 cu.ft (0.491 m³)
- Cement: 3 bags (128 kg)
- Sand: 4.91 cu.ft (0.1391 m³)
- Aggregate: 9.83 cu.ft (0.278 m³)
- Total parts = 13
- Cement = 7.69%
- Sand = 30.77%
- Aggregate = 61.54%
- Total parts = 10
- Cement = 10%
- Sand = 30%
- Aggregate = 60%
- Total parts = 7
- Cement = 14.28%
- Sand = 28.57%
- Aggregate = 57.14%
- Total parts = 5.5
- Cement = 18.18%
- Sand = 27.27%
- Aggregate = 54.54%
- Total parts = 4
- Cement = 25%
- Sand = 25%
- Aggregate = 50%
Concrete Calculation for Beam (10 ft × 9 in × 18 in) – With Density & Mix Ratios
1. Convert All Dimensions Into Feet
2. Wet Concrete Volume
Wet Volume = Length × Width × Depth
= 10 ft × 0.75 ft × 1.5 ft
= 11.25 cubic feet (0.318 m³)
3. Dry Concrete Volume
Standard concrete dry factor = 1.54
Dry Volume = 11.25 × 1.54 = 17.33 cu.ft (0.491 m³)
4. Material Densities Used
5. Mix Ratio Used for Detailed Calculation (M20 – 1:1.5:3)
Total parts = 5.5
Cement
Sand
Aggregate
6. Final Material Summary (M20 Mix) with 4 percent wastage
7. Other Popular Concrete Mix Ratios
M7.5 (1:4:8)
M10 (1:3:6)
M15 (1:2:4)
M20 (1:1.5:3)
M25 (1:1:2)
Note: Beam concrete typically uses M20 or M25 grade with proper reinforcement.
📖 Column Concrete Example
- Height = 10 ft
- Width = 9 in ÷ 12 = 0.75 ft
- Length = 18 in ÷ 12 = 1.5 ft
- Cement: 1440 kg/m³
- Sand: 1600 kg/m³
- Aggregate: 1500 kg/m³
- Volume = (1 / 5.5) × 17.33 = 3.15 cu.ft
- Volume in m³ = 0.0892 m³
- Weight = 0.0892 × 1440 = 128.45 kg
- 50-kg bags = 3 bags
- Volume = 4.72 cu.ft
- Volume in m³ = 0.1336 m³
- Volume = 9.45 cu.ft
- Volume in m³ = 0.267 m³
- Wet Concrete: 11.25 cu.ft (0.318 m³)
- Dry Concrete: 17.33 cu.ft (0.491 m³)
- Cement: 3 bags (128 kg)
- Sand: 4.91 cu.ft (0.139 m³)
- Aggregate: 9.828 cu.ft (0.278 m³)
- Total parts = 13
- Cement = 7.69%
- Sand = 30.77%
- Aggregate = 61.54%
- Total parts = 10
- Cement = 10%
- Sand = 30%
- Aggregate = 60%
- Total parts = 7
- Cement = 14.28%
- Sand = 28.57%
- Aggregate = 57.14%
- Total parts = 5.5
- Cement = 18.18%
- Sand = 27.27%
- Aggregate = 54.54%
- Total parts = 4
- Cement = 25%
- Sand = 25%
- Aggregate = 50%
Concrete Calculation for Column (10 ft × 9 in × 18 in) – With Density & Mix Ratios
1. Convert All Dimensions Into Feet
2. Wet Concrete Volume
Wet Volume = Height × Width × Length
= 10 ft × 0.75 ft × 1.5 ft
= 11.25 cu.ft (0.318 m³)
3. Dry Concrete Volume
Standard dry volume factor = 1.54
Dry Volume = 11.25 × 1.54 = 17.33 cu.ft (0.491 m³)
4. Material Densities Used
5. Mix Ratio Used for Detailed Calculation (M20 – 1:1.5:3)
Total parts = 5.5
Cement
Sand
Aggregate
6. Final Material Summary (M20 Mix) with 4 percent wastage
7. Other Concrete Mix Ratios (Quick Reference)
M7.5 (1:4:8)
M10 (1:3:6)
M15 (1:2:4)
M20 (1:1.5:3)
M25 (1:1:2)
Note: Columns usually require M25 or M30 grade for structural strength.
📖 Slab Concrete Example
- Cement: 1440 kg/m³
- Sand: 1600 kg/m³
- Aggregate: 1500 kg/m³
- Volume = (1/5.5) × 577.5 = 105 cu.ft
- Volume in m³ = 2.97 m³
- Weight = 2.97 × 1440 = 4,282 kg
- 50-kg bags = 86 bags
- Volume = 157.5 cu.ft
- Volume in m³ = 4.46 m³
- Volume = 315 cu.ft
- Volume in m³ = 8.92 m³
- Wet Concrete: 375 cu.ft (10.62 m³)
- Dry Concrete: 577.5 cu.ft (16.35 m³)
- Cement: 86 bags (50 kg each)
- Sand: 163.8 cu.ft (4.63 m³)
- Aggregate: 315 cu.ft (8.92 m³)
- Total parts = 13
- Cement = 1/13 = 7.69%
- Sand = 4/13 = 30.77%
- Aggregate = 8/13 = 61.54%
- Total parts = 10
- Cement = 10%
- Sand = 30%
- Aggregate = 60%
- Total parts = 7
- Cement = 14.28%
- Sand = 28.57%
- Aggregate = 57.14%
- Total parts = 5.5
- Cement = 18.18%
- Sand = 27.27%
- Aggregate = 54.54%
- Total parts = 4
- Cement = 25%
- Sand = 25%
- Aggregate = 50%
Concrete Calculation for a 1000 sq.ft Slab (4.5-inch Thick) M20 (1:1.5:3)
1. Convert Slab Thickness
4.5 inches ÷ 12 = 0.375 ft
2. Wet Concrete Volume
Wet Volume = Area × Thickness
= 1000 sq.ft × 0.375 ft
= 375 cu.ft (10.62 m³)
3. Dry Concrete Volume
Standard dry volume factor = 1.54
Dry Volume = 375 × 1.54 = 577.5 cu.ft (16.35 m³)
4. Material Densities Used
5. Mix Ratio Used for Detailed Calculation (M20 – 1:1.5:3)
Total parts = 5.5
Cement
Sand
Aggregate
6. Final Material Summary for M20 Mix with 4 percent wastage
7.Other Mix Ratios for Quick Reference
M7.5 (1:4:8)
M10 (1:3:6)
M15 (1:2:4)
M20 (1:1.5:3)
M25 (1:1:2)
Note: Slab thickness varies: 125mm for 3m span, 150mm for 4-5m span.
📖 Footing Concrete Example
- Cement: 1440 kg/m³
- Sand: 1600 kg/m³
- Aggregate: 1500 kg/m³
- Volume = (1 / 5.5) × 19.25 = 3.5 cu.ft
- Volume in m³ = 0.0991 m³
- Weight = 0.0991 × 1440 = 142.7 kg
- 50-kg bags = 3 bags
- Volume = 5.25 cu.ft
- Volume in m³ = 0.1487 m³
- Volume = 10.5 cu.ft
- Volume in m³ = 0.2973 m³
- Wet Concrete: 12.5 cu.ft (0.354 m³)
- Dry Concrete: 19.25 cu.ft (0.545 m³)
- Cement: 3 bags (143 kg)
- Sand: 5.46 cu.ft (0.1546 m³)
- Aggregate: 10.92 cu.ft (0.309 m³)
- Total parts = 13
- Cement = 7.69%
- Sand = 30.77%
- Aggregate = 61.54%
- Total parts = 10
- Cement = 10%
- Sand = 30%
- Aggregate = 60%
- Total parts = 7
- Cement = 14.28%
- Sand = 28.57%
- Aggregate = 57.14%
- Total parts = 5.5
- Cement = 18.18%
- Sand = 27.27%
- Aggregate = 54.54%
- Total parts = 4
- Cement = 25%
- Sand = 25%
- Aggregate = 50%
Concrete Calculation for Footing (5 ft × 5 ft × 6 in) – With Density & Mix Ratios
1. Convert Thickness Into Feet
6 inches ÷ 12 = 0.5 ft
2. Wet Concrete Volume
Wet Volume = Length × Width × Thickness
= 5 ft × 5 ft × 0.5 ft
= 12.5 cu.ft (0.354 m³)
3. Dry Concrete Volume
Standard dry factor = 1.54
Dry Volume = 12.5 × 1.54 = 19.25 cu.ft (0.545 m³)
4. Material Densities Used
5. Mix Ratio Used for Detailed Calculation (M20 – 1:1.5:3)
Total parts = 5.5
Cement
Sand
Aggregate
6. Final Material Summary (M20 Mix) with 4 percent wastage
7. Other Concrete Mix Ratios (Quick Reference)
M7.5 (1:4:8)
M10 (1:3:6)
M15 (1:2:4)
M20 (1:1.5:3)
M25 (1:1:2)
Note: Footing size depends on soil bearing capacity and column load.
📖 Lintel Concrete Example
- Length = 3.5 ft
- Width = 4.5 in ÷ 12 = 0.375 ft
- Depth = 6 in ÷ 12 = 0.5 ft
- Cement: 1440 kg/m³
- Sand: 1600 kg/m³
- Aggregate: 1500 kg/m³
- Volume = (1 / 5.5) × 1.01 = 0.1836 cu.ft
- Volume in m³ = 0.0052 m³
- Weight = 0.0052 × 1440 = 7.49 kg
- 50-kg bags = 0.15 bag
- Volume = 0.2754 cu.ft
- Volume in m³ = 0.0078 m³
- Volume = 0.5508 cu.ft
- Volume in m³ = 0.0156 m³
- Wet Concrete: 0.656 cu.ft (0.0186 m³)
- Dry Concrete: 1.01 cu.ft (0.0286 m³)
- Cement: 7.5 kg (0.15 bag)
- Sand: 0.287 cu.ft (0.0081 m³)
- Aggregate: 0.573 cu.ft (0.0162 m³)
- Total parts = 13
- Cement = 7.69%
- Sand = 30.77%
- Aggregate = 61.54%
- Total parts = 10
- Cement = 10%
- Sand = 30%
- Aggregate = 60%
- Total parts = 7
- Cement = 14.28%
- Sand = 28.57%
- Aggregate = 57.14%
- Total parts = 5.5
- Cement = 18.18%
- Sand = 27.27%
- Aggregate = 54.54%
- Total parts = 4
- Cement = 25%
- Sand = 25%
- Aggregate = 50%
Concrete Calculation for Lintel (3.5 ft × 4.5 in × 6 in) – With Density & Mix Ratios
1. Convert All Dimensions Into Feet
2. Wet Concrete Volume
Wet Volume = Length × Width × Depth
= 3.5 ft × 0.375 ft × 0.5 ft
= 0.65625 cu.ft (0.01859 m³)
3. Dry Concrete Volume
Standard dry factor = 1.54
Dry Volume = 0.65625 × 1.54 = 1.01 cu.ft (0.0286 m³)
4. Material Densities Used
5. Mix Ratio Used for Detailed Calculation (M20 – 1:1.5:3)
Total parts = 5.5
Cement
Sand
Aggregate
6. Final Material Summary (M20 Mix) with 4 percent wastage
7. Other Concrete Mix Ratios (Quick Reference)
M7.5 (1:4:8)
M10 (1:3:6)
M15 (1:2:4)
M20 (1:1.5:3)
M25 (1:1:2)
Note: Lintel length = Door/window width + 300mm bearing on each side.
Steel - DiyCivil-Calc | Reinforcement Weight & Bar Estimator
📖 Beam Steel Example
Problem: Calculate steel for beam: Length=5m, Width=300mm, Depth=500mm, Top=2×16mm, Bottom=3×20mm, Stirrups=8mm@150mm.
Solution:
- Top bars: 2 bars × 5m × 1.58 kg/m (16mm) = 15.8 kg
- Bottom bars: 3 bars × 5m × 2.47 kg/m (20mm) = 37.05 kg
- Stirrup length = 2×((300-50-8) + (500-50-8)) + 2×75mm(hooks)-80(bend deduction)= 1438mm = 1.438m
- No. of stirrups = (5000/150) + 1 = 35 nos
- Stirrup weight = 35 × 1.438m × 0.395 kg/m (8mm) = 19.88 kg
- Total = 15.8 + 37.05 + 19.88 = 72.73 kg
- With 5% wastage = 76.35 × 1.05 ≈ 76.36 kg
Note: Use standard unit weight - 8mm: 0.395 kg/m, 16mm: 1.58 kg/m, 20mm: 2.47 kg/m
📖 Column Steel Example
Problem: Calculate steel for column: Height=3m, 400×400mm, 8 bars of 16mm, ties 8mm@200mm spacing.
Solution:
- Main bars: 8 bars × 3m × 1.58 kg/m (16mm) = 37.92 kg
- Tie length = 2×((400-80-8) + (400-80-8)) + 2×75mm(hooks)-80(bend deduction)= 1318mm = 1.318m
- No. of ties = (3000/200) + 1 = 16 nos
- Tie weight = 15 × 1.318m × 0.395 kg/m (8mm) = 8.329 kg
- Total = 37.92 + 8.329 = 46.249 kg
- With 5% wastage = 46.249 × 1.05 ≈ 48.56 kg
Note: Minimum 4 bars for rectangular, 6 for large columns, 8 for heavy loads.
📖 Slab Steel Example
Problem: Calculate steel for slab: 4m×3m×150mm thick, main bars 10mm@150mm, distribution 8mm@200mm.
Solution:
- Main bars (along 4m): No. = (3000/150) + 1 = 21 bars
- Main bars weight = 21 × 4m × 0.617 kg/m (10mm) = 51.8 kg
- Distribution bars (along 3m): No. = (4000/200) + 1 = 21 bars
- Distribution weight = 21 × 3m × 0.395 kg/m (8mm) = 24.9 kg
- Total = 51.8 + 24.9 = 76.7 kg
- With 5% wastage = 76.7 × 1.05 ≈ 80.5 kg
Note: Provide both layers (top & bottom) for slabs > 200mm thick.
📖 Lintel Steel Example
Problem: Calculate steel for lintel: Length=2m, 200×150mm, 4 bars of 12mm, stirrups 8mm@150mm.
Solution:
- Main bars: 4 bars × 2m × 0.888 kg/m (12mm) = 7.1 kg
- Stirrup length = 2×((200-40-8)+(150-20-8)) + 2×75mm hooks-80 = 578mm = 0.578m
- No. of stirrups = (2000/150) + 1 = 14 nos
- Stirrup weight = 14 × 0.85m × 0.395 kg/m (8mm) = 3.196
- Total weigth = 7.1 + 3.196 = 10.296 kg
- With 5% wastage = 10.296 × 1.05 ≈ 10.81 kg
Note: Lintel bars should extend 150-200mm beyond opening on both sides.
📖 Footing Steel Example
Problem: Calculate steel for footing: 1.5m×1.5m×300mm, main bars 12mm@150mm both ways.
Solution:
- Bars in X-direction: No. = (1500/150) + 1 = 11 bars
- X-direction weight = 11 × 1.5m × 0.888 kg/m (12mm) = 14.7 kg
- Bars in Y-direction: No. = (1500/150) + 1 = 11 bars
- Y-direction weight = 11 × 1.5m × 0.888 kg/m (12mm) = 14.7 kg
- Total = 14.7 + 14.7 = 29.4 kg
- With 5% wastage = 29.4 × 1.05 ≈ 30.9 kg
Note: Footing reinforcement placed at bottom with 75mm clear cover.
Brickwork - DiyCivil-Calc | Bricks, Mortar & Cement Estimator
- Wall area = 100 sq.ft
- Wall thickness = 4.5 in = 0.375 ft
- Wall volume = 100 × 0.375 = 37.5 cu.ft
- Brick nominal size = 9" × 4.5" × 3" (common modular brick)
- Mortar joint thickness = 10 mm (≈0.394")
- Mortar mix used = 1 : 6 (cement : sand)
- Wall volume = 37.5 cu.ft
- Brick nominal volume = 9"×4.5"×3" = 0.0703125 cu.ft
- Brick-with-mortar dimensions = (9 + 0.394) × (4.5 + 0.394) × (3 + 0.394) inches
- Brick-with-mortar volume = 0.09028 cu.ft
- Exact number of bricks = 37.5 ÷ 0.09028 = ~415.36 → round up → 416 bricks
- Total brick solid-volume = 416 × 0.0703125 = 29.205 cu.ft
- Mortar (wet) volume = Wall − Brick solid-volume = 37.5 − 29.205 = 8.295 cu.ft
- Dry mortar volume factor ≈ 1.33
- Dry mortar volume = 8.295 × 1.33 = 11.03 cu.ft
- Mortar mix 1:6 → total parts = 7
- Cement volume = (1/7) × 11.03 = 1.576 cu.ft
- Sand volume = (6/7) × 11.03 = 9.456 cu.ft
- Bricks required: 416 bricks (rounded up)
- Mortar (wet): 8.295 cu.ft
- Dry mortar volume: 11.03 cu.ft
- Cement: 1.576 cu.ft → 1.26 bags (50-kg bag) → round up: 2 bags
- Sand: 9.456 cu.ft = 0.268 m³ = 428 kg (≈ 0.43 tonnes)
- The brick count uses 10 mm mortar joints. If you use 8 mm or 12 mm joints the brick-with-mortar volume changes and counts will differ.
- We rounded bricks up to the nearest whole brick and cement bags up — always order a small extra allowance (≈2–5%) for wastage/cuts.
📖 Example Calculations for Brickwork
Brickwork Calculation for 100 sq.ft Wall (4.5" Thick) with Mortar Ratio ( 1:6 )
Method (brief)
1. Compute single brick nominal volume.
2. Add mortar thickness to get brick-with-mortar volume (this gives the volume occupied by one brick + its surrounding mortar).
3. Number of bricks = Wall volume ÷ Brick-with-mortar volume.
4. Brick solid-volume = Number of bricks × Brick nominal-volume.
5. Mortar volume = Wall volume − Brick solid-volume. (This is the correct mortar quantity.)
Key calculations (numbers)
Mortar quantities (convert to dry volume & split by mix)
Practical material summary
Notes & recommendations
Note: 4.5" walls (half brick) used for partition walls and compound walls.
Plastering - DiyCivil-Calc | Cement & Sand Quantity Estimator
📖 Internal Wall Plaster Example
- 12 mm (internal walls)
- 15 mm (external walls)
- 20 mm (uneven surfaces)
- Convert to sq.m: 100 ÷ 10.764 = 9.29 m²
- Thickness = 0.012 m
- 1:4 (Cement : Sand)
- 1:5
- 1:6
- Area: 100 sq.ft (9.29 m²)
- Plaster Thickness: 12 mm
- Dry Mortar Volume: 0.147 m³
- Cement Required: 0.85 bags (≈ 42 kg)
- Sand Required: 0.1176 m³ (4.15 cubic ft.)
- Mix Ratio: 1:4
- Wall unevenness
- Mortar shrinkage
- Wastage
- Plaster thickness variation
- Type of sand used
Calculation of 12 mm thick plaster for 100 sq.ft.
1. Standard Plaster Thickness
Plaster thickness depends on wall leveling and finish requirements. Commonly used thicknesses include:
For this example, we will use a standard 12 mm (0.012 m) thickness.
2. Plaster Area
The basic formula for plaster volume:
Plaster Volume = Area × Thickness
Example: Plastering a wall of 100 sq.ft
Wet Volume = 9.29 × 0.012 = 0.111 m³
3. Dry Mortar Volume
Dry mortar = Wet volume × 1.33 (wastage + bulking)
Dry Volume = 0.111 × 1.33 = 0.147 m³
4. Mortar Mix Ratio
Common plaster mortar ratios:
In this example, we use 1:4 ratio.
5. Cement and Sand Calculation
Total Ratio
1 + 4 = 5
Cement Quantity
Cement = (1/5) × 0.147 = 0.0294 m³
Density of cement = 1440 kg/m³
Weight = 0.0294 × 1440 = 42.3 kg
50 kg bags = 42.3 ÷ 50 = 0.85 bags
Sand Quantity
Sand = (4/5) × 0.147 = 0.1176 m³ (4.15 cubic ft.)
Sand density = 1600 kg/m³
6. Final Summary for 100 sq.ft Plaster (12 mm, 1:4 Mix)
7. Quick Reference Table for Plaster (12 mm Thickness)
Based on 1:4 mortar mix
| Area | Cement (bags) | Sand (m³) |
|---|---|---|
| 100 sq.ft | 0.85 bags | 0.118 m³ |
| 200 sq.ft | 1.70 bags | 0.236 m³ |
| 500 sq.ft | 4.25 bags | 0.589 m³ |
| 1000 sq.ft | 8.50 bags | 1.178 m³ |
8. Factors Affecting Plaster Quantity
9. Conclusion
Plaster calculation becomes simple when area, thickness, mix ratio, and material densities are properly accounted for. This method can be used for internal walls, external walls, ceilings, and all types of plastering work. Using standard formulas ensures accurate and reliable estimations for any construction project.
Note: Internal plastering uses 1:4 or 1:5 ratio with 12mm thickness.
📖 External Wall Plaster Example
Problem: Calculate plaster for external wall: 5m×3m, thickness=15mm, ratio 1:5.
Solution:
- Plaster area = 5m × 3m = 15 m²
- Wet volume = 15 × 0.015 = 0.225 m³
- Dry volume = 0.225 × 1.25 = 0.281 m³
- For 1:5 (total 6 parts): Cement part = 0.281/6 = 0.047 m³
- Cement = 0.047/0.035 = 1.34 bags ≈ 2 bags
- Sand = (0.281/6) × 5 = 0.234 m³
Note: External plastering uses 1:5 or 1:6 ratio with 15-20mm thickness.
📖 Ceiling Plaster Example
Problem: Calculate plaster for ceiling: 4m×3m, thickness=10mm, ratio 1:3.
Solution:
- Ceiling area = 4m × 3m = 12 m²
- Wet volume = 12 × 0.010 = 0.12 m³
- Dry volume = 0.12 × 1.25 = 0.15 m³
- For 1:3 (total 4 parts): Cement part = 0.15/4 = 0.0375 m³
- Cement = 0.0375/0.035 = 1.07 bags ≈ 2 bags
- Sand = (0.15/4) × 3 = 0.1125 m³
Note: Ceiling plastering uses richer mix 1:3 or 1:4 with 10-12mm thickness.
Tiling - DiyCivil-Calc | Tile Boxes & Adhesive Estimator
📖 Floor Tiling Example
- Length = 10 ft
- Width = 10 ft
- Total area = 100 sq.ft
- Tile size = 2 ft × 2 ft = 4 sq.ft/tile
- Wastage = 12%
- Skirting height = 4 inches = 0.33 ft
- Door width = 3 ft (deducted)
- Skirting uses cut strips from 2×2 ft tiles
- Mortar thickness = 1 inch = 0.083 ft
- Ratio = 1:4 (cement : sand)
- Dry volume = +30%
- Total ratio = 1 + 4 = 5
- Cement = 10.79 × (1/5) = 2.158 cu.ft
- Sand = 10.79 × (4/5) = 8.63 cu.ft
- Thickness = 0.5 inch = 0.041 ft
- Area = 12.21 sq.ft
- Ratio = 1:4
- Cement = 0.65 × (1/5) = 0.13 cu.ft
- Sand = 0.65 × (4/5) = 0.52 cu.ft
- 4–5 kg per sq.m for 2×2 ft tiles
- If using larger tiles (600×600, 800×800), change tile area only.
- Mortar thickness varies by site—commonly 10–20 mm.
- Diagonal layout needs 15–18% wastage.
- Adhesive is recommended for vitrified tiles for better bonding.
Floor Tile Calculation for a 10 ft × 10 ft Room – With Skirting, Door Deduction, Floor Mortar, Skirting Mortar & Adhesive
1. Room Dimensions
2. Floor Tile Calculation
Assumptions
Step 1: Tiles Needed (Without Wastage)
Tiles = 100 sq.ft ÷ 4 sq.ft = 25 tiles
Step 2: Add Wastage
12% of 25 = 3 tiles
Total floor tiles = 28 tiles
3. Skirting Tile Calculation
Assumptions
Step 1: Perimeter
Perimeter = 2 × (10 + 10) = 40 ft
Step 2: Deduct Door
Skirting length = 40 − 3 = 37 ft
Step 3: Skirting Area
Area = 37 × 0.33 = 12.21 sq.ft
Step 4: Tiles Required
Tiles = 12.21 ÷ 4 = 3.05 tiles
Add Wastage
3.05 + 12% = 4 tiles
4. Total Tiles Required
| Item | Tiles |
|---|---|
| Floor tiles | 28 |
| Skirting tiles | 4 |
| Total tiles | 32 tiles |
5. Cement Mortar for Floor Tiles
Assumptions
Step 1: Wet Volume
Wet volume = Area × Thickness = 100 × 0.083 = 8.3 cu.ft
Step 2: Dry Volume
Dry volume = 8.3 × 1.30 = 10.79 cu.ft
Step 3: Cement & Sand
Cement Bags
1 bag = 1.25 cu.ft
Cement bags = 2.158 ÷ 1.25 = 1.73 bags
Cement for floor = 1.73 bags
Sand for floor = 8.63 cu.ft
6. Cement Mortar for Skirting
Assumptions
Step 1: Wet Volume
Volume = 12.21 × 0.041 = 0.50 cu.ft
Step 2: Dry Volume
Dry volume = 0.50 × 1.30 = 0.65 cu.ft
Step 3: Cement & Sand
1 bag = 1.25 cu.ft → Cement = 0.13 ÷ 1.25 = 0.10 bag (≈5 kg)
Final (Skirting)
Cement for skirting = 0.10 bag
Sand for skirting = 0.52 cu.ft
7. Tile Adhesive Calculation (Optional)
Industry Standard Consumption
Convert Area
100 sq.ft = 9.29 sq.m
Adhesive Required
Adhesive = 9.29 × 5 = 46.45 kg
Total adhesive ≈ 50 kg (2 bags of 20–25 kg)
8. Final Material Summary
| Material | Quantity |
|---|---|
| Total tiles (floor + skirting) | 32 tiles |
| Cement for floor mortar | 1.73 bags |
| Sand for floor mortar | 8.63 cu.ft |
| Cement for skirting mortar | 0.10 bag |
| Sand for skirting mortar | 0.52 cu.ft |
| Tile adhesive (Optional) | 50 kg |
9. Notes
Note: Floor tiles typically 600×600mm or 800×800mm vitrified tiles.
📖 Wall Tiling Example
Problem: Calculate tiles for bathroom wall: 3m×2.5m, tile 300×450mm, 10% wastage.
Solution:
- Wall area = 3m × 2.5m = 7.5 m²
- Tile area = 0.3m × 0.45m = 0.135 m²
- Number of tiles = 7.5 ÷ 0.135 = 55.56 tiles
- With 10% wastage = 55.56 × 1.10 ≈ 62 tiles
- Tile boxes (if 6 tiles/box) = 62 ÷ 6 = 11 boxes
- Adhesive (5mm bed) = 7.5 × 5 × 1.6 kg/m²/mm = 60 kg
- Grout (2mm joint) = 7.5 × 1.2 kg/m² = 9 kg
Note: Wall tiles usually 300×450mm or 300×600mm ceramic tiles.
📖 Floor with Skirting Example
Problem: Floor 4m×3m + skirting (4" height), tile 600×600mm, door 900mm wide.
Solution:
- Floor area = 4m × 3m = 12 m²
- Floor tiles = (12 ÷ 0.36) × 1.10 ≈ 37 tiles
- Skirting perimeter = 2×(4+3) - 0.9 = 13.1m
- Skirting area = 13.1 × 0.1 = 1.31 m²
- Skirting tiles = (1.31 ÷ 0.36) × 1.10 ≈ 4 tiles
- Total tiles = 37 + 4 = 41 tiles
- Adhesive = (12+1.31) × 4 × 1.6 = 85.6 kg
Note: Skirting height typically 4" (100mm), deduct door opening width.
Painting - DiyCivil-Calc | Paint Quantity & Area Coverage Estimator
📖 Interior Wall Painting Example
Problem: Paint interior wall 5m×3m, 2 coats emulsion, coverage 12 m²/L.
Solution:
- Wall area = 5m × 3m = 15 m²
- Total area (2 coats) = 15 × 2 = 30 m²
- Paint required = 30 ÷ 12 = 2.5 liters
- With 5% wastage = 2.5 × 1.05 = 2.63 liters
- Buy: 1×4L can (or 3×1L cans)
Note: Interior emulsion coverage: 12-14 m²/L, apply 2 coats over primer.
📖 Exterior Wall Painting Example
Problem: Paint exterior wall 6m×3m, 2 coats weatherproof, coverage 10 m²/L.
Solution:
- Wall area = 6m × 3m = 18 m²
- Total area (2 coats) = 18 × 2 = 36 m²
- Paint required = 36 ÷ 10 = 3.6 liters
- With 5% wastage = 3.6 × 1.05 = 3.78 liters
- Buy: 1×4L can
Note: Exterior paint coverage: 10-11 m²/L, apply 2-3 coats over primer.
📖 Ceiling Painting Example
Problem: Paint ceiling 4m×3m, 2 coats emulsion, coverage 11 m²/L.
Solution:
- Ceiling area = 4m × 3m = 12 m²
- Total area (2 coats) = 12 × 2 = 24 m²
- Paint required = 24 ÷ 11 = 2.18 liters
- With 5% wastage = 2.18 × 1.05 = 2.29 liters
- Buy: 1×4L can (or 3×1L cans)
Note: Ceiling coverage slightly lower due to absorption, use flat/matte finish.
📖 Door/Window Painting Example
Problem: Paint 3 doors (2.1m×0.9m each), 2 coats enamel, coverage 14 m²/L.
Solution:
- Area per door = 2.1 × 0.9 × 2 (both sides) = 3.78 m²
- Total door area = 3.78 × 3 = 11.34 m²
- Total area (2 coats) = 11.34 × 2 = 22.68 m²
- Paint required = 22.68 ÷ 14 = 1.62 liters
- With 5% wastage = 1.62 × 1.05 = 1.70 liters
- Buy: 2×1L cans
Note: Enamel paint coverage: 13-15 m²/L for wood/metal surfaces.
Blogs
How to Calculate Cement, Sand, Aggregate and Water for a 1000 sq.ft Slab (4.5-inch Thick) (M20 Grade)
25 November 2025
Accurate concrete calculation is essential for planning, budgeting, and ensuring smooth execution of any construction project. This guide explains the complete calculation for a 1000 sq.ft slab with 4.5-inch thickness, using industry-standard formulas and material densities. Additional mix ratios are also provided for quick reference.
1. Convert Slab Thickness
4.5 inches ÷ 12 = 0.375 ft
2. Wet Concrete Volume
Wet Volume = Area × Thickness
= 1000 sq.ft × 0.375 ft
= 375 cu.ft (10.62 m³)
3. Dry Concrete Volume
Standard dry volume factor = 1.54
Dry Volume = 375 × 1.54 = 577.5 cu.ft (16.35 m³)
4. Material Densities Used
- Cement: 1440 kg/m³
- Sand: 1600 kg/m³
- Aggregate: 1500 kg/m³
5. Mix Ratio Used for Detailed Calculation (M20 – 1:1.5:3)
Total parts = 5.5
Cement
- Volume = (1/5.5) × 577.5 = 105 cu.ft
- Volume in m³ = 2.97 m³
- Weight = 2.97 × 1440 = 4,282 kg
- 50-kg bags = 86 bags
Sand
- Volume = 157.5 cu.ft
- Volume in m³ = 4.46 m³
- Weight = 4.46 × 1600 = 7,136 kg (7.14 tons)
Aggregate
- Volume = 315 cu.ft
- Volume in m³ = 8.92 m³
- Weight = 8.92 × 1500 = 13,380 kg (13.38 tons)
6. Final Material Summary for M20 Mix
- Wet Concrete: 375 cu.ft (10.62 m³)
- Dry Concrete: 577.5 cu.ft (16.35 m³)
- Cement: 86 bags (50 kg each)
- Sand: 157.5 cu.ft (4.46 m³ → 7.14 tons)
- Aggregate: 315 cu.ft (8.92 m³ → 13.38 tons)
7.Other Mix Ratios for Quick Reference
M7.5 (1:4:8)
- Total parts = 13
- Cement = 1/13 = 7.69%
- Sand = 4/13 = 30.77%
- Aggregate = 8/13 = 61.54%
M10 (1:3:6)
- Total parts = 10
- Cement = 10%
- Sand = 30%
- Aggregate = 60%
M15 (1:2:4)
- Total parts = 7
- Cement = 14.28%
- Sand = 28.57%
- Aggregate = 57.14%
M20 (1:1.5:3)
- Total parts = 5.5
- Cement = 18.18%
- Sand = 27.27%
- Aggregate = 54.54%
M25 (1:1:2)
- Total parts = 4
- Cement = 25%
- Sand = 25%
- Aggregate = 50%
8. Conclusion
This guide provides a complete and accurate concrete calculation for a 1000 sq.ft slab with 4.5-inch thickness. Using standard densities and mix ratios ensures reliable results that help in proper planning, cost control, and efficient material management. This method is widely used by engineers, contractors, and quantity surveyors for concrete estimation.
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Brickwork Calculation for 100 sq.ft Wall (4.5" Thick) — Corrected (Mortar Deducted)
20 November 2025
How to estimate bricks and mortar when mortar volume is deducted from the total wall volume (correct method). Parameters used:
- Wall area = 100 sq.ft
- Wall thickness = 4.5 in = 0.375 ft
- Wall volume = 100 × 0.375 = 37.5 cu.ft
- Brick nominal size = 9" × 4.5" × 3" (common modular brick)
- Mortar joint thickness = 10 mm (≈0.394")
- Mortar mix used = 1 : 6 (cement : sand)
Method (brief)
1. Compute single brick nominal volume.
2. Add mortar thickness to get brick-with-mortar volume (this gives the volume occupied by one brick + its surrounding mortar).
3. Number of bricks = Wall volume ÷ Brick-with-mortar volume.
4. Brick solid-volume = Number of bricks × Brick nominal-volume.
5. Mortar volume = Wall volume − Brick solid-volume. (This is the correct mortar quantity.)
Key calculations (numbers)
- Wall volume = 37.5 cu.ft
- Brick nominal volume = 9"×4.5"×3" = 0.0703125 cu.ft
- Brick-with-mortar dimensions = (9 + 0.394) × (4.5 + 0.394) × (3 + 0.394) inches
- Brick-with-mortar volume = 0.09028 cu.ft
- Exact number of bricks = 37.5 ÷ 0.09028 = ~415.36 → round up → 416 bricks
- Total brick solid-volume = 416 × 0.0703125 = 29.205 cu.ft
- Mortar (wet) volume = Wall − Brick solid-volume = 37.5 − 29.205 = 8.295 cu.ft
Mortar quantities (convert to dry volume & split by mix)
- Dry mortar volume factor ≈ 1.33
- Dry mortar volume = 8.295 × 1.33 = 11.03 cu.ft
- Mortar mix 1:6 → total parts = 7
- Cement volume = (1/7) × 11.03 = 1.576 cu.ft
- Sand volume = (6/7) × 11.03 = 9.456 cu.ft
Practical material summary
- Bricks required: 416 bricks (rounded up)
- Mortar (wet): 8.295 cu.ft
- Dry mortar volume: 11.03 cu.ft
- Cement: 1.576 cu.ft → 1.26 bags (50-kg bag) → round up: 2 bags
- Sand: 9.456 cu.ft = 0.268 m³ = 428 kg (≈ 0.43 tonnes)
Notes & recommendations
- The brick count uses 10 mm mortar joints. If you use 8 mm or 12 mm joints the brick-with-mortar volume changes and counts will differ.
- We rounded bricks up to the nearest whole brick and cement bags up — always order a small extra allowance (≈2–5%) for wastage/cuts.
- 4.5 inch wall is not a load bearing wall, it is used as a partition wall. 9 inch and above wall are loading bearing walls.
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