How Many Cement, Sand and Steel Required For 10×10 Feet Room?
44 bags of 50 kg cement and 104.2 cft sand is required for 10×10 feet room by assuming 11 feet height of room with some other requirements and considerations. Read complete to understand the concept.
Calculation For Cement, Sand and Steel Required For 10×10 Feet Room
To calculate the quantity of cement, sand and steel required for 10×10 feet room (1000 sq. feet room), we have some considerations and requirements in the room.
Requirements of the Room
We have the following requirements for 10×10 feet room:
- Four sided walls
- One gate
- One window
- RCC Roof
Considerations of the Room
We are considering the following for 10×10 feet room.
- The wall size is 10 feet height × 10 feet width
- One window of size 4 feet × 3 feet
- One gate of size 7 feet × 3 feet
- Indian standard brick size is 190×90×90 mm
- Mortar thickness 12 mm (cement:sand ratio is 1:4)
- Roof slab thickness is 5 inch (cement:sand:coarse aggregate ratio is 1:1.5:3)
- Wall thickness is 4 inch
- Column size is 10 inch × 12 inch
- Beam size is 10 inch × 12 inch
- Lintel Size 4 inch × 5 inch
- Clear Floor Height is 11 feet
Brick Work Calculation
Face Area of Brick Work
= Face Area of Wall - Face Area of Gate and Windows
= (No. of Walls × Length × Height) - (Length × Width of Gate) - (Length × Width of Window)
= (4 × 10 × 10 × 4/12) - (7 × 3 × 4/12) - (4 × 3 × 4/12)
= 122.33 cft
Volume of One Brick with 12 mm Mortar
= 202 × 102 × 102 = 2101608 mm³ or 0.07422 Cubic Feet
Total Number of Bricks in 10×10 Feet Room
= (Volume of Brick Work/Volume of One Brick With Mortar)
= (122.33/0.07422) = 1648 Bricks
Considering waste factor (10% extra) = 1648 + 10% of 1648 = 1648 = 1812.8, Say 1813 Bricks.
Cement and Sand Calculation in Mortar
- Volume of Brick Work = 122.33 Cubic Feet
- Volume of One Brick Without Mortar is 0.05438 Cubic Feet
- Number of Bricks in 12×12 Feet Room is (122.33/0.05438) = 1813
- Mortar Thickness is 12 mm, Mortar Ratio is 1:4
- Shrinkage Factor is 1.3 (i.e. 30% extra of dry volume) and
- Unit Weight of Cement is 1440 kg/m³
- Unit Weight of Sand is 1600 kg/m³
Volume of Dry Mortar
= Total Volume of Brick Work - Total Volume of Bricks Without Mortar
= 122.33 - 1648 × 0.05438 = 32.712 Cubic Feet
Total Volume of Dry Mortar
= 1.3 × Volume of Dry Mortar
= 1.3 × 32.712 = 42.526 cft or 1.204 m³
Total Mortar Ratio (1:4) = 1+4 = 5
Total Volume of Cement
= (Cement Ratio/Total Mortar Ratio) x Total Dry Volume of Mortar x Unit Weight of Cement
= (1/5) × 1.204 × 1440 kg/m³ = 295 kg or 6 Bags of Cement (Each 50 kg).
Total Volume of Sand
= (Sand Ratio/Total Mortar Ratio) × Total Dry Volume of Mortar × Unit Weight of Sand
= (4/5) × 1.204 × 1600 kg/m³ = 1541 kg or 0.963 m³ or 34 cft.
Calculation For RCC Work
Volume of RCC of Column
= No. of Column × (Total Height of Column × Cross-Section of Column)
= 4 [(11 Feet Height + 5 Feet Below Ground Level) × (10 inch/12) × (12 inch/12)]
= 53.3 cft
Volume of RCC of Beam (Plinth Beam and Floor Beam)
= 2 × [Total Length of Beam × Cross-Section of Beam)]
= 2 × [(4 Sides × 10 feet) × (10 inch/12) × (12 inch/12)]
= 66.67 cft
Volume of RCC of Lintel
= (4 Sides × 10 feet) × (4 inch/12) × (5 inch/12)
= 5.56 cft
Volume of RCC of Roof Slab
= Area of Roof Slab × Thickness of Roof Slab
= 10 Feet × 10 Feet × (5 inch/12)
= 41.67 cft
Total Volume of RCC Work = 53.3+66.67+5.56+41.67 = 167.2 cft (Wet Concrete).
Calculation of Cement, Sand and Coarse Aggregate
Ratio of Concrete (M20) = 1:1.5:3 (cement:sand:coarse aggregate)
Shrinkage Factor = 1.54 (54 % extra for dry volume)
Total Concrete Ratio (1:1.5:3) = 1+1.5+3 = 5.5
Total Volume of Dry Concrete = 1.54 × 167.2 cft = 257.5 cft
Volume of Cement in RCC Work
= (Ratio of Cement/ Total Ratio of Concrete) × Total Volume of Dry Concrete
= (1/5.5) × 257.5 = 46.8 cft or 1.325 m³ or 1908 kg or 38 Bags of 50 kg Cement
Volume of Sand (Fine Aggregate) in RCC Work
= (Ratio of Sand/ Total Ratio of Concrete) × Total Volume of Dry Concrete
= (1.5/5.5) × 257.5 = 70.2 cft
Volume of Coarse Aggregate in RCC Work
= (Ratio of Aggregate/ Total Ratio of Concrete) × Total Volume of Dry Concrete
= (3/5.5) × 257.5 = 140.5 cft
Weight of Steel in RCC Work
Considering volume of Steel is 2% of total volume of concrete.
= 167.2×2 /100 = 3.344 cft or 0.095 m³
Density of Steel is 7850 kg/m³
Weight of Steel = 7850 × 0.095 = 746 kg
Finlay,
Total Cement Required = 295 kg for brick work + 1908 kg for rcc work = 2203 kg or 44 Bag Cement (50 kg each bag)
Total Sand Required = 34 cft for brick work + 70.2 cft for rcc work = 104.2 cft
Total Aggregate Required = 140.5 cft
Total Weight of Steel Required = 746 kg
Summary
Bags of Cement in 10×10 Feet Room is 44 Nos. (Each 50 kg).
Volume of Sand in 10×10 Feet Room is 104.2 cft.
Volume of Aggregate in 10×10 Feet Room is 140.5 cft.
Weight of Steel in 10×10 Feet Room is 4 kg
Number of Bricks in 10×10 Feet Room is 1813 Nos.
Note: This quantity just for reference and it is based on above assumptions and it may vary depending upon material type and other design changes.
How Much Cement Required For 100 Sq Feet Area?
As per above calculation, cement, sand, coarse aggregate and steel quantity is given below.
Bags of Cement in 100 Sq Feet Area is 44 Nos. (Each 50 kg).
Volume of Sand in 100 Sq Feet Area is 104.2 cft.
Volume of Aggregate in 100 Sq Feet Area is 140.5 cft.
Weight of Steel in 100 Sq Feet Area is 4 kg
Number of Bricks in 100 Sq Feet Area is 1813 Nos.
Note: This quantity just for reference and it is based on above assumptions and it may vary depending upon material type and other design changes.
