# How Many Cement and Sand Required For 12x12 Feet Room?

54 bags of 50 kg cement and 125 cft sand is required for 12x12 feet room by assuming 11 feet height of room with some other requirements and considerations. Read complete to understand the concept.

## Calculation For Cement and Sand Required For 12x12 Feet Room

To calculate the quantity of cement, sand and steel required for 12x12 feet room, we have some considerations and requirements in the room.

### Requirements of the Room

We have the following requirements for 12x12 feet room:

- Four sided walls
- One gate
- One window
- RCC Roof

### Considerations of the Room

We are considering the following for 12x12 feet room.

- The wall size is 10 feet height x 12 feet width
- One window of size 4 feet x 3 feet
- One gate of size 7 feet x 3 feet
- Indian standard brick size is 190x90x90 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 x 12 inch
- Beam size is 10 inch x 12 inch
- Lintel Size 4 inch x 5 inch
- Clear Floor Height is 11 feet

### Brick Work Calculation

**Volume of Brick Work**

= Volume of Wall - Volume of Gate and Windows

= (No. of Walls x Length x Height x Thickness of Wall) - (Length x Width x Thickness of Gate + Length x Width x Thickness of Window)

= (4 x 12 x 10 x 4/12) - (7 x 3 x 4/12 + 4 x 3 x 4/12)

= 149 Cubic Feet

**Volume of one Brick with 12 mm Mortar**

= 202 x 102 x 102 = 2101608 mm³ or 0.07422 Cubic Feet

**Total Number of Bricks in 12x12 Feet Room**

= (Volume of Brick Work/Volume of One Brick With Mortar)

= (149/0.07422) =

**2008 Bricks**

**Cement and Sand Calculation in Mortar**

- Volume of Brick Work = 149 Cubic Feet
- Volume of One Brick Without Mortar is 0.05438 Cubic Feet
- Number of Bricks in 12x12 Feet Room is (149/0.05438) = 2008
- 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

= 149 - 2008 x 0.05438 = 39.805 Cubic Feet

**Total Volume of Dry Mortar**

= 1.3 x Volume of Dry Mortar

= 1.3 x 39.805 = 51.747 Cubic Feet or 1.465 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) x 1.465 x 1440 kg/m³ =

**422 kg**or 8.5 Bags of Cement (Each 50 kg).**Total Volume of Sand**

= (Sand Ratio/Total Mortar Ratio) x Total Dry Volume of Mortar x Unit Weight of Sand

= (4/5) x 1.465 x 1600 kg/m³ =

**1875 kg**or**1.172 m³**or**41.4 Cubic Feet**.### Calculation For RCC Work

**Volume of RCC of Column**

= No. of Column x (Total Height of Column x Cross-Section of Column)

= 4 [(11 Feet Height + 5 Feet Below Ground Level) x (10 inch/12) x (12 inch/12)]

=

**53.3 Cubic Feet****Volume of RCC of Beam (Plinth Beam + Floor Beam)**

= 2 x [Total Length of Beam x Cross-Section of Beam)]

= 2 x [(4 Sides x 12 Feet) x (10 inch/12) x (12 inch/12)]

=

**80 Cubic Feet****Volume of RCC of Lintel**

= (4 Sides x 12 Feet) x (4 inch/12) x (5 inch/12)

=

**6.67 Cubic Feet****Volume of RCC of Roof Slab**

= Area of Roof Slab x Thickness of Roof Slab

= 12 Feet x 12 Feet x (5 inch/12)

=

**60 Cubic Feet**Total Volume of RCC Work = 53.3+80+6.67+60 =

**200 Cubic Feet (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 x 200 Cubic Feet =

**308 cft****Volume of Cement in RCC Work**

= (Ratio of Cement/ Total Ratio of Concrete) x Total Volume of Dry Concrete

= (1/5.5) x 308 =

**56 cft**or 1.586 m³ or 2284 kg or 46 Bags of 50 kg Cement

**Volume of Sand (Fine Aggregate) in RCC Work**

= (Ratio of Sand/ Total Ratio of Concrete) x Total Volume of Dry Concrete

= (1.5/5.5) x 308 =

**84 cft****Volume of Coarse Aggregate in RCC Work**

= (Ratio of Aggregate/ Total Ratio of Concrete) x Total Volume of Dry Concrete

= (3/5.5) x 308 =

**168 cft****Weight of Steel in RCC Work**

Considering volume of Steel is 2% of total volume of concrete.

= 200 x 2/100 = 4 cft or 0.113 m³

Density of Steel is 7850 kg/m³

Weight of Steel = 7850 x 0.113 =

**887 kg**

**Finlay,**

Total Cement Required = 422 kg + 2284 kg = 2706 kg or

**54 Bag**cement (50 kg each bag)Total Sand Required = 41.4 cft for brick work + 84 cft for rcc work =

**125.4 cft**Total Aggregate Required =

**168 cft**Total Weight of Steel Required =

**887 kg****Summary**

- Number of Bags of Cement in 12x12 Feet Room is 54 (each 50 kg).
- Volume of Sand in 12x12 Feet Room is 125 cft
- Volume of Aggregate in 12x12 Feet Room is 168 cft
- Weight of Steel in 12x12 Feet Room is 887 kg
**.** - Number of Bricks in 12x12 Feet Room is 2008 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.