**Calculating number of blocks for a 3 bedroom flat. **

Most asked question from a would be house owner is always how many blocks do I need to get my building completed. So therefore this topic is written with illustrations to make it easy for you. The quality to estimate is 6 inches block(without holes) to complete the project.

Lets use this plan for our demonstration.

The data from the building is shown below;

**DOOR TYPES**

(D1) = 1 piece (Dimension = 2100 × 1200)

(D2) = 6 pieces (Dimension = 2100 × 900)

(D3) = 4 pieces (Dimension = 2100 × 700)

**WINDOWS**

Window Type 1 (W1) = 12 pieces (Dimension = 1200 × 1200)

Window Type 2 (W2) = 4 pieces (Dimension = 600 × 600)

Storey height = 3000mm

### Three main phases of constructing a bungalow

- Foundation to DPC
- DPC to lintel level, and
- Lintel level to overhead level

**Estimation of number blocks**

The simplest approach is to estimate the number of blocks required in a metre square of wall.

The planar dimensions of a standard block in Nigeria is (450mm × 225mm).

Therefore the planar area of 1 block = (0.225 × 0.45) = 0.10125 m^{2}

Therefore one metre square of wall will contain;

1/0.10125 = 9.876

Therefore, one metre square of wall will contain approximately 10 blocks.

**PHASE 1: FOUNDATION TO DPC**

At this stage, the total length of the walls of the building and in particular the height of the walls are considered. There are no openings, and also the entire foundation excavation length is considered. For the building we are considering, the foundation layout is as shown below;

Using any method of your choice, you can calculate the total length of the walls at the foundation level.

This is approximately equal to 100256 mm = 100.256m from my calculations.

The height of the wall from foundation to DPC (ignoring thickness of mortar) = 900mm = 0.9m

Therefore, the total area of wall = 100.256m × 0.9m = 90.2304m^{2}

^{
}If 10 blocks is required for 1m^{2} of wall, therefore (10 × 90.2304) = 903 blocks is required for 90.2304m^{2} of wall.

The number of blocks required to raise the building from foundation to DPC = 903 blocks (disregarding wastage and damages).

**PHASE 2: DPC TO LINTEL LEVEL**

At this level, we will consider the effects of openings in the buildings which includes the doors, windows, and also other openings as specified in the drawing. These openings are the verandah areas, and some of the room entrances. We will use the same process to calculate the total length of the wall and then minus the area of the openings. This process is as shown below;

Total Length of walls (excluding areas with no walls like verandahs) = 87420 mm = 87.42m

Height of wall = 2100 = 2.1m

Therefore area of wall = (87.42 × 2.1) = 183.582 m^{2}

^{
}Total area of all doors from building data = 19.74 m^{2}

Total area of all windows from building data = 18.72 m^{2}

^{
}Therefore the net area of walls = 183.582 m^{2} – 19.74 m^{2} – 18.72 m^{2} = 145.122 m^{2}

^{
}Hence, the total number of blocks required to take the building from DPC to lintel level = 145.122 × 10 = 1451.22 = 1452 blocks (disregarding damages and wastage)

**PHASE 3: LINTEL LEVEL TO OVERHEAD LEVEL**

Here, we are going to assume that the building is chained at the lintel level (this means that the entire level of the building will be casted and reinforced with steel at the lintel level. This offers the advantage of helping the building behave as a unit, and also helps in reducing cracking in the building).

Obviously, we will employ the length of wall without considering openings, so we are going to make use of 100.256m as the foundation level.

Length of wall = 100.256m

Height of wall = 675mm = 0.675m

Area of wall = (100.256 × 0.675) = 67.6728 m^{2}

^{
}Hence, the total number of blocks required to take the building from lintel level to roof level = 67.6728 × 10 = 676.728 = 677 blocks (disregarding damages and wastage).

**Ultimately, the total number of blocks required to complete the construction = 903 + 1452 + 677 = 3032 blocks (disregarding wastage).**

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