In this article, we will learn to calculate the number of Bricks, Cement, sand required for a Brick Wall having **7 m Long**, **3.2 m Height**, and **0.23 Width**. The brick wall consists of has a mortar joint with a cement-sand ratio of 1:6. The Bricks of Size 230 mm x 114 mm x 76 mm.

Let’s First Calculate for 1 cum Brickwork.

Volume of Brickwork = 1 cum.

Size of Bricks = 230 mm x 114 mm x 76 mm

**Bricks Cement Sand Calculation**

Volume of One Brick without mortar = 0.23 m x 0.114 m x 0.076 m = 0.00199272 cum.

Number of Bricks Required without Mortar = (Volume of Brick Work)/ (Volume of One Brick without Mortar)

= 1/ (0.00199272) **= 502 Nos.**

Add 10 mm thickness mortar joint, Hence

Size of Bricks = 240 mm x 124 mm x 86 mm.

Volume of one Bricks with mortar = 0.24 m x 0.124 m x 0.086 m

= 0.00255936 cum

Number of Bricks Required with Mortar = (Volume of Brick Work)/ (Volume of One Brick with Mortar)

= 1/ (0.00255936) = **391 Nos**.

**Hence. The numbers of Bricks have reduced from 502 to 391 due to the presence of Mortar.**

**Calculation of Cement & sand**

Quantity of Mortar Volume = Volume of Brickwork – (Volume of Brick without mortar x Number of brick required with mortar)

= 1 – (0.00199272 x 391) = **0.22084648 cum**

Therefore we need 391 Bricks and 0.23 cum Cement Mortar for 1 Cum Brickwork.

Now we also have to find how much cement and how much quantity of sand is needed for 0.23 cum of Mortar.

As per my Particle experience, the ratio of Cement and Sand used in 230 mm thick wall is 1:6.

Wet volume of Cement Mortar = 0.23 cum.

Dry Volume of Mortar = 0.23 x 1.33 = 0.3059 cum.

C: S Ratio = 1:6.

**Cement Calculation**

Cement in Cum = (Volume of Mortar x Ratio of Cement) / (Sum of Ratio)

= (0.3059 x 1) (1+6) = 0.3059/7 = **0.0437 m ^{3}**

Cement in Kilogram = 0.0437 x 1440 = **62.928 Kg.**

Cement in Bag = 62.928 / 50 = **1.26 Bags.**

**Sand Calculation**

Sand in cum = (Volume of Mortar x Ratio of sand) / (Sum of Ratio)

= (0.3059 x 6) (1+6) = 0.3059/7 **= 0.2622 m ^{3}**

In Short for 1 Cum Brickwork we need;

Bricks = 391 Nos.

Cement = 1.26 Bags

Sand = 0.2622 cum.

Now Consumption of Material for the brick wall of Size 7 m Long, 3.2 m Height and 0.23 Width

Volume of Brick wall = 7 x 0.23 x 3.2 **= 5.152 cum.**

**Consumption of Material**

Therefore, the consumption of Material shall be such that;

Cement = 1.26 Bags x 5.152 **= 6.5 Bags.**

Sand = 0.2622 cum x 5.152 **= 1.35 cum.**

Bricks = 391 Bricks x 5.152 **= 2015 Nos.**

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