## How To Calculate Brick, Cement And Sand In Brick Masonry

Contents

Quantity estimation of materials is essentially required in any construction works and the quantity of materials depends on the mix proportions of the concrete. In this article, I will discuss how to calculate bricks, cement, and sand in brick masonry. So let’s get started.

**Assuming**,

- Volume of brickwork = 1 m
^{3} - Grade of mortar = 1:6 (cement : sand)
- First class brick (190 mm x 90 mm x 90 mm)
- Thickness of mortar = 10 mm = 0.01 m

### No. Of **Bricks:**

No. of bricks = (Volume of brickwork / Volume of one brick with mortar)

Volume of one brick without mortar = 0.19×0.09×0.09 = 0.001539 m^{3}

*V*olume of brick with mortar = (0.19+0.01) x (0.09+0.1)x (0.09+0.1)

= 0.2×0.1×0.1 = 0.002** ** m

^{3}

No.of bricks = 1.0/ (0.002)** = **500

**Consider 10% to 15 % bricks as wastage.**

∴ Total no. of bricks = 500 + (10 x 500 )/100 = 550

**Quantity Of Mortar:**

**Volume occupied by bricks = No. of bricks x Volume of one brick**

Volume of bricks = 500 x 0.001539 = 0.7695 m^{3}

**Volume of mortar = Volume of brickwork – Volume of bricks**

∴ Volume of Mortar = 1.0 – 0.7695 = 0.2305** ** m^{3}

**Quantity Of Cement:**

**Cement = (Dry volume of mortar x Cement ratio)/ Sum of the ratio**

Dry volume of Mortar = 1.54 x 0.2305 = 0.35497 m^{3} (54% increment due to volume shrink after adding water)

Cement= (0.35497 x 1) / (1+6) = 0.35497/7= 0.05071 m^{3}

Cement = 0.043795 x 1440 = 73.0224 kg

∴ No. of cement bags = 73.0224 / 50 =* *1.45 bags (1 bag cement contains 50 kg cement)

**Quantity of Sand:**

**Sand =** **(Dry volume of mortar x Sand ratio)/Sum of the ratio**

Sand = (0.35497 x 6)/7 = 2.12982/7 = 0.30426 m^{3}

∴ Sand = 0.30426 x 35.3147 = 10.7448 cft.

### Summery:

**Number of Bricks = 550**

**Cement = 1.45 bags of 50 Kg**

**Sand = 0.30426 ** **m ^{3}**

**or10.7448 cft.**

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