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

**Contents**show

### 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}

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*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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