For an area of 2000 ft² with wall thickness 20 cm find

(i) No. of bricks

(ii) C:M required for Brick work with ratio 1:5

Solution:

Assuming:-

- Brick size = 19cm x 19cm x 19cm
- Thickness of mortar = 10 mm

Before we get into the calculation part, Let’s convert ft² to m² for convenient of calculations.

2000 ft² = ? m²

= 2000/3.28² (since 1 m = 3.28 ft)

= 185.901 m²

**Volume of brickwork = Area x Thickness of wall**

= 185.901 x 0.20

= 37.1802 m³

### Quantity of Bricks:-

**No. of bricks = Volume of brickwork
**

**volume of 1 brick with mortar**

= 37.1802/0.002

∴ No. of bricks = 18590 bricks

Volume of 1 Brick with mortar = 0.20×0.10×0.10=0.002 m³

**CM required for brickwork:-**

Volume of total no.of bricks = (0.19×0.09×0.09) x 18590

= 0.001539 x 18590

= 28.61 m³

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

Quantity of mortar = 37.1802 – 28.61

= 8.5702 m³

**Mix Ratio –> 1:5**

**Dry volume of mortar = Wet volume x 1.33**

∴Dry volume of mortar = 8.5702 m³x 1.33

= 11.3984 m³

**Quantity of cement:-**

**Quantity of Cement = Dry Volume of mortar x Cement ratio
**

**Sum of the ratio**

∴Quantity of cement = (11.3984 x 1) /(1+5)

= 1.8997 m³

**Density of Cement = 1440 kg/m****³**

∴ Weight of Cement = 1440 x 1.8997

= 2735.616 Kg

**1 bag of cement contains 50 kg of cement**

∴ Number of bags = 2735.616 Kg / 50 kg

= **57.712**** No’s**

**Quantity of Sand:-**

**Cement : Sand :: 1:5**

**Quantity of Sand = Quantity of Cement x 5**

∴ Quantity of Sand = 1.8997 m³x 5

= **9.4985 ****m³**

**1 m³=35.3147 Cubic Feet (CFT)**

∴ Quantity of sand = 9.4985 x 35.3147

= 335.436 CFT

**Density ****of sand = 1650 kg/m³**

∴ Weight of the sand = 9.4985 x 1650

= 15672.53 kg

=15.6725 tonnes

**Watch the video given in the link below, for better understanding and detailed explanation**

Easy method to CALCULATE NO OF BRICKS AND QUANTITY OF CEMENT & SAND in BRICKWORK

Click here to download pdf of “How to calculate brick, sand and cement in brick masonry?”

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