# How to calculate brick, sand and cement in brick masonry?

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:-

1. Brick size = 19cm x 19cm x 19cm
2. 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/m³

∴ 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

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

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