Register Now

Login

Lost Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Login

Register Now

Lorem ipsum dolor sit amet, consectetur adipiscing elit.Morbi adipiscing gravdio, sit amet suscipit risus ultrices eu.Fusce viverra neque at purus laoreet consequa.Vivamus vulputate posuere nisl quis consequat.

Quantity of Brick, Cement, Sand And Aggregates Required To Build 10ft x 10ft Room

Before calculating the number of bricks and quantity of concrete, we need to know some data. Let us assume


The length of room = 10 ft

breadth of room = 10 ft

Height of room = 10 ft.

For easier calculation we will use M15 grade concrete. For M15 concrete

Cement : Sand : Aggregates = 1:2:4

Calculation For Floor And Ceilings:

Volume of floor/ceiling = 10x10x1 (assuming thickness of slab =1ft)= 100 cft

1 bag of cement contains 50 kg cement = 1.71cft

For M15 concrete (1:2:4),

Sand required for 1 bag cement = 1.71 x 2 = 3.42 cft

Aggregates required for 1 bag cement = 1.71 x 4v = 6.84 cft

Total Volume = Cement+Sand+Aggregates = 1.71 + 3.42 + 6.84 = 11.97 cft

Cement required for 100 cft volume = (100/11.97) x 1.71 = 14.19 cft

Sand required for 100 cft volume = (100/11.97) x 3.42 = 28.38 cft

Aggregates for 100 cft volume = (100/11.97) x 6.84 = 56.76 cft

For ceiling, the abouve quantity will be doubled.

Calculation For Walls:

Assuming load bearing structure (ignoring beams and columns) assume 20% of floor area is required for openings (door, windows, ventilation).

Openings need to be provided = 20sq.ft

Thickness of wall = 9 inch

Rough volume of wall = [ (9In x 9In x 4) + (8.5ft x 9In x 4) ] x 10 ft = 257.25 cuft

Opening subtractions = 257.25 – 20% of 257.25 = 205.8 cuft

Volume of one Brick = 0.24 ft x 0.37 ft x 0.75 ft = 0.06 cuft

Total no. of bricks = 205.8 cuft / 0.06 cuft = 3430 Bricks

Mortar to be applied on bottom and one end of brick. Volume of mortar for 1 brick = (0.24 x 0.75 x 0.03) + (0.75 x 0.37 x 0.03) = 0.013 cuft.

Total number of bricks = 3430; Thus total Mortar required = 3430 x 0.013 = 44.59 cuft; Mortar ratio normally used = 1 : 5 ( Cement : Sand ); Thus total Cement = 11.84 cuft; Total Sand = 32.75 cuft.

Now For the Final Calculations, The total material required is –

a. Cement = 14.19 x 2 + 11.84 = 40.22 cuft Or 1.14 cum; (Density of cement is 1440 kg/cum); Thus, Mass = 1,641.6 Kg or 33 bags of 50 Kg.

b. Sand = 28.38 x 2 + 32.75 = 117.89 cuft Or 3.3 cum; (Density of sand is 1350 kg/cum); Thus, Mass = 1,455 Kg or 90 Bags of 50 Kg.

c. Coarse Aggregate = 56.72 x 2 = 113.44 cuft Or 1.6 cum; (Density of coarse aggregate is 1300 kg/cum); Thus, Mass = 2,080 Kg or 42 bags of 50 Kg.

d. Bricks 3,430 Bricks.



Comments ( 2 )

  1. Pls, let your unit be in metres (m)

  2. 1 bag of cement = 1.71 CFT????? 1.25 CFT is the right answer! Right??

Leave a reply

Captcha Click on image to update the captcha .


error: Content is protected !!