Materials Calculation for 1 Cum Concrete




Hello Friends, In this article today I will show you the method for Materials Calculation for 1 Cum Concrete




Materials Calculation for 1 Cum Concrete

For calculation, We consider a nominal concrete mix proportion of 1:2:4 (~M20).

Considering a concrete mix proportion (by volume) of 1:2:4

Cement: Fine aggregate (Sand) : Coarse Aggregate is in the ratio of 1:2:4 by volume.

Cement required = 01 bag = 50 kg ~ 36 liters or 0.036 cum

Fine Aggregate required = 2*0.036 ~ 0.07 cum = 0.072*1600 = 115 kg

Coarse Aggregate required = 4*0.036 ~ 0.14 cum =0.144*1450 = 209 kg

Considering water – cement ratio (w/c) of 0.55 , water = 0.55*50 kgs = 27.5 kg




Calculate Material Requirement For Producing 1 Cum Concrete

The volume proportion in weight turns out to be

Cement : Sand : Aggregate (in Kgs) is 50 kgs : 115 kgs : 209 kgs (by weight)

Water required for the mixture = 27.5 kgs

Total weight of concrete ingredients = 50+115+209+27.5 = 401.5 say 400 kg

Density of concrete = 2400 kg/cum

So, 1 bag of cement produces = 400/2400 = 0.167 cum

No. of bags required for 01 cum of concrete = 1/0.167 = 5.98 bags ~ 6 bags



From above, if the concrete mix is 1:2:4 , to get a cubic meter of concrete we require

1.Cement = 6 bags = 300 kgs.

2.Fine Aggregate = 115/0.167 = 689 kg

3.Coarse Aggregate = 209/0.167 = 1252 kg.

4.Water = 300/0.55 = 165 kg.

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Engr Sami Ullah Author

Founder & Admin of civilengineeringinfo.com I am Civil Engineer working as a Site Engineer. With a good subject knowledge in civil Engineering. And I have started the blog to share my valuable information with civil engineer students. Civil Global is the Civil Engineering learning Home around all over the world.

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