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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