Tuesday, 29 November 2016

HOW TO CALCULATE CEMENT, SAND AND AGGREGATE QUANTITY IN CONCRETE

Consider Volume (Wet) = 1 m3
Grade of Concrete = M15 (Cement - 1 : sand - 2 : Aggregate - 4)
Sum Of Ratio = 1+2+4 = 7

Dry Volume = Wet Volume + 54% of Wet Volume = 1 + 0.54 = 1.54 m3

CEMENT
1. CUM : (1.54 x 1)/7 = 0.22 CUM
2. Kg : (1.54 x 1 x 1440)/7 = 316.8 Kg
3. Bags : We know 1 bag of cement weight = 50 Kg
                Then, 316.8 Kg contains, 316.8/50 = 6.336 bags
4. CFT : If we multiply No. of bags by 1.226 then we get the value in CFT.
               i.e, 6.336 x 1.226 = 7.77 CFT

SAND
1. CUM : (1.54 x 2)/7 = 0.44 CUM
2. Kg : (1.54 x 2 x 1450)/7 = 638 Kg [Density Varies in between 1450-1500 Kg/m3]
3. CFT : (1.54 x 2 x 35.3147)/7 = 15.54 CFT
     Alternative : No. of Cement Bags x Value of 1 bag of cement in CFT x ratio of sand
                       = 6.336 x 1.226 x 2 = 15.54 CFT

AGGREGATE
1. CUM : (1.54 x 4)/7 = 0.88 CUM
2. Kg : (1.54 x 4 x 1500)/7 = 1320 Kg [Density Varies in between 1450-1550 Kg/m3]
3. CFT : (1.54 x 4 x 35.3147)/7 = 31.07 CFT
     Alternative : No. of Cement Bags x Value of 1 bag of cement in CFT x ratio of sand
                       = 6.336 x 1.226 x 4 = 31.07 CFT


DERIVATION : HOW MUCH CFT IN ONE BAG CEMENT

We Know Density of Cement = 1440 Kg/m3

That means for 1 m3 Cement required is 1440 Kg
We Know, 1 Bag of Cement = 50 Kg
For,  1 m3 Cement required = 1440/50 = 28.8 Bags

We know, 1 m3 = 35.3147 CFT
or, 28.8 bags = 35.3147 CFT
or, 1 Bag = 35.3147/28.8 = 1.226 CFT