What are the quantities of cement, sand, and aggregate in concrete?


Consider grade of Concrete is M20 (1:1.5:3)

Consider the volume of concrete (wet volume) = 1 Cum

The dry volume of concrete is usually considered to be 54% more than it’s wet volume.

Therefore, Dry volume of concrete = wet volume + (54% of wet volume) = 1 + (0.54 x 1) = 1.54 Cum

Hence, we can say that the dry volume of concrete is 1.54 times it’s wet volume.

Mix ratio for M20 grade concrete = 1 : 1.5 : 3 (cement : sand : aggregate)

Sum of mix ratio = 1 + 1.5 + 3 =5.5

 

CEMENT :

Volume of cement = dry volume of concrete x (proportion of cement in mix ratio / sum of mix ratio) = 1.54 x (1 / 5.5) = 0.28 Cum.

Quantity of cement in kg. = volume of cement x density of cement = 0.28 x 1440 = 403.2 kg

Quantity of cement in no. of bags = 403.2 / 50 = 8.064 ~ 8 bags

Since, 1 bag has 50 kg cement.

SAND :

Volume of sand = dry volume of concrete x (proportion of sand in mix ratio / sum of mix ratio) = 1.54 x (1.5 / 5.5) = 0.42 Cum.

Quantity of sand in kg. = volume of sand x density of sand = 0.42 x 1450 = 609 kg

Density of sand varies between 1440 kg/Cum. to 1600 kg/Cum. Here, I’ve considered 1450 kg/Cum.

AGGREGATE :

Volume of aggregate = dry volume of concrete x (proportion of aggregate in mix ratio / sum of mix ratio) = 1.54 x (3 / 5.5) = 0.84 Cum.

Quantity of aggregate in kg. = volume of aggregate x density of aggregate = 0.84 x 1550 = 1302 kg

Density of aggregate varies between 1600 kg/Cum. to 1880 kg/Cum. Here, I’ve considered 1550 kg/Cum.

WATER :

Assuming w/c ratio = 0.5 (for details, refer IS 456 : 2000 Table 5)

Therefore, Water = 0.5 x cement = 0.5 x 403.2 = 201.6 kg. i.e., 201.6 lit.

Since, 1 lit. of water weighs exactly 1 kg.

                                    Thank you...................................