We've already discussed Nominal Combine of Concrete and Blend Proportions.

For those who have skipped that please go on and read that. We will wait for you.

Itâs time for a few calculations.

In this article, we will explain âHow to Calculate Cement Volume and its own ingredients?â?*************)

If it's a volume, twe understand there must end up being 3 dimensions then,Length, Height, Width or even Breadth.

We covered some Simple SURFACE and Volume Formulas.

For **Slab**,Â If we have to calculate the concrete quantity,

Volume = Size X Breadth X Width/Thickness = 5 X 3 X 0.125 = 3.75 m^{3}

For **Beam**,Â If we have to calculate the concrete quantity,

Volume = Duration X Breadth X Width = 5 X 0.6 X 0.3 = 3.75 m^{3}

For **Column**,Â If we have to calculate a concrete quantity for the Â below Columns,

**Rectangular Column**,Â Volume = Elevation X Breadth X Width = 5 X 0.6 X 0.3 = 0.9 m^{3}

**Circular Column**,Â Quantity = Ïr^{2}Â h= 3.141256 X (0.15)^{2} X 5 = 0.35 m^{3}

## Concrete Ingredients Calculation

For Cement, Coarse and sand Aggregate.Â

It is a Volumetric Calculation.

Assuming we are in need of 2 m^{3} of concrete for M20 Cement Mix, (Combine Ratio, M20 = 1 : 1.5 : 3)

Total Area of the Concrete = 1+1.5+3 = 5.5 Parts

Therefore, **Cement Volume** = (Cement Part / Concrete Components ) * Concrete Volume

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (1/5.5)* Concrete Volume = (1/5.5)*2 = 0.3636 m^{3}

Density of Cement = 1440 kg/ m^{3} = 0.3636 X 1440 = 523 kg = **10.5 Approx. Luggage **

**Sand Quantity** = ( Sand Component / Concrete Parts ) * Concrete Quantity = (1.5/5.5) * 2 = **0.5454 m ^{3}**

**Coarse Aggregate** = (Coarse Aggregate Part / Concrete Parts ) * Cement Volume

Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â = (3/5.5) * 2 = **1.09 m ^{3}**

**Water Cement Ratio**

In accordance with IS 10262 (2009), AssumingÂ Water-Cement Ratio for the Concrete since 0.45

Required Amount of Drinking water = W/C Ratio X Cement Quantity

Therefore, Drinking water = 0.45 X 0.3636 m^{3Â }= 0.16362 m^{3}

Unit Weight of Drinking water = 1000 litres/m^{3}

**Required Amount of Drinking water** = 0.16362 X 1000 = **163.62 litres**

Therefore, we are in need of 10.5 bags of cement, 0.5454 m^{3} of sand, 1.09 m^{3} of Coarse aggregates and 163.62 litres of water.

## Concrete ingredients Calculator

Hope you enjoyed this content.

Happy Learning ð

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