# How to find out the volume & weight of cement, sand & aggregate in concrete

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The purpose of this civil engineering tutorial is to teach you how to find out the quantity of cement, sand and aggregates in concrete or the weight of these materials used in 1 m^{3} concrete. The process is simple and easy.

1 m^{3} means the height, width and length of concrete are 1 meter and the volume will be obtained by multiplying them.

Generally, the we select wet volume of concrete but with the addition of cement, sand and aggregates, we should select dry volume of concrete.

Before adding water, when we create dry mix, it is known as dry volume. The dry volume remains in excess of 54% of wet volume.

So, dry volume = wet volume x 1.54 = 1m^{3} x 1.54 = 1.54 m^{3}

The mix proportion of concrete is taken as 1:3:6 i.e. one part cement, three parts sand and 6 parts aggregates.

Now, total parts = 1+3+6 = 10 parts

The quantity of cement, sand and aggregate will be calculated as follow :-

**CEMENT**:

= 1/10 x 1.54 = 0.154 m^{3}

Now, it is required to convert the quantity to bags.

We know 1 bag of cement = 50 kg and its volume is 0.035 m^{3}

So, total numbers of cement bags = 0.154 / 0.035 = 4.4 bags

So, the weight of the cement = 4.4 x 50 = 220 kg

### Also Read: Some useful tips to work out the quantity of cement, sand & aggregate in concrete

**SAND**:

= 3/10 x 1.54 = 0.462 m^{3}

The density of sand = 1450 kg/m^{3} (approx)

So, the weight of the sand = volume x density = 0.462 x 1450 = 670 kg

**AGGREGATE**:

= 6/10 x 1.54 = 0.924 m^{3}

The density of aggregate = 1650 kg/m^{3} (it differs from 1550 to 1750)

So, the weight of the aggregates = volume x density = 0.924 x 1650 = 1524 kg

Video Source: Civiconcepts

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