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Quantities of materials for concrete such as cement, sand and aggregates for production of required quantity of concrete of given mix proportions such as 1:2:4 (M15), 1:1.5: 3 (M20), 1:1:2 (M25) can be calculated by absolute volume method.

This method is based on the principle that the volume of fully compacted concrete is equal to the absolute volume of all the materials of concrete, i.e. cement, sand, coarse aggregates and water.

A Concrete structure may consists of beams, slabs, columns and foundations etc. based on type of structure. The volume of concrete required for concrete structure can be calculated by summing up the volumes of each structural member or each parts of members.

The volume of a rectangular cross sectional member can be calculated as length x width x height (or depth or thickness). Suitable formula shall be used for different cross-sectional shapes of members.

The formula for calculation of materials for required volume of concrete is given by:

Where, Vc = Absolute volume of fully compacted fresh concrete

W =Mass of water

C = Mass of cement

Fa = Mass of fine aggregates

Ca = Mass of coarse aggregates

Sc, Sfa and Sca are the specific gravities of cement, fine aggregates and coarse aggregates respectively.

The air content has been ignored in this calculation.

This method of calculation for quantities of materials for concrete takes into account the mix proportions from design mix or nominal mixes for structural strength and durability requirement.

Now we will learn the material calculation by an example.

## Calculating Quantities of Materials for per cubic meter or cubic feet or cubic yards concrete

Consider concrete with mix proportion of 1:1.5:3 where, 1 is part of cement, 1.5 is part of fine aggregates and 3 is part of coarse aggregates of maximum size of 20mm. The water cement ratio required for mixing of concrete is taken as 0.45.

Assuming bulk densities of materials per cubic meter, cubic feet and cubic yards as follows:

Cement = 1500 kg/m3 = 93.642 lb/ft3 = 2528.332 lb/cubic yards

Sand = 1700 kg/m3 = 105 lb/ft3 = 2865.443 lb/cubic yards

Coarse aggregates = 1650 kg/m3 = 103 lb/ft3 = 2781.166 lb/cubic yards

Specific gravities of concrete materials are as follows:

Cement = 3.15

Sand = 2.6

Coarse aggregates = 2.6.

The percentage of entrained air assumed is 2%.

The mix proportion of 1:1.5:3 by dry volume of materials can be expressed in terms of masses as:

Cement = 1 x 1500 = 1500

Sand = 1.5 x 1700 = 2550

Coarse aggregate = 3 x 1650 = 4950.

Therefore, the ratio of masses of these materials w.r.t. cement will as follows =

= 1 : 1.7 : 3.3

The water cement ratio = 0.45

Now we will calculate the volume of concrete that can be produced with one bag of cement (i.e. 50 kg cement) for the mass proportions of concrete materials.

Thus, the absolute volume of concrete for 50 kg of cement =

Thus, for the proportion of mix considered, with one bag of cement of 50 kg, 0.1345 m3 of concrete can be produced.

We have considered an entrained air of 2%. Thus the actual volume of concrete for 1 cubic meter of compacted concrete construction will be = 1 -0.02 = 0.98 m3.

Thus, the quantity of cement required for 1 cubic meter of concrete = 0.98/0.1345 = 7.29 bags of cement.

The quantities of materials for 1 m3 of concrete production can be calculated as follows:

The weight of cement required = 7.29 x 50 = 364.5 kg.

Weight of fine aggregate (sand) = 1.5 x 364.5 = 546.75 kg.

Weight of coarse aggregate = 3 x 364.5 = 1093.5 kg.

## Concrete Calculator: Calculate Quantities of Materials for Concrete

 Select Concrete Mix Proportion Cement : Sand : Aggregates : W/C Ratio :
 Select Concrete Member*: Select OptionsSlabFootingRectangular ColumnCircular ColumnBeam
 Dimension of Concrete Member Length /Height: mm Width : mm Thickness / Depth : mm

## Calculation Results: Quantities of Materials

Volume of Concrete = m3 = Yards = CFT = Pounds

Quantity of Cement = kg = Bags of 50 kg = Pounds

Quantity of Sand = kg = m3 = Yards = CFT = Pounds

Quantity of Coarse Aggregates = kg = m3 = Yards = CFT = Pounds

Quantity of Water = Liter = Gallons

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1. #### premend01

Can u please tell me how this W/C ratio decided?

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I want to know Following things:
1. For 0.5 Cum Pan Mixer what will be Compacted Volume ?

2. What all Aggregates Sizes Required with respect to their Size and Quantity?

3. Amount of Water and Cement Max can be used?

4. Amount of Sand ?

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I understood most of the article, but I am confused about the way you calculate weight of each materials that produce 1 m3 of concrete at the end of this article , where you use volumetric proportion instead of weight proportion

Weight of fine aggregate (sand) = 1.5 x 364.5 = 546.75 kg. (I think it should be 1.7 instead of 1.5)

Weight of coarse aggregate = 3 x 364.5 = 1093.5 kg. ( I think it should be 3.3 instead of 3 )

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4. #### dennis howell

I love this. Make things so easy.

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5. #### Jai A_mitt

in 6″ slab how we find the sand and aggregate amount retio=1.2.4

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6. But this is'nt actual formula …bcoz we r using grade M20 (1:1.5:3)..then cement is 8 bags/m3,sand 14.672 cft, nd coarse 29.672 cft..

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7. How much ( M15 – 1 2 4 ) wet concrete do we get with the above mix. Or what is the exact quantity of materials used in 1 cum ready mix concrete.

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• #### Jigar Sonani

In ready mix concrete quantity of material depend on the grade of concrete we can not say any exact quantity without the proportion of concrete

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8. I want to find how to determine the aggregate amount in concrete mix. Suppose-M25A20. Any one is having idea.

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9. any short method?

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• #### Boniankom

Some engineers using an emperical method:
If you want 250 kg/cm2
Cement = 250+(50 to 100)
Cement = 250+100 = 350
Cement = 350/50 = 7 bags
W/c = 0.5
Water 175 liters
Use sand 0.4 cum and coarse aggregates 0.8 cum
Remember, use this method for small jobs.
These method components will not give accurate 1 cum.

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