Quantities of Materials for Concrete

QUANTITIES OF MATERIALS FOR CONCRETE

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.

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.

The formula for calculation of materials for required volume of concrete

Where,

V_c = Absolute volume of fully compacted fresh concrete 

W =Mass of water 

C = Mass of cement

F_a = Mass of fine aggregates

C_a = Mass of coarse aggregates

S_c, S_fa and S_ca 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.

 

Illustrative Example

Concrete with mix proportion of 1:2:4

where,

1 is part of cement,

2 is part of fine aggregates and

4 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 as follows:

Cement = 1500 kg/m³

Sand = 1700 kg/m³

Coarse aggregates = 1650 kg/m³

Specific gravity 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 1:2:4 by dry volume of materials can be expressed in terms of masses as:

Cement = 1 x 1500 = 1500

Sand = 2 x 1700 = 2400

Coarse aggregate = 4 x 1650 = 6600.

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

=1: (2400/1500) : (6600/1500)

= 1 : 1.6 : 4.4

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 =

(0.45x50/1000) + (1x50/1000x3.15) + (1.6x50/1000x2.6) + (4.4x50/1000x2.6)

= 0.0225 + 0.015 + 0.0307 + 0.084 = 0.1528 m³

Thus, for the proportion of mix considered, with one bag of cement of 50 kg, 0.1528 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 = 0.98/0.1528 = 6.41  bags of cement.

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

The weight of cement required = 6.41  x 50 = 320.5 kg.

Weight of fine aggregate (sand) = 2 x 320.5 = 641.0 kg.

Weight of coarse aggregate = 3 x 641 = 1282.0 kg.

Some Important Conversions

1 cum = 35.3147 cft

1 cft = 0.0283 cum

Volume of 50kg Bag in cft

Unit Weight (Density) of Cement in loose condition = 1440kg/m3 …….. (1)

We know that, Density = Mass/Volume

Therefore, Volume = Mass/Density …….. (2)

Mass of 1 Bag cement = 50Kg …….. (3)

Dividing equation (3) by (1) ,

Volume in Cum = 50/1440

= 0.034722 Cum

Volume in Liters = 0.034722 x 1000 = 34.722 Lit.

i.e., 35 Liters

Converting Cum to Cft by multiplying 35.3147

We get,

Volume in Cft = 0 .0347 x 35.314 = 1.22539