How to Calculate the Materials Quantities for Concrete
Calculate the Materials Quantities for Concrete
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 (M-15), 1:1.5: 3
(M-20), 1:1:2 (M-25) 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 consist
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.
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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 gravitates 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 considers the mix proportions from
design mix or nominal mixes for structural strength and durability requirement.
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 20 mm. 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/cum =
93.642 lb/cft = 2528.332 lb/cubic yards
Sand = 1700 kg/cum = 105 lb/cft = 2865.443 lb/cubic yards
Coarse aggregates = 1650
kg/cum = 103 lb/cft = 2781.166 lb/cubic yards
Specific
gravitates of concrete materials are as follows:
Cement = 3.14
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 follow =
= 1: 1.7 : 3.3
The water cement ratio = 0.45
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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 cum 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 cum.
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 cum 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.
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