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CALCULATE QUANTITIES OF MATERIALS FOR CONCRETE

QUANTITIES OF MATERIALS FOR CONCRETE

Quantities of materials for the production of required quantity of concrete of given mix proportions 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.

QUANTITIES OF MATERIALS FOR CONCRETE

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

Vc = (W / 1000) + (C / 1000Sc) + (Fa / 1000Sfa) + (Ca / 1000Sca) = 

Where, V = 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.

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 as follows:

Cement = 1500 kg/m
Sand = 1700 kg/m
Coarse aggregates = 1650 kg/m

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 : (2550/1500) : (4950/1500) = 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 =

Vc = 

{(0.45 x 50) / 1000} = 0.0225

{(1 x 50) / (1000 x 3.15)} = 0.0158

{(1.7 x 50) / (1000 x 2.6)} = 0.0326

{(3.3 x 50) / (1000 x 2.6)} = 0.0634

Vc = 0.0225 + 0.0158 + 0.0326 + 0.0634 = 0.1343 Cum.

Thus, for the proportion of mix considered, with on 3 bag of cement of 50 kg, 0.1343 m 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 m .

Thus, the quantity of cement required for 1 cubic meter of concrete = 0.98/0.1343 = 7.29 bags of cement.

The quantities of materials for 1 Cum of concrete 1:1.5:3 Ratio 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.

QUANTITIES OF MATERIALS FOR CONCRETE

рдХंрдХ्рд░ीрдЯ рдХे рд▓िрдП рд╕ाрдордЧ्рд░ी рдХी рдоाрдд्рд░ा
рджिрдП рдЧрдП рдоिрд╢्рд░рдг рдЕрдиुрдкाрдд рдХे рдХंрдХ्рд░ीрдЯ рдХी рдЖрд╡рд╢्рдпрдХ рдоाрдд्рд░ा рдХे рдЙрдд्рдкाрджрди рдХे рд▓िрдП рд╕ाрдордЧ्рд░ी рдХी рдоाрдд्рд░ा
рдкूрд░्рдг рдоाрдд्рд░ा рд╡िрдзि рдж्рд╡ाрд░ा рдЧрдгрдиा рдХी рдЬाрдПрдЧी। рдпрд╣ рд╡िрдзि рдЗрд╕ рд╕िрдж्рдзांрдд рдкрд░ рдЖрдзाрд░िрдд рд╣ै рдХि рдкूрд░ी рддрд░рд╣ рд╕े рдаोрд╕ рдХंрдХ्рд░ीрдЯ рдХी рдоाрдд्рд░ा рдХंрдХ्рд░ीрдЯ рдХी рд╕рднी рд╕ाрдордЧ्рд░िрдпों, рдпाрдиी рд╕ीрдоेंрдЯ, рд░ेрдд, рдоोрдЯे рд╕рдоुрдЪ्рдЪрдп рдФрд░ рдкाрдиी рдХी рдкूрд░्рдг рдоाрдд्рд░ा рдХे рдмрд░ाрдмрд░ рд╣ै।
рдХंрдХ्рд░ीрдЯ рдХी рдЖрд╡рд╢्рдпрдХ рдоाрдд्рд░ा рдХे рд▓िрдП рд╕ाрдордЧ्рд░ी рдХी рдЧрдгрдиा рдХा рд╕ूрдд्рд░ рдиिрдо्рдиाрдиुрд╕ाрд░ рд╣ै: -

Vc = (W / 1000) + (C / 1000Sc) + (Fa / 1000Sfa) + (Ca / 1000Sca) =

рдЬрд╣ां, рд╡ी = рдкूрд░ी рддрд░рд╣ рд╕े рддाрдЬा рдХंрдХ्рд░ीрдЯ рдХे рдкूрд░्рдг рдоाрдд्рд░ा
рдбрдм्рд▓्рдпू = рдкाрдиी рдХा рдж्рд░рд╡्рдпрдоाрди
рд╕ी = рд╕ीрдоेंрдЯ рдХा рдж्рд░рд╡्рдпрдоाрди
Fa = рдаीрдХ рд╕рдоुрдЪ्рдЪрдп рдХा рдж्рд░рд╡्рдпрдоाрди
рд╕ीрдП = рдоोрдЯे рд╕рдоुрдЪ्рдЪрдп рдХा рдж्рд░рд╡्рдпрдоाрди
Sc, Sfa рдФрд░ Sca рдХ्рд░рдорд╢ः рд╕ीрдоेंрдЯ, рдаीрдХ рд╕рдоुрдЪ्рдЪрдп рдФрд░ рдоोрдЯे рд╕рдоुрдЪ्рдЪрдп рдХे рд╡िрд╢िрд╖्рдЯ рдЧुрд░ुрдд्рд╡ рд╣ैं।
рдЗрд╕ рдЧрдгрдиा рдоें рд╡ाрдпु рд╕ाрдордЧ्рд░ी рдХी рдЕрдирджेрдЦी рдХी рдЧрдИ рд╣ै।
рдХंрдХ्рд░ीрдЯ рдХे рд▓िрдП рд╕ाрдордЧ्рд░ी рдХी рдоाрдд्рд░ा рдХे рд▓िрдП рдЧрдгрдиा рдХी рдпрд╣ рд╡िрдзि рд╕ंрд░рдЪрдиाрдд्рдордХ рд╢рдХ्рддि рдФрд░ рд╕्рдеाрдпिрдд्рд╡ рдХी рдЖрд╡рд╢्рдпрдХрддा рдХे рд▓िрдП рдбिрдЬाрдЗрди рдоिрд╢्рд░рдг рдпा рдиाрдордоाрдд्рд░ рдоिрд╢्рд░рдг рд╕े рдоिрд╢्рд░рдг рдЕрдиुрдкाрдд рдХो рдз्рдпाрди рдоें рд░рдЦрддी рд╣ै।

рдЕрдм рд╣рдо рдПрдХ рдЙрджाрд╣рд░рдг рдж्рд╡ाрд░ा рд╕ाрдордЧ्рд░ी рдЧрдгрдиा рд╕ीрдЦेंрдЧे।

1: 1.5: 3 рдХे рдоिрд╢्рд░рдг рдЕрдиुрдкाрдд рдХे рд╕ाрде рдХंрдХ्рд░ीрдЯ рдкрд░ рд╡िрдЪाрд░ рдХрд░ें рдЬрд╣ां, 1 рд╕ीрдоेंрдЯ рдХा рд╣िрд╕्рд╕ा рд╣ै, 1.5 рдаीрдХ рдХा рд╣िрд╕्рд╕ा рд╣ै
рд╕рдоुрдЪ्рдЪрдп рдФрд░ 3 20 рдоिрдоी рдХे рдЕрдзिрдХрддрдо рдЖрдХाрд░ рдХे рдоोрдЯे рд╕рдоुрдЪ्рдЪрдп рдХा рд╣िрд╕्рд╕ा рд╣ै। рдХंрдХ्рд░ीрдЯ рдХे рдоिрд╢्рд░рдг рдХे рд▓िрдП рдЖрд╡рд╢्рдпрдХ рдЬрд▓ рд╕ीрдоेंрдЯ рдЕрдиुрдкाрдд 0.45 рдХे рд░ूрдк рдоें рд▓िрдпा рдЬाрддा рд╣ै।

рдиिрдо्рдиाрдиुрд╕ाрд░ рд╕ाрдордЧ्рд░ी рдХे рдеोрдХ рдШрдирдд्рд╡ рдХो рдоाрдирддे рд╣ुрдП:

рд╕ीрдоेंрдЯ = 1500 рдХिрдЧ्рд░ा / рдоी
рд░ेрдд = 1700 рдХिрдЧ्рд░ा / рдоी
рдоोрдЯे рд╕рдоुрдЪ्рдЪрдп = 1650 рдХिрдЧ्рд░ा / рдоी

рдХंрдХ्рд░ीрдЯ рд╕ाрдордЧ्рд░ी рдХी рд╡िрд╢िрд╖्рдЯ рдЧुрд░ुрдд्рд╡ाрдХрд░्рд╖рдг рдиिрдо्рдиाрдиुрд╕ाрд░ рд╣ैं:

рд╕ीрдоेंрдЯ = 3.15
рд░ेрдд = реи.рем
рдоोрдЯे рд╕рдоुрдЪ्рдЪрдп = 2.6।
рдк्рд░рд╡ेрд╢िрдд рд╡ाрдпु рдХा рдк्рд░рддिрд╢рдд 2% рд╣ै।
1: 1.5: 3 рдХे рдоिрд╢्рд░рдг рдХी рд╕ूрдЦी рдоाрдд्рд░ा рдХे рдЕрдиुрдкाрдд рдХो рдж्рд░рд╡्рдпрдоाрди рдХे рд░ूрдк рдоें рд╡्рдпрдХ्рдд рдХिрдпा рдЬा рд╕рдХрддा рд╣ै:
рд╕ीрдоेंрдЯ = 1 x 1500 = 1500
рд░ेрдд = 1.5 x 1700 = 2550
рдоोрдЯे рдХुрд▓ = 3 x 1650 = 4950।

рдЗрд╕рд▓िрдП, рдЗрди рд╕ाрдордЧ्рд░िрдпों рдХे рдж्рд░рд╡्рдпрдоाрди рдХा рдЕрдиुрдкाрдд w.r.t. рд╕ीрдоेंрдЯ рдиिрдо्рдиाрдиुрд╕ाрд░ рд╣ोрдЧा =

1: (2550/1500): (4950/1500) = 1: 1.7: 3.3
рдкाрдиी рд╕ीрдоेंрдЯ рдЕрдиुрдкाрдд = 0.45
рдЕрдм рд╣рдо рдХंрдХ्рд░ीрдЯ рдХी рдоाрдд्рд░ा рдХी рдЧрдгрдиा рдХрд░ेंрдЧे рдЬो рдХंрдХ्рд░ीрдЯ рд╕ाрдордЧ्рд░ी рдХे рдмрдб़े рдЕрдиुрдкाрдд рдХे рд▓िрдП рд╕ीрдоेंрдЯ рдХे рдПрдХ рдмैрдЧ (рдпाрдиी 50 рдХिрд▓ोрдЧ्рд░ाрдо рд╕ीрдоेंрдЯ) рдХे рд╕ाрде рдЙрдд्рдкाрджिрдд рдХी рдЬा рд╕рдХрддी рд╣ै।
рдЗрд╕ рдк्рд░рдХाрд░, 50 рдХिрд▓ो рд╕ीрдоेंрдЯ рдХे рд▓िрдП рдХंрдХ्рд░ीрдЯ рдХी рдкूрд░्рдг рдоाрдд्рд░ा =

Vc =

{(0.45 x 50) / 1000} = 0.0225

{(1 x 50) / (1000 x 3.15)} = 0.0158

{(1.7 x 50) / (1000 x 2.6)} = 0.0326

{(3.3 x 50) / (1000 x 2.6)} = 0.0634

Vc = 0.0225 + 0.0158 + 0.0326 + 0.0634 = 0.1343 Cum.

рдЗрд╕ рдк्рд░рдХाрд░, рдоिрд╢्рд░िрдд рдоिрд╢्рд░рдг рдХे рдЕрдиुрдкाрдд рдХे рд▓िрдП, 50 рдХिрд▓ो рд╕ीрдоेंрдЯ рдХे 3 рдмैрдЧ рдкрд░, 0.1343 рдоीрдЯрд░ рдХंрдХ्рд░ीрдЯ рдХा рдЙрдд्рдкाрджрди рдХिрдпा рдЬा рд╕рдХрддा рд╣ै।

рд╣рдордиे 2% рдХी рдПрдХ рдЫिрдж्рд░िрдд рд╣рд╡ा рдкрд░ рд╡िрдЪाрд░ рдХिрдпा рд╣ै। рдЗрд╕ рдк्рд░рдХाрд░ рдаोрд╕ рдХंрдХ्рд░ीрдЯ рдиिрд░्рдоाрдг рдХे 1 рдШрди рдоीрдЯрд░ рдХे рд▓िрдП рдХंрдХ्рд░ीрдЯ рдХी рд╡ाрд╕्рддрд╡िрдХ рдоाрдд्рд░ा = 1 0.02 = 0.98 рдоीрдЯрд░ рд╣ोрдЧी।

рдЗрд╕ рдк्рд░рдХाрд░, 1 рдШрди рдоीрдЯрд░ рдХंрдХ्рд░ीрдЯ рдХे рд▓िрдП рдЖрд╡рд╢्рдпрдХ рд╕ीрдоेंрдЯ рдХी рдоाрдд्рд░ा = 0.98 / 0.1343 = рд╕ीрдоेंрдЯ рдХे 7.29 рдмैрдЧ।

The quantities of materials for 1 Cum of concrete 1:1.5:3 Ratio production can be calculated as follows:-

рд╕ीрдоेंрдЯ рдХा рд╡рдЬрди рдЖрд╡рд╢्рдпрдХ = 7.29 x 50 = 364.5 рдХिрд▓ोрдЧ्рд░ाрдо।
рдаीрдХ рд╕рдоुрдЪ्рдЪрдп (рд░ेрдд) рдХा рд╡рдЬрди = 1.5 x 364.5 = 546.75 рдХिрд▓ोрдЧ्рд░ाрдо।
рдоोрдЯे рдХुрд▓ рдХा рд╡рдЬрди = 3 x 364.5 = 1093.5 рдХिрд▓ोрдЧ्рд░ाрдо।

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