What will be Volume of Concrete for Given Cement Sand and Aggregate Quantity?

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What will be Volume of Concrete for Given Cement Sand and Aggregate Quantity?

Question

Cement volume is 0.25m^{3}, Sand volume is 0.25m^{3}, Coarse aggregate volume is 0.5m^{3}. Concrete grade is M25.(mix proportion is 1:1:2). What will be the concrete volume in m^{3}?

Volume of fully compacted concrete is given by following formula:

where V_{c} is the volume of concrete. W, C, F_{a} and C_{a} are the masses of water, cement, fine aggregates and coarse aggregates respectively.

Density of cement = 1500 kg/m^{3}, fine aggregates = 1700 kg/m^{3} and coarse aggregates = 1650 kg/m^{3}.

So, Weight of cement C = 0.25 x 1500 = 375kg

Weight of Fine aggregates F_{a} = 0.25 x 1700 = 425 kg

Weight of Coarse aggregates C_{a} = 0.5 x 1650 = 825 kg

Consider w/c ratio as 0.45, so weight of water = 0.45×0.25×1000 = 112.5 L or kg

S_{c}, S_{fa} and S_{ca} are the specific weight of cement, fine aggregates and coarse aggregates which are 3.15 for cement and 2.6 for fine and coarse aggregates.

Therefore, putting these values in the formula above, we get

## Answers ( 2 )

Given,

M25, mix proportion would be- (1:1:2)

VC=0.25m3

VS=0.25m3

Va=0.5m3

Adding the mix ratios=1+1+2

=4

VConcrete=1/4* VC+1/4* VS+2/4* Va

=1/4*0.25m3+1/4*0.25m3+2/4*0.5m3

=0.375 m3

Volume of fully compacted concrete is given by following formula:

where V

_{c}is the volume of concrete. W, C, F_{a}and C_{a}are the masses of water, cement, fine aggregates and coarse aggregates respectively.Density of cement = 1500 kg/m

^{3}, fine aggregates = 1700 kg/m^{3}and coarse aggregates = 1650 kg/m^{3}.So, Weight of cement C = 0.25 x 1500 = 375kg

Weight of Fine aggregates F

_{a}= 0.25 x 1700 = 425 kgWeight of Coarse aggregates C

_{a}= 0.5 x 1650 = 825 kgConsider w/c ratio as 0.45, so weight of water = 0.45×0.25×1000 = 112.5 L or kg

S

_{c}, S_{fa}and S_{ca}are the specific weight of cement, fine aggregates and coarse aggregates which are 3.15 for cement and 2.6 for fine and coarse aggregates.Therefore, putting these values in the formula above, we get

V

_{c}= (112.5/1000)+(375/(1000×3.15))+(425/(1000×2.6))+(825/(1000×2.6)) = 0.712m^{3}So, Volume of concrete = 0.712m

^{3}