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

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Asked: November 18, 20172017-11-18T22:55:06+05:30 2017-11-18T22:55:06+05:30In: Concrete
Gopal Mishra
Gopal Mishra

Gopal Mishra

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Gopal Mishra

Cement volume is 0.25m3, Sand volume is 0.25m3, Coarse aggregate volume is 0.5m3. Concrete grade is M25.(mix proportion is 1:1:2). What will be the concrete volume in m3?

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  1. henok tefera
    2017-11-20T17:02:40+05:30Added an answer on November 20, 2017 at 5:02 pm

    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

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  2. Gopal Mishra

    Gopal Mishra

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    Gopal Mishra
    2017-11-21T19:19:45+05:30Added an answer on November 21, 2017 at 7:19 pm

    Volume of fully compacted concrete is given by following formula:

    volume of concrete formula

    where Vc is the volume of concrete. W, C, Fa and Ca are the masses of water, cement, fine aggregates and coarse aggregates respectively.

    Density of cement = 1500 kg/m3, fine aggregates = 1700 kg/m3 and coarse aggregates = 1650 kg/m3. 

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

    Weight of Fine aggregates Fa = 0.25 x 1700 = 425 kg

    Weight of Coarse aggregates Ca = 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

    Sc, Sfa and Sca 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

    Vc = (112.5/1000)+(375/(1000×3.15))+(425/(1000×2.6))+(825/(1000×2.6)) = 0.712m3

    So, Volume of concrete = 0.712m3

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