Answer / suresh

let the total volume of the concrete is = a
and let the given mix proportion is 1:1.5:3
then the quantity of the
cement required = a*(1.52/(1+1.5+3))m^3 = x m^3
we know that one cubic metre = 1450kg = 29 bags
1 cubic metre = 29 bags
x cubic metre = ?
=(x * 29)/1 bags

now, fine agg = 1.5*x
coarse agg = 3*x

Answer / brajesh kaurav

let the total volume of the concrete is = a
and let the given mix proportion is =1:1.5:3
then the quantity of the
cement required = a*(1.57/(1+1.5+3))m^3 = X m^3
where 1.57 is loose volume of 1 cum concret.
we know that density of cement = 1440kg/cum
quantity of cement in per cum in bag=1440/50=28.8bag/cum(in
loose volume it consider 30bag)
1 cubic metre = 30 bags
x cubic metre = ?
=(X*30)bag/cum

now, fine agg = 1.5*X(+5% of voids)
coarse agg = 3*X(+1% of voids)

Answer / jumrik ngomdir

let the total volume of the concrete is = a
and let the given mix proportion is =1:2:4
then the quantity of the
cement required = a*(1.52/(1+2+4))m^3 = (X) m^3
where 1.52 is loose volume of 1 cum concret.
we know that density of cement = 1440kg/cum
quantity of cement in per cum in bag=1440/50=28.8bag/cum(in
loose volume it consider 30bag)
1 cubic metre = 30 bags
x cubic metre = ?
=(X*30)bag/cum

now, fine agg = 2*X(+5% of voids)
coarse agg = 4*X(+1% of voids)

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