Answer / guruprasad

1000mmx1000mmx1000mm =1 mx1mx1m = 1cum ok
M20 Grade =1:11/2:3
summation of proportion=1+1.5+3=5.5
1.54cum of dry materials of concrete will yield only 1 cum
concrete after adding water.
1cum x1.54/5.5= 0.28 cum
so cement=0.28cum= 0.28/.0347=8.07 bags=403.5 kg
sand=4.2cum=15.036cft
aggregate=8.4cum=30.072cft

Answer / rajendra

1000mmx1000mmx1000mm =1 mx1mx1m = 1cum ok
M20 Grade =1:11/2:3
summation of proportion=1+1.5+3=5.5
1.54cum of dry materials of concrete will yield only 1 cum
concrete after adding water.
1cum x1.54/5.5= 0.28 cum
so cement=0.28cum= 0.28/.0347=8.07 bags=403.5 kg
sand=0.42cum= 14.82 cft
aggregate=0.84cum= 29.65 cft

Answer / reethesh.m.p.

1000mmx1000mmx1000mm =1 mx1mx1m = 1cum ok
M20 Grade =1:1.5:3
summation of proportion=1+1.5+3=5.5
1.54cum of dry materials of concrete will yield only 1 cum
concrete after adding water.So the constant is 1.54
Quantity of cement for 1m3=1 x1/5.5x1.54 = 0.28 cum
1m3 cement=1440kg. So cement for 1m3=0.28x1440=403.2 kg
sand=easy method=1.5xcement=1.5x0.28=0.42 m3x35.313=14.83cft
or sand=1x1.5/5.5x1.54=0.42 m3x35.313=14.83 cft
Coarse aggregate = 3xcement or 2xsand=(3x0.28) or (2x0.42)
=0.84cum x 35.313 = 29.66 cft
or coarse aggregate= 1x3/5.5x1.54=0.84 m3x35.313=29.66 cft

Answer / atish kumar

what is .0347 ????

is it constant

Answer / vj

the ratio for M20 1:1.5:3 is basically by weight
if we convert this into by volume then the ratio becomes 1:2:4.

thus for 1 m3 concrete cement  = 0.12
sand = 0.22 and aggregate = 0.46 and water 0.18 = .98 m3 (.2 considered to be entrapped air)

Answer / k.raja

Correction in Question we made:
What is the quantity of cement, sand, and gravel to be used
in M20 grade of concrete for 1000 x 1000 x 1000 mm.
(for cement please mention in kilos, and for sand and
gravel please mention it in CFT)

1000mmx1000mmx1000mm =1 mx1mx1m = 1cum ok
M20 Grade =1:1 1/2:3
summation of proportion=1+1.5+3=5.5
1.54cum of dry materials of concrete will yield only 1 cum
concrete after adding water.
1cum x1.54/5.5= 0.28 cum
so cement = 0.28cum x 1440 = 403.5 kg
sand = 0.42cum x 10.47 = 4.40 cft
aggregate = 0.84cum x 10.47 = 8.80 cft

Answer / sjtbehera

1000mmx1000mmx1000mm =1 mx1mx1m = 1cum ok
M20 Grade =1:11/2:3
summation of proportion=1+1.5+3=5.5
1cum x15.4/5.5= 2.8 cum
so cement=2.8cum= 2.8/.0347=80.7 bags=4035 kg
sand=4.2cum=150.36cft
aggregate=8.4cum=300.72cft

relation bet Coefficient consolidation Cv, time factor Tv, drainage path d, & time t is Cv=d2Tv/t Cv=d2t/Tv Cv=tTv/d2 Cv=Tv/td2

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