Answer / vishnu soni

For 1 cum concrete 1:4:8, then we reqd following
quantities:-
1.54/sum of ratio
1.54/13= .11/.034=3.48 bags of cement
Fine Aggregate:-- .11x4=.44 cum
Coarse aggregate:-- .11x8=.88cum

Answer / ajith

1 Cement 4 Sand and 8 Coarse aggregate
One M3 of PCC hence requires 0.076923 M3 of Cement
0ne M3 Cement =1440 Kg
0.076923 X 1440 = 110.7692
Or 2.22 Bag of Cement

Answer / suresh

for ex 1 m3 of concrete
conc 1;4;8
cement = 0.45 m3 of mix =1440x0.45/4=162kgs or 3.24 bag
sand = 0.45m3
40mm aggregate = 0.90m3

Answer / shailesh

For 100 cum

Cement should be in kg=(152/sum of proportion)*cement density
Sand should be cum=(152/sum of proporation)*sand ratio
percentage

Aggregate shuid be in cum=(152/sum of proporation)*
percentage of aggreagate
For PCC 1:4:8 (100CUM)

Cement=(152/13)*1*1440=16836.92kg
Sand=(152/13)*446.769 cum
Aggregate=(152/13)*8=93.53cum

Answer / sumathi lakshmanan

eg:1:3:6(M10)nominal mix ratio(bcoz size of coarse aggregate is 45mm) for 2 cub.m of PCC
Req:
1)PCC unit weight = 2400kg/cub.m
2)sum of ratio = 1+3+6=10
Procedure:
(1/10)*2*2400 (i.e) (ratio of cement/sum of ratio)*cub.m of PCC*unit weight of PCC =480kg of cement (9.6 or 10bags)
Therefore sand=480*3=1440kgs
Coarse agg(45mm) = 480*6=2880kgs

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