for M20 grade ratio is (1:1.5:3)
we consider for 10m3
add 50% for shrinkage & wastage
then it is 10+5=15m3
cement= 15/1+1.5+3 *1
for no. of cement bags = 2.72/0.0347
= 78.38 no.
= 79bags for(10m3)
so that for 1 m3 it is around 8 bags..........
For M20 concrete, The ratio of the mix is 1:1.5:3 (Nominal mix)
The total parts is 1+1.5+3 =5.5
Part of the cement is = 1/5.5
The total volume raw material for 1m3 of mixed concrete practically =1.57m3
Unit weight of cement =1440kg
Weight of cement per bag = 50kg
Now the calculation is =1/5.5X1.57X1440
In terms bags per m3 of conctrete is 411/50 = 8.22bags
While designing the mix, we shall take various paramenters
such as; workability, exposure, quality control etc., into
cognizance. If we consider the workability as 0.8, exposure
is moderate, QC as good, FA is in Zone-III, then quantum of
Cement required for 1cum of M20 concrete is; 53Gr OPC-465
kg, CA-1184.82 kg,FA-495.77 kg and water 186 lt.
for M20 grade of concrete
(i.e) cement : coarse aggregate : Fine Aggregate.
( jally ) ( sand )
1cubic metre cement = 1440KG
In the concrete 0.45% of the sand ratio is utilised for
cement.so for M20
= 0.45/3 X 1440
=0.15 X 1440
= 4.32Bags of cement/cubicmetre of concrete
Is This Answer Correct ?
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