Answer / raghuram

if 1:2:4 is the ratio the cement require = (1/(4+2+1))x 1440

9density of cememnt)

sand required = (2/(4+2+1))x density of sand
similarly for agg

Answer / renupandey

1:2:4
1+2+4=7
cement- 1/7=.143*1.52=0.217cum
no. of bag 0.217/0.0347=6.26
sand- 2*0.217=0.434cum
c.a.-4*0.217=0.868cum

Answer / dheeraj deshmukh

Mr. Ravsaheb,
The calculation is like below.

1/1+2+4=.142
.142x1.52x(1440/50)=.22
Hence, .22/.035=6.29bag/m3

Answer / sourabh

1:2:4=7
1 cu m
cement=1*1*28.8/7 = 4.11 bags
coresand=1*2/7= 0.28 cu m
agreegates=1*4/7= 0.57 cu m

Answer / ravsaheb

1/1+2+4=0.1428571
0.1428571 this is the Quantity of cement in Cum
So cement qty in no of bags = 0.1428571/.035
= 4.08 bags
Then for Sand =0.1428571*2
= .2857143 CuM
For Agg. = 0.1428571*4
= 0.5714286CuM

As per Kandal suggestions, we have implemented rolling starts on 140 (+/- 5) degrees and we kept 3 rollars to complete the rolling before falling 125 degrees and the Aggregate AIV is below 22. But When the DBM bed is being rolled 3rd high viabration, (Before 1 plain & low ,2 high Viabrations), the aggregate is getting sharp edge agg. projections are damaged. As the literature is said that, the VG-30 Grade having better control the temp. susceptibility than Penetration grade. if so we could not get this problem at site. But we are facing on site. let me know.

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