Answer Posted / himanth kumar

Consider for 1CUM of concrete is to be prepare,For that

M15 is a nominal mix, its proportion is 1:2:4

for 1cum of final mix(Dence) is required, 1.50 to 1.55 Cum
of loose ingredients with the required proportion we want
to add.

Based on that take avg of 1.50 & 1.55 is 1.52

Total proportion = 1+2+4 = 7

For Cement Qty = 1.52/7=0.217 Cum of cement required to
prepare 1Cum of M15 concrete.But cement is always taken in
Weight.For that the unit weight of cement is 1440 KG/Cum
Total cement required = 0.217x1440= 312.68 Kgs.

Sand = 0.217 x 2 =0.434 Cum

Metal or Aggregate = 0.217 x 4 =0.868 Cum.

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