How many cement bags are used in 1:1.5:3 concrete?
Cement 1 part Sand 1.5 part Aggregate 3 part Total dry volume of Material Required = 1.57 cu.m 1+1.5+3 =5.5 Volume of cement needed = Ratio of cement x 1.57/(1+1.5+3) = 1 x 1.57/5 .5 = 0.285 cu.m X 1440 kg/m3 =410 kg =410/50 =8.2 bags Cement density =1440 kg/m3 Volume of sand needed = Ratio of Sand x 1.57/(1+1.5+3) = 1.5 x 1.57/5 .5 = 0.427 cu.m X 1680 kg/m3 =717.36 kg/m3 Average density of sand =1680 kg/m3 Volume of aggregate needed = Ratio of agg x 1.57/( 1+1.5+3) = 3 X 1.57/5.5 = 0.854 cu.m X 1800 kg/M3 =1537.27 kg/M3 Aggregate density valu 1600–1800 kg/M3 Density value as site metarial available Cement required of bag =8.2 bag Sand required =717.36 kg/M3 Aggregate required =1537.27 kg/m3 Below blog question answer for civil engineer https://civilbilling.blogspot.com/2023/03/civil-engineering-gpsc-questions-paper.html