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
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