How many cement bags are used in 1:1.5:3 concrete?

 Cement 1 part 

Sand 1.5 part 

Aggregate 3 part

  1. Total dry volume of Material Required = 1.57 cu.m
  2. 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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