Consider Volume (Wet) = 1 m3

Grade of Concrete = M15 (Cement - 1 : sand - 2 : Aggregate - 4)

Sum Of Ratio = 1+2+4 = 7

Dry Volume = Wet Volume + 54% of Wet Volume = 1 + 0.54 = 1.54 m3

1.

2.

3.

Then, 316.8 Kg contains, 316.8/50 = 6.336 bags

4.

i.e, 6.336 x 1.226 = 7.77 CFT

1.

2.

3.

= 6.336 x 1.226 x 2 = 15.54 CFT

1.

2.

3.

= 6.336 x 1.226 x 4 = 31.07 CFT

Grade of Concrete = M15 (Cement - 1 : sand - 2 : Aggregate - 4)

Sum Of Ratio = 1+2+4 = 7

Dry Volume = Wet Volume + 54% of Wet Volume = 1 + 0.54 = 1.54 m3

__CEMENT__1.

*CUM*: (1.54 x 1)/7 = 0.22 CUM2.

*Kg*: (1.54 x 1 x 1440)/7 = 316.8 Kg3.

*Bags*: We know 1 bag of cement weight = 50 KgThen, 316.8 Kg contains, 316.8/50 = 6.336 bags

4.

*CFT*: If we multiply No. of bags by 1.226 then we get the value in CFT.i.e, 6.336 x 1.226 = 7.77 CFT

__SAND__1.

*CUM*: (1.54 x 2)/7 = 0.44 CUM2.

*Kg*: (1.54 x 2 x 1450)/7 = 638 Kg [Density Varies in between 1450-1500 Kg/m3]3.

*CFT*: (1.54 x 2 x 35.3147)/7 = 15.54 CFT*Alternative*: No. of Cement Bags x Value of 1 bag of cement in CFT x ratio of sand= 6.336 x 1.226 x 2 = 15.54 CFT

__AGGREGATE__1.

*CUM*: (1.54 x 4)/7 = 0.88 CUM2.

*Kg*: (1.54 x 4 x 1500)/7 = 1320 Kg [Density Varies in between 1450-1550 Kg/m3]3.

*CFT*: (1.54 x 4 x 35.3147)/7 = 31.07 CFT*Alternative*: No. of Cement Bags x Value of 1 bag of cement in CFT x ratio of sand= 6.336 x 1.226 x 4 = 31.07 CFT

how much Quantity of water ?

ReplyDelete1.54

Deletew/c=.40

Deletethen 50*.40=20 ltr per bag of cement

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