Civil Engineering - Concrete Technology - Discussion

Discussion Forum : Concrete Technology - Section 2 (Q.No. 22)
22.
For batching 1:3:6 concrete mix by volume, the ingredients required per bag of 50 kg cement, are:
70 litres of sand and 120 litres of aggregates
70 kg of sand and 140 litres of aggregates
105 litres of sand and 140 litres of aggregates
105 litres of sand and 210 litres of aggregates
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
22 comments Page 2 of 3.

Dhanashree said:   9 years ago
Thank you all for providing the clear solution.

Dikshant said:   1 decade ago
How the quantities are converted into liters?

Shrikant said:   10 years ago
Its a density of cement.

Xai said:   10 years ago
@Rohit the density of concrete is 2400 kg/m3. Where did you get 1400 kg/m3?

Bhavyasree said:   10 years ago
What its's bulk density?

Ramesh said:   1 decade ago
Density = Mass/Volume.

Weight = Density*Volume*9.81.

Why should we multiply with 9.81?

Rohit said:   1 decade ago
1:3:6 is the proportion that he has asked.

Density = Mass/Volume.

Density of concrete = 1400 kg/m3.

Mass of cement = 50 kg (given).

1440 = 50/v, hence v = 0.035 m3.

Water = 0.035*3 = 0.105 m3.

Aggregates = 0.035*6 = 0.21 m3.

Cement:Sand:Aggregates = 0.035:0.105:0.21 (in m3).

Cement:Sand:Aggregates = 35:105:210 (in lts if multiplied above result with 1000 as 1 m3 = 1000 lts).

Vaibhav S said:   1 decade ago
Density = Mass/Volume.

Weight = Density*Volume*9.81.

Niharika said:   1 decade ago
How to convert volume to weight?

SURESH said:   1 decade ago
Kg and liter are equal, one gram = one milliliter.


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