Civil Engineering - Concrete Technology - Discussion

Discussion Forum : Concrete Technology - Section 2 (Q.No. 37)
37.
The internal dimensions of a ware house are 15 m x 5.6 m, and the maximum height of piles is 2.70 m, the maximum number of bags to be stored in two piles, are
1500 bags
2000 bags
2500 bags
3000 bags
4000 bags
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
45 comments Page 2 of 5.

Sanjay said:   8 years ago
Answer is 4200 bags. It's a perfect answer.

Manesh PB said:   7 years ago
@All.

I don't think the above answers are correct because according to the guidelines for cement warehouses.

1. It is advisable to never stack more than 10 cement bags over each other as this will lead to lumping under pressure.
2. Should provide 600mm space away from the wall.
3. The width of the stack shall not be more than 4 bag width or 3m
4. A passage of 600mm should provide in between each stack for easy access and inspection.

Tanveer said:   7 years ago
Then, what is the answer? Please explain it correctly.

Deepu Maury said:   7 years ago
Area-(15-0.3) *(5.6-0.4) =77.91.
1.2m space leaves between two piles 1.2 * 14.7 = 17.69.

Redution 77.91-17.91 = 60.27.
Volume-60. 27 * 2.7 = 162.72 ÷ 054 = 3013.5bag.

Sabin said:   7 years ago
Give the explanation of the correct answer, please.

Saugat Oli said:   7 years ago
Right, thanks for the Answer @Ayon Som.

Asif wazir said:   7 years ago
(5.6-0.6-0.6) x (15-0.6-0.6) x 2.7/(0.3x.18) = 3036.

Achhami baddo said:   6 years ago
Total volume of room= 15*5.6*2.7 =226.8 m3,
Space between two piles = 1.6m,
Total volume of space = 15*1.6*2.7 =64.8 m3,
Effective volume = 226.8-64.8 m3= 162 m3,
Volume of one bag cement = 0.054m3.
No's of bag to be stored = 162/0.054= 3000bags.

So the correct answer is option "C" 3000 bags.

Tiger said:   6 years ago
={(1500-60)x(560-80-60)x270}/(3000x18).
= 3024.
= 3000 is the Answer.

Anjali said:   6 years ago
14.7*5.3*2.7 ÷ 035 = 3005.


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