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 4 of 5.

Sufazo said:   8 years ago
Hi guys, you have a room dimension, but nobody doing deduction properly, between the piles diff should be 1.2 m and 0.30 m from both side wall, but in width. Both side wall deduction. After the deduction, you have the answer, approximately 3600.

Monu kumar said:   8 years ago
Length = 15-0.6 = 14.4m.
Breadth=5.6-0.6-0.8=4.2m.
0.8 less for gap b/w two piles.
Area=14.4*4.2=60.48sq.m.
No.of bag in one layer=60.48/0.3=201.6bags.
0.3 area of one bag cement.
2.7/0.18 = 15.
0.18 is the height of one bag cement.
Then, 201.6 * 15=3024 bags -> 3000 bags.

Sai said:   8 years ago
7 bags for 1 cum concrete.

15 * 5.6 * 2.7 = 226.8 cum.
226.8 * 7 = 1587.6 cum for 1 pile.

For 2 piles 2 * 1587.6 = 3175.2 bags.

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

Sharad said:   1 decade ago
How this answer is possible please explain with size of cement bag or volume of cement bag?

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.


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