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
Discussion:
45 comments Page 3 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.
Beewaek Mandal said:
6 years ago
The answer is 3000 Bags.
Internal dim.Area=15*5.6*2.70=226.8m3.
Ext. Dim area=15*1.6*2.70=64.8m3 (1.6 is bcoz one piles spacing 0.8m for two piles=1.6m)
Effective= Int-Ext= 162.
For one bag cement= 0.3 sq.m Ht= 0.18m.
No = 162/0.18 =3000.
Internal dim.Area=15*5.6*2.70=226.8m3.
Ext. Dim area=15*1.6*2.70=64.8m3 (1.6 is bcoz one piles spacing 0.8m for two piles=1.6m)
Effective= Int-Ext= 162.
For one bag cement= 0.3 sq.m Ht= 0.18m.
No = 162/0.18 =3000.
Er. Bhim Prasad said:
5 years ago
Effective dimensions of Warehouse:
=> (15 -.3*2)= 14.4m.
=> (5.6 -..3*2-1.6) = 3.4m. where 1.6 gap between piles.
Effective area of ware house =14.4*3.4=49.64m^2.
No of bag in two pile = (49.64*2.70)/(.3*.18)= 2482 bag~=2500bag.
=> (15 -.3*2)= 14.4m.
=> (5.6 -..3*2-1.6) = 3.4m. where 1.6 gap between piles.
Effective area of ware house =14.4*3.4=49.64m^2.
No of bag in two pile = (49.64*2.70)/(.3*.18)= 2482 bag~=2500bag.
Kapil Acharya said:
5 years ago
No of bags in single layer= 15 * 5.6 * 0.8/0.3= 224.
No of bags along with height = 270/18 =15.
Total no of bags= 224*15 = 3360 max.
Area for storage of cement can max be only 80% of the room area. So, here, the correct option is 3000.
No of bags along with height = 270/18 =15.
Total no of bags= 224*15 = 3360 max.
Area for storage of cement can max be only 80% of the room area. So, here, the correct option is 3000.
Krishna said:
10 years ago
I don't think all the answers are correct. As per guideline we can stack max of 10 bags vertically and we need to keep 600 mm space away from walls. And also need to keep a distance of 600mm between two rows.
Bhimcha said:
5 years ago
Length=15-.6=14.4m.
The gap between pile 1.6 m.
So breath= 5.6-.6-1.6=3.4m.
area for cement=14.4*3.4=49.64m2.
No of bag stored= (49.64*2.7)/(.3*.18) = 2482~2500 is the correct answer.
The gap between pile 1.6 m.
So breath= 5.6-.6-1.6=3.4m.
area for cement=14.4*3.4=49.64m2.
No of bag stored= (49.64*2.7)/(.3*.18) = 2482~2500 is the correct answer.
(1)
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.
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.
Er. soumya choudhury said:
1 decade ago
But volume of one cement bag is 0.035 cum.
Volume of ware house is 15*5.6*2.7 = 226.8 cum.
No of bags = 226.8/0.035 = 6480 bags. In two rows means 3240 bags.
Volume of ware house is 15*5.6*2.7 = 226.8 cum.
No of bags = 226.8/0.035 = 6480 bags. In two rows means 3240 bags.
Ravi said:
4 years ago
Width of gallery = .6m,
So, width of warehouse = 5.6 - .6 = 5.
Now volume=15 * 5 * 2.7 = 202.5.
Now, the volume of one bag=.054.
So, 202.5/.054 = 3750 bags.
So, width of warehouse = 5.6 - .6 = 5.
Now volume=15 * 5 * 2.7 = 202.5.
Now, the volume of one bag=.054.
So, 202.5/.054 = 3750 bags.
(1)
Ravi said:
4 years ago
Width of gallery = .6m,
So, width of warehouse = 5.6 - .6 = 5.
Now volume=15 * 5 * 2.7 = 202.5.
Now, the volume of one bag=.054.
So, 202.5/.054 = 3750 bags.
So, width of warehouse = 5.6 - .6 = 5.
Now volume=15 * 5 * 2.7 = 202.5.
Now, the volume of one bag=.054.
So, 202.5/.054 = 3750 bags.
(1)
Post your comments here:
Quick links
Quantitative Aptitude
Verbal (English)
Reasoning
Programming
Interview
Placement Papers