Civil Engineering - Concrete Technology - Discussion

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.

What amount of floor space will one bag of cement occupy? Tell me.

(15-0.6-0.6)*(5.6-0.6-0.6)*1.8÷(.3*18) = 2024 bags.

0.6m is the distance from the external wall from all the four sides.

1.8 m is taken instead of 2.7 m because maximum of 10 bags of cement can be stacked in a stack.

0.18 m = 18 cm Height of cement bag.
0.3m^2 = 3000cm^2 = Surface area of cement bag.

But if the height is taken as 2.7m then the answer becomes 3036 bags.

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.

@Pratik.

0.3 m^2 floor space of 1 bag of cement.

@All.

*As per IS code 4082, the spacing of 600mm between the pile and the exterior wall should be provided.

*And passes or spacing of 600 mm should be provided between two piles.

* Max height of pile shouldn't exceed 10 bags

*Entrance will be provided where there is less area for opening means along shorter side i.e. 5.6 m wall.

* Volume of cement bag=2000cm^2*18cm=.054 m^3.

The general meaning of piles: A number of things lying on top of one another, or an amount of something lying in a mass.

Solution: Effective Length = 15-(2* spacing ) = 15-2*.6=13.8m.
Effective breadth= 5.6-(.6+.6+.6) = 3.8 m.
.6 from boths wall and one from passes of .6m between two piles.
Total area= 13.8*3.8=52.44 m^2.
Total volume= 52.55* 2.7=141.588 m^3.

Number of cement that can be stored = Total volume / gross volume of cement bags
N= 141.588/(.054)=2622 cement bags.

If max height of the pile is taken to be 10 bags (1.8m) as per code height cant exceed 10 bags then;

N= (52.44*1.8)/.054=1748.

The maximum thickness of floor in cement warehouse is 25 cm hence we have to deduct this value from pile height.

2.7-0.25=2.45.

Then the total volume =15*5.6*2.45=205.8cum.

As we know that cement per bag volume =0.035.

Hence no.of the bag of cement =205.8/0.035=5880.

And it is stored in two piles the one pile capacity 5880/2=2940.

Hence the nearest option is 3000.

Correct answer.

Effective dimentions are,

Horizontal clearance 0.3 m
15 - 2*0.3=14.4m,
5.6 - 2*0.3= 5m.

Now vertical clearance 0.2 m from top and bottom.
Therefore 2.7-2*0.2 = 2.3m.
Therefore volume of warehouse= 14.4 * 5 * 2.3 = 165.6 m3.
Area of 1 bag= 0.3 m2.

And the height of 1 bag = 0.18m.
So,the volume of bag= 0.3 * 0.18 = 0.054 m3.
Now, max no of the bag can be stored= volume of warehouse ÷ volume of the bag.
= 165.6 ÷ 0.054 = 3066.6667 bags.

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.

The answer should be 30 bags. Because It's not possible for 2 pile to contain 3000 bags.

Height of pile=2.7.
Plain area for 1 bag=0.3,
Total piles=2.
Volume of 2 piles=2*0.3*2.7=1.62,

Volume of 1 bag for storage =0.3*0.18=0.054.

Number of the bag in 2 piles=1.62/0.054=30.