Civil Engineering - Concrete Technology - Discussion

@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.

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.

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.

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.

It's overall dimensions of the warehouse, so we need to convert it to the effective area by subtracting lengths from sidewalls.
Effective dimensions of Warehouse:
=> (15 -.6 -.6)= 13.8m.
=> (5.6 -.6 -.6) = 4.4m.
> Hight of pile = 2.7m.
The effective dimensions of Cement Bag:
Area of 1 cement beg = .3 sqm,
Height of 1cement beg = .18 sqm.

Calculation:
=>(13.8 * 4.4* 2.7)/(.3*.18)= 3036 bags.

14.7*5.3*2.7 ÷ 035 = 3005.

={(1500-60)x(560-80-60)x270}/(3000x18).
= 3024.
= 3000 is the Answer.

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.

(5.6-0.6-0.6) x (15-0.6-0.6) x 2.7/(0.3x.18) = 3036.

Right, thanks for the Answer @Ayon Som.