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.

As per IS 4082:1996, (i) minimum clear space around or between the exterior wall face and the stacks should be 600 mm, (ii) Maximum width of the stack should not be more than 3.0 m and (iii) the maximum height of stacks should not be more than 1.40 m.

No of piles stack or row or column = 2 nos.
Effective length of the stack = 15-2 x 0.3 = 14.4m,
Effective width of the stack = 5.6-0.3 x 2-0.75 = 4.25m,
Height of pile stacks = 2.70m,
We know that, plan area of each bag of cement = 0.3 m^2.
Height of each bag of cement = 0.18 m.
The Maximum number of bags of cement that can be stored in two piles = (14.4x4.25x2.7)/(0.3x0.18)
= 3060 bags ~3000 bags.

The answer should be 3600 bags.

Internal dimension of warehouse = 15m x 5.6m

Effective storage floor area which is available:
=> (15m - 300mm - 300mm) x (5.6m - 300 mm - 300 mm) = 14.4m x 5m

(300mm distance should be maintained from each exterior wall)

Therefore, No. of bag in one layer = Available area / Area occupied by one bag:
=> 14.4m x 5m / 0.3 sq.m = 240 bags in one layer

No. of layer of bags = Max. height of pile allowed / Height of one bag:
=> 2.7m / 0.18 m = 15 layers.

Therefore, total no. bags can be stored = Total bag in one layer X Total no. of layers:
=> 240 X 15 = 3600 Bags of cement

Hence, The answer should be 3600 bags.

Concept :

Generally, 0.6m is taken as the distance from the external walls and 0.8 is the gap between piles.

Calculation :

Given :

Length = 15 m, Breadth = 5.6 m, and Height of Pile = 2.70 m.
Considering the distance from the external wall and the gap between piles,
Length = 15 - 0.6 - 0.6 = 13.8 m,
Breadth = 5.6 - 0.6 - 0.6 - 0.8 = 3.6 m,
Area = 13.8 * 3.6 = 49.68 m2.
Area of one cement bag = 0.3 m2.
Height of one cement bag = 0.15 m.

No. of bags = Space for cement/Volume of one cement bag

Number of Bags = (49.68 * 2.7)/(0.3 * 0.15) = 2980 bags = 3000 Bags.

SOLUTION

Generally, 0.6m is taken as the distance from the external walls
And 0.8 is the gap between piles.

Calculation :

Given :

Length = 15 m, Breadth = 5.6 m, and Height of Pile = 2.70 m

Considering the distance from the external wall and the gap between piles,

Length = 15 - 0.6 - 0.6 = 13.8 m.
Breadth = 5.6 - 0.6 - 0.6 - 0.8 = 3.6 m.
Area = 13.8* 3.6 = 49.68 m2.
Area of one cement bag = 0.3 m2.
Height of one cement bag = 0.18 m.

No. of bags = Space for cement/Volume of one cement bag

Number of Bags = (49.68 *2.7)/(0.3 * 0.18) = 2484 bags ~~2500 Bags.

Concept :
Generally, 0.6m is taken as the distance from the external walls and 0.8 is the gap between piles.

Calculation :
Given :

Length = 15 m, Breadth = 5.6 m, and Height of Pile = 2.70 m

Considering the distance from external wall and gap between piles,

Length = 15 - 0.6 - 0.6 = 13.8 m,
Breadth = 5.6 - 0.6 - 0.6 - 0.8 = 3.6 m,
Area = 13.8 * 3.6 = 49.68 m2,
Area of one cement bag = 0.3 m2,
Height of one cement bag = 0.15 m,
No. of bags = Space for cement/Volume of one cement bag.
Number of Bags = (49.68 *2.7)/(0.3* 0.15) = 2980 bags = 3000 Bags.

@All.

Solution is

Height of cement = 18cm
Area = ( 40*75) cm2.
Space between 2 pile 1.5m .
Pile space from wall= 30cm.
Maximum height of cement bags = 2.7 m = 15 bags vertically.
Floor area of the warehouse for 2 piles will be.
Length= (15- 0.6)m.
= 14.4.

Breath = ( 5.6 - 0.6 - 1.5)m.
= 3.5m.

So,
Area is =14.4 * 3.5m2.
= 50.4m2.
Area of cement = 0.4 * 0.75m2 = 0.3m2.

No. Of bags on horizontal line= 50.4/0.3= 168 bags per horizontal line,
Maximum 15 bags vertically place so,
168*15= 2520.
Nearly= 2500.

So, answer will be C .

Density of cement is 0.035 cum.
And height of pile means the depth of ware house.
Length and breadth of the ware house is given,

For the volume of the ware house (length x breadth x height).
= 15 x 5.6 x 2.7.
= 226.8 cum.

Total quantity of cement = 226.8/0.035.
= 6480cum.
In two row it contains = 3240cum cement total.
So 3000 is suitable option for that because it can contain more than 2000 and less than 4000.

Solution is;

Height of cement = 18cm
Area = ( 40*75) cm2.
Space between 2 pile 1.5m .
Pile space from wall= 30cm.
Maximum height of cement bags = 2.7 m = 15 bags vertically.
The floor area of the warehouse for 2 piles will be.
Length= (15- 0.6)m.
= 14.4.

Breath = ( 5.6 - 0.6 - 1.5)m.
= 3.5m.

So,
Area is =14.4 * 3.5m2.
= 50.4m2.
Area of cement = 0.4 * 0.75m2 = 0.3m2.

No Of bags on horizontal line= 50.4/0.3= 168 bags per horizontal line,
Maximum 15 bags vertically place so,
168 * 15= 2520 (Nearly 2500).

So, the answer will be C.

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.