Civil Engineering - Concrete Technology - Discussion

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.

Effective area of hall = 15 x (5.6- 0.6 * 3) = 57m
Area of a cement bag = 0.3 sqm.
No. of cement bag in one layer = 57/.3 = 190,
Height of pile = 0.7.
No. of layer Height of one cement bag = 2.7/0.18 = 15 layer,
Total no. of bags 15 * 90 = 2850 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.

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.

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.

The Answer is 4200.

@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 .

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.

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.