Civil Engineering - Concrete Technology - Discussion

The total volume of room = 54 * 2.7 = 145.8 cu.m.
The volume of 1 cement bag = 0.0347 (This is the wet Volume and We've to calculate at site in dry Volume).
Dry Volume= Wet Vol * 1.54 to 1.57,
Dry Volume= 0.0347 * 1.5,
Dry Vol= 0.053 cu.m.

The Number of cement bags = 145.5cum/0.053 = 2745.
Hence, 2745 is near to 2700.

Here,

In first case: (standard):

Area required for 1 bag of cement = 0.3 sq.m,
Height of pile required for 1 bag = 0.18 m,
Volume of 1 bag = 0.3 * 0.18,
= 0.054 cubic metre.

In second case:(Given):
Plan area of warehouse = 54 sq.m,
Max.height of piles = 270 cm =2.7 m,
Volume of warehouse = 54*2.7.
= 145.8 cubic metre.

We know that;
No.of bags = volume of warehouse/volume of 1 bag of cement.
= 145.8/0.054.
= 2700 bags.(Answer).

Simply,

No. Of bags = Effective area* Max.ht of piles/ Area coverd by one bag cement is 0.3 sq.m & ht= 0.18.

No= 54*2.7/0.3*0.18 = 2700 bags you must get.

Area = 54m^2.
height = 270cm = 2.7m,
volume = 54 * 2.7 = 145.8m^3
The Area required for one bag of cement = 0.3m^2,
The height required for one bag of cement = 0.18m.
The volume required for one bag of cement = 0.3 * 0.18 = 0.054m^3.
So, no.of bags = 145.8/0.054 = 2700.

As per rule the distance between two piles is 1.6m and distance from wall is 30cm and 20cm above the ground level is not included in this calculation why?

According to rule, it is a bad way to store cement in ware house.

270cm = 2.7m.

Yeah, you are right, agree @Darshan H A.

The weight of 1 kg of cement is 50 kg.
50×57 = 2700.

We know, surface area and height of pile required for storing of 1 bag of cement are 0.3 sq. m. and 0.18 m respectively.

Therefore, maximum nos of bags can be stored into the given stored house = (5.4 x 2.7)/(0.3 x 0.18) = 2700 bags.

2.7 How come?

Please explain.