Civil Engineering - Concrete Technology - Discussion

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.

L = 15-0.6-0.6 = 13.8
B = 5.6 - 0.6 - 0.6 = 4.4
H = 2.7
13.8 × 4.4 × 2.7/0.3 × 0.18 = 3000 bag.

The Width of Pile = 1.6m,
So, width of warehouse = 5.6 - 1.6 = 4
Now volume=15 * 4* 2.7 = 162
Now, the volume of one bag = .054.
So, 162/.054 = 3000 bags.

2100 for a single pile.

So, the right answer is 2500.

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.

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.

@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 .

The Answer is 4200.

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.