All about civil construction knowledge- PARAM VISIONS

Material estimation for M15 (1:2:4 mix ) grade concrete./Calculating the quantity of materials in 100 cu ft. & 1 cum. of M15 (1:2:4 ) grade concrete.

 Let us consider 100 cu ft. of M15 grade concrete for the calculation purpose. 




Here, 

cement:sand: aggregates = 1:2:4 (a:b:c)

The wet volume of concrete = 100 cu ft.

The  dry volume of concrete 

                 = 1.54 × wet volume of concrete.

                = 1.54 × 100 cu ft. = 154 cu ft.

              

1. Volume of cement required / 100 cu ft.

    =[{ a ÷ (a + b + c)} × dry volume of concrete.]

   = [{1÷ ( 1 + 2 + 4 )} × 154 cu ft.]

   = [{ 1 ÷ 7 } × 154]

   = [0.142857 × 154]

   =  22 cu ft.

   = 0.6229 cum.

(1cu ft = 0.028317 cum.)


As you know, 1 bag of cement = 0.03472 cum.

The number of cement bags required for 100 cu ft. of concrete

       = [0.6229 ÷ 0.03472]

       = 17.942 bags  


The weight of cement

     = [17.942 × 50 kgs.]

( 1 bag of cement = 50 kgs. )

   = 897.1 kgs.


2. Volume of sand required / 100 cu ft of concrete

    =( b ÷ (a + b + c)) × dry volume of concrete.

   = (2 ÷ ( 1 + 2 +4 )) × 154 cu ft.

   = ( 2 ÷ 7 ) × 154

   = 0.2857 × 154

   =  44 cu ft


3. Volume of aggregates required / 100 cu ft. of concrete

    = [{ c ÷ (a + b + c)} × dry volume of concrete.]

   = [{4÷ ( 1 + 2 +4 )} × 154 cu ft.]

   = [{ 4 ÷ 7 }× 154]

   = [0.5714 × 154]

   =  88 cu ft.

   

4. Volume of water required.

  Let us consider the w/c ratio for M15 grade concrete as 0.5.

i.e. wt. of cement/wt. of water = 0.5

 So, the wt. of water 

       = [0.5 × wt. of cement] 

       = [0.5 × 897.1 kgs]

      = 448.55 kgs.

      = 448.55 liters 

      ( 1kg = 1 liter.)


Alternate method:

1. Volume of cement ( from above) = 22 cu ft.


2. Volume of sand     

         = [volume of cement × b]

         =   [22 cu ft. × 2]

         = 44 cu ft.


3.  Volume of aggregates

      = [volume of cement × c]

      = [ 22 cu ft. × 4]

      = 88 cu ft.


Quantity of materials in 1 cum of M15 (1:2:4) mix concrete:

As you know, 1 cum. = 35.3147 cu ft.

So, multiplying the above-derived quantity of cement / cum of concrete with (35.3147 cu ft  ÷ 100 cu ft. ), we will get,

1. Volume of cement in 1 cum. of M15 concrete 

    = [0.6229 cum. × (35.3147   ÷ 100  )]

    = [0.6229 cum  × 0.353147]

   = 0.22 cum.

   

The density of cement = 1440 kg/ cum.

The wt. of cement in kgs

   = [0.22 cum. × 1440 kg/cum]

   = 316.8 kgs


The number of cement bags

   = [316.8 ÷ 50]

(1 bag of cement = 50kgs.)

  = 6.336 bags.


2. Volume of sand in 1 cum of M15 concrete

       = [volume of cement × b]

       =  [ 0.22 cum. × 2]

      = 0.44 cum.

       = 15.538 cu ft.

      ( As 1CUM = 35.3147 cu ft.)


3.  Volume of aggregates in 1cum of M15 concrete

      = [volume of cement × c]

      =  [0.22 cum. × 4]

      = 0.88 cum.

      = 31.077 cu ft.    

                         

4. Volume of water in 1cum of M15 concrete

   w/c ratio for M15 grade concrete is taken as 0.5.

   wt. of cement/wt. of water = 0.5

   So, the wt. of water 

       = [ 0.5 × wt. of cement ]

       = [0.5 × 316.8 kgs.]

      = 158.4 kgs.

      = 158.4 liters 

      ( 1kg = 1 liter.)

Now, let us put these material quantities in a table format for easy reference.

Materials in 1CUM of M15 grade concrete.

Sl. No.

Item

 Unit

In CUM.

In cu ft.

1.

Cement

bags

6.336 bags

-

2.

Sand

-

0.44 CUM

15.54

3.

Aggregates

-

0.88 CUM

31.08

4.

Water

litres

159 litres

-

 

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