Material estimation for M20 grade concrete./ Estimating cement, sand, & aggregates in 1 CUM. of 1:1.5:3 mix ratio (M20)concrete.

 Let us consider 1 CUM. of  M20 grade concrete for the calculation purpose.

Here, 

Cement:sand: aggregates = 1:1.5:3 (a:b:c)

Wet volume of concrete = 1 cum.

Now, dry volume of concrete 

                 = 1.54 × wet volume of concrete.

                 = [1.54 × 1 cum]

                 = 1.54 cum.

              

1. Volume of  cement required

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

   = [{1÷ ( 1 + 1.5 +3 )} × 1.54]

   = [{ 1 ÷ 5.5 } × 1.54]

   =[ 0.18182 × 1.54]

   =  0.28 cum.

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

So, the number of cement bags required

       = [0.28 ÷ 0.03472]

       = 8.06 bags  

Weight of cement 

= [8.06 × 50 kgs.]

(1 bag of cement = 50 kgs.)

= 403 kgs.

2. Volume of sand required

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

   = [{1.5 ÷ ( 1 + 1.5 +3 )} × 1.54]

   = [{ 1.5 ÷ 5.5 } × 1.54]

   = [0.27273 × 1.54]

   =  0.42 cum.

3. Volume of aggregates required

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

   = [{3÷ ( 1 + 1.5 +3 )} × 1.54]

   = [{ 3 ÷ 5.5 } × 1.54]

   = [0.54545 × 1.54]

   =  0.84 cum.

   

4. Volume of water required.

  Let us consider the w/c ratio for M20 grade concrete as 0.45.

i.e. wt. of cement / wt. of water = 0.45

 So, the wt. of water 

       = [0.45 × wt. of cement] 

       = [0.45 × 403 kgs]

      = 181.35 kgs

     = 181.35 liters 

      ( 1kg= 1 liter)

Note: This w/c ratio holds good if you use plasticizers. Practically for good workability, the water we add for the concrete is 200 to 220 liters.

Alternate method:

1. Volume of cement ( from above) = 0.28 cum.

2. Volume of sand     

         = [volume of cement × b]

         =   [0.28 cum × 1.5]

         = 0.42 cum.

3.  Volume of aggregates

      = [volume of cement × c]

      =  [0.28 cum × 3]

      = 0.84 cum.

Quantity of materials in cubic ft.:

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

So, multiplying the above-derived quantity of materials with 35.3147, we will get,

1. Volume of cement 

      = [0.28 × 35.3147 cu ft.]

      = 9.888 cu ft.

2. Volume of sand      

     = [0.42 × 35.3147 cu.ft.]

    = 14.832 cu ft.

3. Vol. of aggregates 

      = [0.84 × 35.3147cu.ft.]

     = 29.664 cu ft.

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

Materials in 1CUM of M20 grade concrete.
Sl. No.	Item	Unit	In CUM.	In cu ft.
1.	Cement	bags	8.06 bags	-
2.	Sand	-	0.42 CUM	14.83
3.	Aggregates	-	0.84 CUM	29.66
4.	Water	litres	182 litres	-

 

                             

Similarly, for the different grades of concrete, you have to replace a, b, and c respectively for the given mix ratio, to get the answer.

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