# PARAM VISIONS

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 × 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 -