Let us consider 100 cu ft. of M15 grade concrete, for calculating the quantity of cement, sand, & water in its mix proportion.

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

=( 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

= 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

=( 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

=( 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 in cum. 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 × 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.**

3. Volume of aggregates in 1cum of M15 concrete

= volume of cement × c

= 0.22 cum. × 4

=** 0.88 cum.**

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.)