Let us consider 100 cu ft. of M15 grade concrete for calculation purposes.
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 |
- |