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Method of findingthe dry ingredient quantity of 1 m3
concrete.
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Answer Posted / khuram shahzad
mix proportions are
M15 = 1: 2 : 4
M20 = 1: 1.5 : 2
M25 = 1: 1 : 2
Ok lets go to the calculaton part
Now we are going to calculate the quantity of materials required for 1m3 Concrete (M15)
Quantity of Cement required for 1m3
[1/(1+2+4)] * 1.52 = 0.217m3
0.217 * 1440 = 312.48 kg/m3
To calculate in bags
312.48/50 = 6.24 bags/m3
Here 1.52 is the dry co-efficient and 1440 is the unit weignt of cement.
Quantity of Fine aggregate( sand) for 1m3
[2/(1+2+4)] * 1.52 = 0.434m3
Quantity of course aggregate for 1m3
[4/(1+2+4)] * 1.52 = 0.868m3
So this is the Quantity for 1m3. For example, If you want 6m3 concrete of M15 grade
for 6m3
cement = 312.48 * 6 = 1874.88 kg for 6m3 (37.5 bags )
fine aggregate = 0.434 * 6 = 2.604 m3 ( 2.604 m3 sand required for 6m3 concrete )
course aggregate = 0.868 * 6 = 5.208 m3 ( 5.208 m3 aggregate required for 6m3 concrete)
The same procedures should be followed for M20, M25 concrete
For example M20
[1/(1+1.5+2)]* 1.52
For M25
[1/(1+1+2)] * 1.52
| Is This Answer Correct ? | 121 Yes | 20 No |
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