Our aim is to mix the fine aggregate & coarse aggregate in such a ratio so that the final combined aggregate gives the specified or desirable surface index. Follow the below mentioned steps to find the proportion of fine to coarse aggregate.
Procedure
- Calculate the total surface index of fine aggregate (let it be x)
- Calculate the total surface index of coarse aggregate (let it be y)
- Let z be the total surface index of combined aggregate and a is the proportion of fine to coarse aggregate.
- Then a = (z-y)/(x-z)
Worked out Example
Calculating total surface index of fine aggregate
Sieve sizes within which particles lie | Percentage of particles within sieve size | Surface index for particles for within sieve size | Calculation of total surface index |
4.75 mm to 2.36 mm | 10 | 4 | 40 |
2.36 mm to 1.18 mm | 20 | 7 | 140 |
1.18 mm to 600 micron | 20 | 9 | 180 |
600 micron to 300 micron | 30 | 9 | 270 |
300 micron to 150 micron | 15 | 7 | 105 |
Total = | 735 | ||
Adding constant of 330 | 1065 |
Total surface index of coarse aggregate (i.e. x) = 1065/1000 = 1.065
Calculating total surface index of coarse aggregate
Sieve sizes within which particles lie | Percentage of particles within sieve size | Surface index for particles for within sieve size | Calculation of total surface index |
20 mm to 10 mm | 65 | -1 | -65 |
10 mm to 4.75 mm | 35 | 1 | 35 |
Total = | -30 | ||
Adding a constant of 330 | 300 |
Total surface index of coarse aggregate (i.e. y) = 300/1000 = 0.30
Let the total surface index of combined aggregate (i.e. z) = 0.60
So a = (z-y)/(x-z)
= (0.60-0.30)/(1.065-0.60)
= 1/1.55
So fine aggregate : Coarse aggregate = 1 : 1.55
how to get adding constant of 330?