Surface Index of Aggregates
Surface index of aggregates is another way to represent the specific surface of aggregate. This method was suggested by Murdock. Murdock provided some empirical numbers to represent the specific surface of different aggregates grading (as shown in the table below).
Surface Index Values for Single Graded Aggregate
|
TABLE-1 |
|
|
Sieve Size Within Which Particles Lie |
Surface Index For Particles Within Sieve Size Indicated |
| 80 mm to 40 mm | -2.5 |
| 40 mm to 20 mm | -2 |
| 20 mm to 10 mm | -1 |
| 10 mm to 4.75 mm | 1 |
| 4.75 mm to 2.36 mm | 4 |
| 2.36 mm to 1.18 mm | 7 |
| 1.18 mm to 600 micron | 9 |
| 600 micron to 300 micron | 9 |
| 300 micron to 150 micron | 7 |
| < 150 micron | 2 |
From the above table it is clear that aggregate larger than 10 mm has less surface index value, and the value goes on increasing for aggregates within range 10 mm to 300 micron and surface index value starts to decrease with decrease in aggregate size.
The values mentioned in the above table only represent the aggregate particles which fall in the sieve size as shown on column-1. But in actual practice the aggregate (whether fine or aggregate) consist of particles of different sizes. Therefore it is necessary to find a way to calculate the surface index of aggregate consisting particles of different sizes. And this is done by calculating total surface index of aggregate.
How to Calculate Total Surface Index
Step-1
Sieve the combined aggregates as per the sieves mentioned below and record the amount of material retained by corresponding sieves.
Sieves required for sieving = 80 mm, 40 mm, 20 mm, 10 mm, 4.75 mm, 2.36 mm, 1.18 mm, 600 micron, 300 micron & 150 micron.
Step-2
Find out the corresponding surface index values for different aggregate grading (refer table-1)
Step-3
Multiply the percentage of material retained on its sieve by the corresponding surface index. Then add all the multiplied values and to their sum add a constant 330 and divide the result by 1000. And the answer is the total surface index of the aggregate.
Example Calculation for Total Surface Index
| Sieve size within which particle lie | % of particles within sieve sizes | Surface index for particles within sieve size | Calculating Total Surface Index |
| 20 mm – 10 mm | 55 | -1 | -55 |
| 10 mm – 4.75 mm | 15 | 1 | 15 |
| 4.75 mm – 2.36 mm | 7 | 4 | 28 |
| 2.36 mm – 1.18 mm | 7 | 7 | 49 |
| 1.18 mm – 600 micron | 7 | 9 | 63 |
| 600 micron – 300 micron | 7 | 9 | 63 |
| Total = | 164 | ||
| Adding constant 330 | 494 | ||
So Total Surface Index = 494/1000 = 0.494
Blending of Aggregate Using Total Surface Index Value
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 for Aggregate Blending
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 the ratio of Fine aggregate: Coarse aggregate = 1:1.55