# HOW TO BLEND AGGREGATE USING SURFACE INDEX VALUE?

## Surface Index of Aggregates

The Surface Index of an aggregate is another way of representing the specific surface of an aggregate. This method was suggested by Murdock. Murdock provided some empirical numbers to represent the typical surface grading of various aggregates (as shown in the table below).

## Surface Index Values ​​for Single Graded Aggregate

 Table 1 sieve size The surface index for particles within the sieve size is indicated 80mm to 40mm -2.5 40mm to 20mm -2 20mm to 10mm -1 10mm to 4.75mm 1 4.75mm to 2.36mm 4 2.36mm to 1.18mm 7 1.18 mm to 600 µm 9 600 micron to 300 micron 9 300 micron to 150 micron 7 <150 µm 2

It is clear from the above table that aggregates larger than 10 mm have a lower surface index value, and for aggregates within the range of 10 mm to 300 µm, the value increases and the surface index value decreases with the decrease of aggregate size. .

The values ​​mentioned in the above table represent only the total particles which fall in the sieve size as shown in column-1. But in actual practice the aggregate (whether fine or aggregate) consists of particles of different sizes. It is therefore necessary to find a way to calculate the surface index of an aggregate consisting of particles of different sizes. And this is done by calculating the total surface index of the aggregate.

## How to Calculate Total Surface Index

### step 1

Filter the combined aggregate according to the sieve mentioned below and record the amount of material held by the respective sieve.

Sieve required for sieve = 80 mm, 40 mm, 20 mm, 10 mm, 4.75 mm, 2.36 mm, 1.18 mm, 600 μm, 300 μm and 150 μm.

### step 2

Find the corresponding surface index value for different composite grading (see Table-1)

### step 3

Multiply the percentage of material placed on its sieve by the corresponding surface index. Then add up all the multiplied values ​​and add a constant 330 to their sum and divide the result by 1000. And the answer is the total surface index of the total.

## Example calculation for total surface index

 the size of the sieve within which the particle is located % of particles within sieve size Surface index for particles within the sieve size Calculation of Total Surface Index 20mm – 10mm 55 -1 -55 10mm – 4.75mm 15 1 15 4.75mm – 2.36mm 7 4 28 2.36mm – 1.18mm 7 7 49 1.18 mm – 600 µm 7 9 63 600 micron – 300 micron 7 9 63 total = 164 continuously 330. Add 494

So total surface index = 494/1000 = 0.494

## Composite Aggregate Using Total Surface Index Value

We aim to mix fine aggregates and coarse aggregates in such a ratio that the final composite aggregate gives the specified or desirable surface index. Follow the steps below to find the ratio of fine to coarse aggregate.

### Process

1. Calculate the total surface index of the fine aggregate (let it be x,
2. Calculate the total surface index of the coarse aggregate (let it be You,
3. Let it happen jade be the total surface index of the composite set and a The ratio of fine to coarse aggregate is
4. Then a = (z)/(xz)

## O . Work doneUT example for total blending

### Calculating the total surface index of fine aggregates

 Filter out shapes that contain particles Percentage of particles within the size of the sieve Surface index for particles within the sieve size Calculation of Total Surface Index 4.75mm to 2.36mm 10 4 40 2.36mm to 1.18mm 20 7 140 1.18 mm to 600 µm 20 9 180 600 micron to 300 micron 30 9 270 300 micron to 150 micron 15 7 105 total = 735 330. adding a constant of 1065

Total surface index of coarse aggregate (ie.) x) = 1065/1000 = 1.065

### Calculation of Total Surface Index of Coarse Aggregates

 Filter out shapes that contain particles Percentage of particles within the size of the sieve Surface index for particles within the sieve size Calculation of Total Surface Index 20mm to 10mm 65 -1 -65 10mm to 4.75mm 35 1 35 total = -30 330. adding a constant of 300

Total surface index of coarse aggregate (ie.) You) = 300/1000 = 0.30

Let the total surface index (ie.) of the composite set be jade) = 0.60

therefore a = (z)/(xz)

, (0.60-0.30)/(1.065-0.60)

= 1/1.55

so the ratio Foffline Aggregates: = Coarse Aggregates 1:1.55

Er. Mukesh Kumar

Er. Mukesh Kumar is Editor in Chief and Co-Funder at ProCivilEngineer.com Civil Engineering Website. Mukesh Kumar is a Bachelor in Civil Engineering From MIT. He has work experience in Highway Construction, Bridge Construction, Railway Steel Girder work, Under box culvert construction, Retaining wall construction. He was a lecturer in a Engineering college for more than 6 years.