What is fineness modulus of aggregate?
The fineness modulus of aggregate is simply a measurement of the average size of the aggregate. It can be calculated with the help of different sizes of the sieve. Now, let us see some numerical Examples to calculate the fineness modulus of aggregate.
Q.1) Find the fineness modulus of coarse aggregates for which the sieve analysis is given it the table below:
Weight of sample = 10 kg
Solution:-
Coarse aggregate
| IS sievi (mm) | Weight retained 9(kg) | Cumulative weight retained(kg) | Cumulative% retained |
| 75 mm | 0 | 0 | 0 |
| 40 mm | 0 | 0 | 0 |
| 20 mm | 2.5 | 2.5 | (2.5/10)*100=25 |
| 10 mm | 4.5 | 7 | (7/10)*100=70 |
| 4.75 mm | 2 | 9 | 90 |
| 2.36 mm | 1 | 10 | 100 |
| 1.18 mm | – | – | 100 |
| 600 micron | – | – | 100 |
| 300 micron | – | – | 100 |
| 150 micron | – | – | 100 |
| Total | 685 |
So,
Fineness modules(F.M.)
=685/100
=6.85
Q.2) Find the F.M of coarse aggregate and fine aggregate form the sieve analysis data as given in the table below:
Weight of sample = 15 kg
Solution:-
Coarse aggregate
| IS sieve (mm) | Weight retained(kg) | Cumulative weight retained (kg) | Cumulative% retained |
| 75mm | 0 | 0 | 0 |
| 40mm | 0 | 0 | 0 |
| 20mm | 7 | 7 | (7/15)*100=46.67 |
| 10mm | 6 | 13 | (13/15)*100=86.67 |
| 10mm | 2 | 15 | 100 |
| 4.75mm | – | – | 100 |
| 2.36 | – | – | 100 |
| 1.18 | – | – | 100 |
| 600micron | – | – | 100 |
| 300micron | – | – | 100 |
| 150micron | – | – | 100 |
| total | 733.34 | ||
Fineness modules (F.M.) = 733.34/100 =7.33
Total weight of fine aggregate = 500gm
Fine aggregate
| IS sieve(mm) | Weight retained (kg) | Cumulative weight retained (kg) | Cumulative% retained |
| 75mm | – | – | – |
| 40mm | – | – | – |
| 20mm | – | – | – |
| 10mm | 0 | 0 | 0 |
| 4.75mm | 5 | 5 | (5/500)*100=1 |
| 2.36 | 55 | 60 | (60/500)*100=12 |
| 1.18 | 55 | 115 | 23 |
| 600micron | 90 | 205 | 41 |
| 300micron | 170 | 375 | 75 |
| 150micron | 90 | 465 | 93 |
| Lower than 150 micron | 35 | 500 | 100(not unit) |
| total | 245 |
NOTE*
Weight lower than 150 micron is not considered in calculation :-
Fineness modules (F.M.) = 2.45
Q.3) A sieve analysis was carried out for 5 kg aggregate in the laboratory with available sieves. The weights retained in sieve of 40,32,20,16,10,4.75,are 0,1.2,1.8,0.5,1.5and 0 kg respectively. Obtain fineness modules for the sample.
Solution:
| IS sieve(mm) | Weight retained(kg) | Cumulative weight retained (kg) | Cumulative% retained |
| 40mm | 0 | 0 | 0 |
| 32mm | 1.2 | 1.2 | 24 |
| 20mm | 1.8 | 3 | 60 |
| 16mm | 0.5 | 3.5 | 70 |
| 10mm | 1.5 | 5 | 100 |
| 4.75 | 0 | 5 | 100 |
| total | 354 | ||
COMULATIVE % weight retained = (commutative weight retained/total weight)*100%
Fineness modules (F.M.) = (total cumulative % weight retained)/100
=354/100
= 3.54
I hope this article on “Fineness modulus of aggregate” remains helpful to you.
Happy Learning – Civil Concept
Contributed by,
Civil Engineer – Ranjeet Sahani
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