# Civil Engineering - Concrete Technology - Discussion

Discussion Forum : Concrete Technology - Section 4 (Q.No. 2)
2.
If aggregates completely pass through a sieve of size 75 mm and are retained on a sieve of size 60 mm, the particular aggregate will be flaky if its minimum dimension is less than
20.5 mm
30.5 mm
40.5 mm
50.5 mm
none of these.
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

60+75 = 135.

135/2 = 67.5.

3/5*67.5 = 40.5.

(75+60)/2 = 67.5.

67.5*3/5 = 40.5.

I didn't get it why did you multiply by 3.

ITAF HUSSAIN said:   1 decade ago
@prahumac we multiply it by 3/5 of its mean dimension

G.V.S. Bhargav said:   10 years ago
Flaky aggregate is a aggregate having a size less than 60% of average size of the two aggregate.

So now (75+60)/2 = 67.5.

Now 60% of 67.5 is,

That is 67.5*(60/100) = 40.5.

Sankalp kadam said:   8 years ago
If avg has be taken into account then according to me, as agg are retained on 60mm sieve size so it means that most of the agg are retained at 60 so we have to consider that value only rather than taking commutative of two sieve sizes isn't it?

Means (3/5) * 60 that comes 36mm.

Abhi said:   8 years ago
Flakiness means 3/5 of mean size of aggregate, therefore it is 40.5.
(1)

Sandeep kumar said:   7 years ago
Since we know.

Flakiness index= .6(3/5)* average size of aggregate((passing size+retained size)/2).
So it is 40.5mm.
(1)

Avijit mirdda said:   7 years ago
Average of 75mm and 60 mm sieve is (75+60)/2=67.5.

For flaky if its minimum dimension is less than 3/5 of 67.5 i.e. 67.5*(3/5) = 40.5mm.
(1)

Yasir said:   6 years ago
Flakiness means 60%(3/5) of avg size of particles.