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
Discussion:
14 comments Page 1 of 2.
Lalit said:
6 years ago
Flaky aggregate = 3/5 of there's mean dimension.
= (3/5) *{(75+60) /2}.
= (3/5) *(67.5).
= 40.5mm.
= (3/5) *{(75+60) /2}.
= (3/5) *(67.5).
= 40.5mm.
(18)
Sankalp kadam said:
9 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.
Means (3/5) * 60 that comes 36mm.
(1)
Abhi said:
8 years ago
Flakiness means 3/5 of mean size of aggregate, therefore it is 40.5.
(1)
Sandeep kumar said:
8 years ago
Since we know.
Flakiness index= .6(3/5)* average size of aggregate((passing size+retained size)/2).
So it is 40.5mm.
Flakiness index= .6(3/5)* average size of aggregate((passing size+retained size)/2).
So it is 40.5mm.
(1)
Avijit mirdda said:
8 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.
For flaky if its minimum dimension is less than 3/5 of 67.5 i.e. 67.5*(3/5) = 40.5mm.
(1)
Anurag said:
6 years ago
An aggregate is said to be flaky if its least dimension is less that 3-5.
That's why we use 3÷5.
That's why we use 3÷5.
(1)
Reddy said:
1 decade ago
60+75 = 135.
135/2 = 67.5.
3/5*67.5 = 40.5.
135/2 = 67.5.
3/5*67.5 = 40.5.
ARUNA said:
1 decade ago
(75+60)/2 = 67.5.
67.5*3/5 = 40.5.
67.5*3/5 = 40.5.
Prahumac said:
1 decade ago
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
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