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.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
13 comments Page 1 of 2.

Lalit said:   5 years ago
Flaky aggregate = 3/5 of there's mean dimension.
= (3/5) *{(75+60) /2}.
= (3/5) *(67.5).
= 40.5mm.
(15)

Abhi said:   7 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)

Anurag said:   4 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.
(1)

Reddy said:   1 decade ago
60+75 = 135.

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.

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

G.V.S. Bhargav said:   9 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.


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