Civil Engineering - Concrete Technology - Discussion

Discussion Forum : Concrete Technology - Section 1 (Q.No. 29)
29.
If 20 kg of coarse aggregate is sieved through 80 mm, 40 mm, 20 mm, 10 mm, 4.75 mm, 2.36 mm, 1.18 mm, 600 micron, 300 micron and 150 micron standard sieves and the weights retained are 0 kg, 2 kg, 8 kg, 6 kg, 4 kg respectively, the fineness modulus of the aggregate, is
7.30
7.35
7.40
7.45
none of these.
Answer: Option
Explanation:
No answer description is available. Let's discuss.
Discussion:
44 comments Page 2 of 5.

Santosh said:   7 years ago
Can't understand it, please explain it clearly.

Munir said:   7 years ago
Thanks for the answers.

Ganga said:   7 years ago
Well explained @Shailendra.

Shailendra said:   7 years ago
FM= Mi/M=148/20=7.4.

Where M1=m1=0 kg.
M2=m1+m2=0+2=2 kg,
M3=m1+m2+m3=2+8=10 kg,
M4=m1+m2+m3+m4=10+6=16 kg,
M5=m1+m2+......+m5=16+4=20 kg,
M6=m1+m2+......+m6=20+0=20 kg,
M7=m1+m2+......+m7=20+0=20 kg,
M8=m1+m2+......+m8=20+0=20 kg,
M9=m1+m2+......+m9=20+0=20 kg,
M10=m1+m2+......+m10=20+0=20 kg.
(3)

Rahul said:   7 years ago
The fm of coarse aggregate is C+500/100.
The fm of fine aggregate is f/100.

So, what we try on sieve 4.75 lower all are fine particles as per code.

Priyanka said:   7 years ago
How to get 7.40?

Kavita said:   8 years ago
Please explian clearly.

Ritesh said:   8 years ago
Please explain this. I can't understand,

How get 740?

Pranit Vitekar said:   8 years ago
Seive. Weight % weight. Cumulative % wt
size. Retained retained. Retained
80. 0. 0. 0
40. 2 10. 10
20. 8 40. 50
10. 6 30. 80
4.75. 4 20. 100
2.36. 0. 0. 100
1.18. 0 0 100
600. 0 0 100
300. 0 0 100
150. 0. 0 100

Total cumulative % weight retained= 740.
Fineness modulus = total cumulative % weight retained/ 100.
FM=740/100.
7.40 is the correct answer.

Kavita said:   8 years ago
@ALL.

Please, Refer Is code 2386.


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