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
Discussion:
44 comments Page 2 of 5.
Nirav Bhatt said:
1 decade ago
Pass Wt Cum Wt. %Cum %cum.
Retained % passing.
0 0 0 100.
2 2 10 90.
8 10 40 50.
6 16 30 20.
4 20 20 0.
--- --- --- ---.
20 48 100 260.
Sieves used are 10.
Hence FM = (10*100-260)/100 = 7.4.
In method suggested by @Kishor, the total is done individually for each sieve where as here it is done by using the nos. sieve.
Both the methods are correct. So the Answer is 7.4.
Retained % passing.
0 0 0 100.
2 2 10 90.
8 10 40 50.
6 16 30 20.
4 20 20 0.
--- --- --- ---.
20 48 100 260.
Sieves used are 10.
Hence FM = (10*100-260)/100 = 7.4.
In method suggested by @Kishor, the total is done individually for each sieve where as here it is done by using the nos. sieve.
Both the methods are correct. So the Answer is 7.4.
Jagadeesh said:
10 years ago
Sieve weight % weight cumulative.
80 0 0 0.
40 2 10 10.
20 8 40 50.
10 6 30 80.
7.5 4 20 100.
4.75 0 0 100.
2.36 0 0 100.
1.18 0 0 100.
600 0 0 100.
300 0 0 100.
150 0 0 100.
-----------------
Total = 740.
-----------------
Fineness modulus of given course aggregate is 740/100 = 7.4.
80 0 0 0.
40 2 10 10.
20 8 40 50.
10 6 30 80.
7.5 4 20 100.
4.75 0 0 100.
2.36 0 0 100.
1.18 0 0 100.
600 0 0 100.
300 0 0 100.
150 0 0 100.
-----------------
Total = 740.
-----------------
Fineness modulus of given course aggregate is 740/100 = 7.4.
Sanu sonkamble said:
9 years ago
Thanks Kishorn & Jagadeesh.
Vinod said:
9 years ago
Fineness modulus of given course aggregate is 740/100 =7.4.
Bapureddy said:
9 years ago
I didn't get it. Please, Can anyone explain it clearly?
Naman said:
9 years ago
I didn't get this, Please explain briefly.
Sanjana said:
9 years ago
Can anyone explain last column?
Sudip said:
9 years ago
Sieve.size wt. retained(kg) %wt.retained(%) cumulative%
80 0 0 0.
40 2 10 10.
20 8 40 50.
10 6 30 80.
7.5 4 20 100.
4.75 0 0 100.
2.36 0 0 100.
1.18 0 0 100.
600 0 0 100.
300 0 0 100.
150 0 0 100.
-----------------
Total = 740.
-----------------
Fineness modulus of given course aggregate is 740/100 = 7.4.
For wt. retained in %, for 40mm sieve size, (2/20)*100% where 20 is the total wt.
80 0 0 0.
40 2 10 10.
20 8 40 50.
10 6 30 80.
7.5 4 20 100.
4.75 0 0 100.
2.36 0 0 100.
1.18 0 0 100.
600 0 0 100.
300 0 0 100.
150 0 0 100.
-----------------
Total = 740.
-----------------
Fineness modulus of given course aggregate is 740/100 = 7.4.
For wt. retained in %, for 40mm sieve size, (2/20)*100% where 20 is the total wt.
Abi said:
9 years ago
Can't understand. Please explain this.
Umang said:
9 years ago
Thanks for the explanation @Maha.
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