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 1 of 5.
Tetsu said:
3 years ago
Fineness modulus = (Total Cumulative % retained)/100.
Here 10 nos of sieves and total weight retained is W=(2+8+6+4) = 20kg,
Now
80mm sieve - 0kg retained.
40mm - 2kg,
20mm - 8kg,
10mm - 6kg,
4.75mm - 4kg,
2.36- 0,
1.18 - 0,
600U - 0,
300U - 0,
150U - 0,
Now wt retained on 40mm is 2kg so the cumulative of weight retained is 2kg since on the 80mm sieve 0 kg is retained.
And so,
80-0kg.
40-(2+0)kg=2kg.
20-(8+2)kg=10kg.
10mm-(8+2+6)kg=16kg.
4.75-(8+2+6+4)kg=20kg.
2.36- (20+0)kg=20kg.
1.18 - (20+0)kg=20kg.
600U - (20+0)kg=20kg.
300U - (20+0)kg=20kg.
150U - (20+0)kg=20kg.
Now cumulative %(say C%) is ({Cumulative weight of correspinding Sieve}/Total weight retained)x100.
C%=
40mm-(2/20)x100 = 10.
20mm- (10/20)x100 = 50.
10mm-(16/20)x100 = 80.
4.75mm-(20/20)x100 = 100.
2.36- (20/20)x100 = 100.
1.18 - (20/20)x100 = 100.
600U - (20/20)x100 = 100.
300U - (20/20)x100 = 100.
150U - (20/20)x100 = 100.
Now the total of cumulative % retained is (10+50+80+100+100+100+100+100+100) = 740.
Now Fineness Modulus = (Total Cumulative% retained )/100.
= 740/100 = 7.4.
Here 10 nos of sieves and total weight retained is W=(2+8+6+4) = 20kg,
Now
80mm sieve - 0kg retained.
40mm - 2kg,
20mm - 8kg,
10mm - 6kg,
4.75mm - 4kg,
2.36- 0,
1.18 - 0,
600U - 0,
300U - 0,
150U - 0,
Now wt retained on 40mm is 2kg so the cumulative of weight retained is 2kg since on the 80mm sieve 0 kg is retained.
And so,
80-0kg.
40-(2+0)kg=2kg.
20-(8+2)kg=10kg.
10mm-(8+2+6)kg=16kg.
4.75-(8+2+6+4)kg=20kg.
2.36- (20+0)kg=20kg.
1.18 - (20+0)kg=20kg.
600U - (20+0)kg=20kg.
300U - (20+0)kg=20kg.
150U - (20+0)kg=20kg.
Now cumulative %(say C%) is ({Cumulative weight of correspinding Sieve}/Total weight retained)x100.
C%=
40mm-(2/20)x100 = 10.
20mm- (10/20)x100 = 50.
10mm-(16/20)x100 = 80.
4.75mm-(20/20)x100 = 100.
2.36- (20/20)x100 = 100.
1.18 - (20/20)x100 = 100.
600U - (20/20)x100 = 100.
300U - (20/20)x100 = 100.
150U - (20/20)x100 = 100.
Now the total of cumulative % retained is (10+50+80+100+100+100+100+100+100) = 740.
Now Fineness Modulus = (Total Cumulative% retained )/100.
= 740/100 = 7.4.
(46)
Bilal said:
5 years ago
Fineness modulus = cumulative retained/sample size.
= 148/20,
= 7.4.
= 148/20,
= 7.4.
(11)
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.
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)
Aamir said:
5 years ago
@Shalindra. Thanks, you explained so easily and scientific I near seen such in whole academic.
(2)
Ravi said:
6 years ago
Thanks @Shailendra.
(2)
Anees ur Rehman said:
1 year ago
The correct answer for the fineness modulus is 7.4.
(1)
Kaisiki said:
6 years ago
There are 10 serial numbers.
No 7.6 mm sieve is there still its added by someone that's why it's addition is coming wrong
Removing calculation for 7.5 n shifting it to 4.75 you will get correct solution with 740 as total.
Thanks.
No 7.6 mm sieve is there still its added by someone that's why it's addition is coming wrong
Removing calculation for 7.5 n shifting it to 4.75 you will get correct solution with 740 as total.
Thanks.
(1)
Bhavin said:
7 years ago
As per my knowledge the answer is 7.35.
(1)
Virendra kumar said:
8 years ago
How to total 740 and what is 740? please explain.
HARSHA G said:
9 years ago
SIEVE WEIGHT RET CUMULATIVE WT CUM.%
SIZE
80 0 0 0
40 2 2 10
20 8 10 50
10 6 16 80
4.75 4 20 100
2.36 0 20 100
1.18 0 20 100
600M 0 20 100
300M 0 20 100
150M 0 20 100
ADD CUM % RET AND DIVIDE BY 100 740/100 = 7.4.
Fineness modulus is an empirical factor obtained by adding the cumulative percentages of aggregate retained on each of the standard sieves ranging from 80 mm to 150 micron and dividing this sum by 100.
CUM% = (CUM WT / TOTAL WT )* 100.
SIZE
80 0 0 0
40 2 2 10
20 8 10 50
10 6 16 80
4.75 4 20 100
2.36 0 20 100
1.18 0 20 100
600M 0 20 100
300M 0 20 100
150M 0 20 100
ADD CUM % RET AND DIVIDE BY 100 740/100 = 7.4.
Fineness modulus is an empirical factor obtained by adding the cumulative percentages of aggregate retained on each of the standard sieves ranging from 80 mm to 150 micron and dividing this sum by 100.
CUM% = (CUM WT / TOTAL WT )* 100.
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