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 5 of 5.
Aamir said:
6 years ago
@Shalindra. Thanks, you explained so easily and scientific I near seen such in whole academic.
(2)
Bilal said:
5 years ago
Fineness modulus = cumulative retained/sample size.
= 148/20,
= 7.4.
= 148/20,
= 7.4.
(11)
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)
Anees ur Rehman said:
1 year ago
The correct answer for the fineness modulus is 7.4.
(1)
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