Civil Engineering - Concrete Technology - Discussion

The correct answer for the fineness modulus is 7.4.

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.

Fineness modulus = cumulative retained/sample size.
= 148/20,
= 7.4.

@Shalindra. Thanks, you explained so easily and scientific I near seen such in whole academic.

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.

Answer Will be 7.4.

Thanks @Shailendra.

Thank you all.

Thanks @Shailendra.

As per my knowledge the answer is 7.35.