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)
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.
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.
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.
KISHORE said:
1 decade ago
% RETAINED CUMULATIVE % RETAINED
80 MM 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 740.
FINENESS MODULUS = 740/100 = 7.4.
80 MM 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 740.
FINENESS MODULUS = 740/100 = 7.4.
Maha said:
1 decade ago
2 out of 20 kg retained in that particular sieve, so for 100 how much is retained give % retained hence.
(2/20) * 100 = 10
(8/20) * 100 = 40
(6/20)* 100 = 30
(4/20)*100 = 20. And so on, if u have to find cumulative % retained u have to add the presently available value with all previous values hence first value is 10.
Second value is 10 + 40 = 50.
Third value is 10+ 40 +30 = 80.
Fourth value is 10+40+30+20 = 100 and so on.
(2/20) * 100 = 10
(8/20) * 100 = 40
(6/20)* 100 = 30
(4/20)*100 = 20. And so on, if u have to find cumulative % retained u have to add the presently available value with all previous values hence first value is 10.
Second value is 10 + 40 = 50.
Third value is 10+ 40 +30 = 80.
Fourth value is 10+40+30+20 = 100 and so on.
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)
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.
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.
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)
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