Solution :
<u>Sieve Size</u> (in) <u>Weight retain</u><u>(g)</u>
3 1.62
2 2.17
3.62
2.27
1.38
PAN 0.21
Given :
Sieve weight % wt. retain % cumulative % finer
size retained wt. retain
No. 4 59.5 10.225% 10.225% 89.775%
No. 8 86.5 14.865% 25.090% 74.91%
No. 16 138 23.7154% 48.8054% 51.2%
No. 30 127.8 21.91% 70.7154% 29.2850%
No. 50 97 16.6695% 87.3849% 12.62%
No. 100 66.8 11.4796% 98.92% 1.08%
Pan <u> 6.3 </u> 1.08% 100% 0%
581.9 gram
Effective size = percentage finer 10% (
)
0.149 mm, N 100, % finer 1.08
0.297, N 50 , % finer 12.62%
x , 10%
![$y-1.08 = \frac{12.62 - 1.08}{0.297 - 0.149}(x-0.149)$](https://tex.z-dn.net/?f=%24y-1.08%20%3D%20%5Cfrac%7B12.62%20-%201.08%7D%7B0.297%20-%200.149%7D%28x-0.149%29%24)
![$(10-1.08) \times \frac{0.297 - 0.149}{12.62 - 1.08}+ 0.149=x$](https://tex.z-dn.net/?f=%24%2810-1.08%29%20%5Ctimes%20%5Cfrac%7B0.297%20-%200.149%7D%7B12.62%20-%201.08%7D%2B%200.149%3Dx%24)
x = 0.2634 mm
Effective size, ![$D_{10} = 0.2643 \ mm$](https://tex.z-dn.net/?f=%24D_%7B10%7D%20%3D%200.2643%20%5C%20mm%24)
Now, N 16 (1.19 mm) , 51.2%
N 8 (2.38 mm) , 74.91%
x, 60%
![$60-51.2 = \frac{74.91-51.2}{2.38-1.19}(x-1.19)$](https://tex.z-dn.net/?f=%2460-51.2%20%3D%20%5Cfrac%7B74.91-51.2%7D%7B2.38-1.19%7D%28x-1.19%29%24)
x = 1.6317 mm
![$\therefore D_{60} = 1.6317 \ mm$](https://tex.z-dn.net/?f=%24%5Ctherefore%20D_%7B60%7D%20%3D%201.6317%20%5C%20mm%24)
Uniformity co-efficient =
![$Cu= \frac{1.6317}{0.2643}$](https://tex.z-dn.net/?f=%24Cu%3D%20%5Cfrac%7B1.6317%7D%7B0.2643%7D%24)
Cu = 6.17
Now, fineness modulus = ![$\frac{\Sigma \text{\ cumulative retain on all sieve }}{100}$](https://tex.z-dn.net/?f=%24%5Cfrac%7B%5CSigma%20%5Ctext%7B%5C%20cumulative%20retain%20on%20all%20sieve%20%7D%7D%7B100%7D%24)
![$=\frac{\Sigma (10.225+25.09+48.8054+70.7165+87.39+98.92+100)}{100}$](https://tex.z-dn.net/?f=%24%3D%5Cfrac%7B%5CSigma%20%2810.225%2B25.09%2B48.8054%2B70.7165%2B87.39%2B98.92%2B100%29%7D%7B100%7D%24)
= 4.41
which lies between No. 4 and No. 5 sieve [4.76 to 4.00]
So, fineness modulus = 4.38 mm