Respuesta :
Solution :
Sieve Size (in) Weight retain(g)
3 1.62
2 2.17
[tex]$1\frac{1}{2}$[/tex] 3.62
[tex]$\frac{3}{4}$[/tex] 2.27
[tex]$\frac{3}{8}$[/tex] 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 6.3 1.08% 100% 0%
581.9 gram
Effective size = percentage finer 10% ([tex]$D_{20}[/tex])
0.149 mm, N 100, % finer 1.08
0.297, N 50 , % finer 12.62%
x , 10%
[tex]$y-1.08 = \frac{12.62 - 1.08}{0.297 - 0.149}(x-0.149)$[/tex]
[tex]$(10-1.08) \times \frac{0.297 - 0.149}{12.62 - 1.08}+ 0.149=x$[/tex]
x = 0.2634 mm
Effective size, [tex]$D_{10} = 0.2643 \ mm$[/tex]
Now, N 16 (1.19 mm) , 51.2%
N 8 (2.38 mm) , 74.91%
x, 60%
[tex]$60-51.2 = \frac{74.91-51.2}{2.38-1.19}(x-1.19)$[/tex]
x = 1.6317 mm
[tex]$\therefore D_{60} = 1.6317 \ mm$[/tex]
Uniformity co-efficient = [tex]$\frac{D_{60}}{D_{10}}$[/tex]
[tex]$Cu= \frac{1.6317}{0.2643}$[/tex]
Cu = 6.17
Now, fineness modulus = [tex]$\frac{\Sigma \text{\ cumulative retain on all sieve }}{100}$[/tex]
[tex]$=\frac{\Sigma (10.225+25.09+48.8054+70.7165+87.39+98.92+100)}{100}$[/tex]
= 4.41
which lies between No. 4 and No. 5 sieve [4.76 to 4.00]
So, fineness modulus = 4.38 mm
