Using the following data, determine the percentage retained, cumulative percentage retained, and percent passing for each sieve. Sieve size Weight retained (g) No. 4 59.5 No. 8 86.5 No. 16 138.0 No. 30 127.8 No. 50 97.0 No. 100 66.8 Pan 6.3

Solution :

Sieve Size (in)                   Weight retain(g)

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            6.3               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%

x = 0.2634 mm

Effective size,

Now, N 16 (1.19 mm)  ,  51.2%

N 8 (2.38 mm)  ,  74.91%

x,  60%

x = 1.6317 mm

Uniformity co-efficient =

Cu = 6.17

Now, fineness modulus =

= 4.41

which lies between No. 4  and No. 5 sieve [4.76 to 4.00]

So, fineness modulus = 4.38 mm

answer:

(b)

explanation: