Question

The following are the results of a sieve analysis. U.S. sieve no. Mass of soil retained (g) 4 0 10 18.5 20 53.2 40 90.5 60 81.8 100 92.2 200 58.5 Pan 26.5 a. Determine the percent ner than each sieve and plot a grain-size distribution curve. b. Determine D10, D30, and D60 for each soil. c. Calculate the uniformity coefcient Cu. d. Calculate the coefcient of gradation Cc

Answer 1

a.)

US Sieve no. % finer (C₅ )

4 100

10 95.61

20 82.98

40 61.50

60 42.08

100 20.19

200 6.3

Pan 0

b.) D10 = 0.12, D30 = 0.22, and D60 = 0.4

c.) Cu = 3.33

d.) Cc = 1

Step-by-step explanation:

As given ,

US Sieve no. Mass of soil retained (C₂ )

4 0

10 18.5

20 53.2

40 90.5

60 81.8

100 92.2

200 58.5

Pan 26.5

Now,

Total weight of the soil = w = 0 + 18.5 + 53.2 + 90.5 + 81.8 + 92.2 + 58.5 + 26.5 = 421.2 g

⇒ w = 421.2 g

As we know that ,

% Retained = C₃ = C₂×
`(100)/(w)`

∴ we get

US Sieve no. % retained (C₃ ) Cummulative % retained (C₄)

4 0 0

10 4.39 4.39

20 12.63 17.02

40 21.48 38.50

60 19.42 57.92

100 21.89 79.81

200 13.89 93.70

Pan 6.30 100

Now,

% finer = C₅ = 100 - C₄

∴ we get

US Sieve no. Cummulative % retained (C₄) % finer (C₅ )

4 0 100

10 4.39 95.61

20 17.02 82.98

40 38.50 61.50

60 57.92 42.08

100 79.81 20.19

200 93.70 6.3

Pan 100 0

The grain-size distribution is :

b.)

From the diagram , we can see that

D10 = 0.12

D30 = 0.22

D60 = 0.12

c.)

Uniformity Coefficient = Cu =
`(D60)/(D10)`

⇒ Cu =
`(0.4)/(0.12) = 3.33`

d.)

Coefficient of Graduation = Cc =
`(D30^(2))/(D10 . D60)`

⇒ Cc =
`(0.22^(2))/((0.4) . (0.12))` = 1