Engineering, 10th Edition Braja M. Das
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, Chapter 2
D60 0.41
2.1 Cu
= = = 5.13
D10 0.08
( D30 ) 2 (0.22) 2
=Cc = = 1.48
( D10 )( D60 ) (0.08)(0.41)
D60 1.81
2.2 Cu
= = = 7.54
D10 0.24
( D30 ) 2 (0.82) 2
=Cc = = 1.55
( D10 )( D60 ) (0.24)(1.81)
2.3 a.
Mass of soil retained Percent retained
Sieve no. on each sieve (g) on each sieve Percent finer
4 0.0 0.0 100.0
10 18.5 4.4 95.6
20 53.2 12.6 83.0
40 90.5 21.5 61.5
60 81.8 19.4 42.1
100 92.2 21.9 20.2
200 58.5 13.9 6.3
Pan 26.5 6.3 0
Σ 421.2 g
The grain-size distribution is shown in the figure.
1
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, b. D60 = 0.4 mm; D30 = 0.2 mm; D10 = 0.095 mm
D60 0.4
Cu
c. = = = 4.21
D10 0.095
( D30 ) 2 (0.2) 2
d. Cc
= = = 1.05
( D10 )( D60 ) (0.4)(0.095)
2.4 a.
Mass of soil retained Percent retained Percent
Sieve no. on each sieve (g) on each sieve finer
4 0.0 0.0 100
6 30 6.0 94.0
10 48.7 9.74 84.26
20 127.3 25.46 58.80
40 96.8 19.36 39.44
60 76.6 15.32 24.12
100 55.2 11.04 13.08
200 43.4 8.68 4.40
Pan 22 4.40 0
Σ 500 g
The grain-size distribution is shown in the figure.
2
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, b. D10 = 0.13 mm; D30 = 0.3 mm; D60 = 0.9 mm
D60 0.9
Cu
c. = = = 6.923 ≈ 6.92
D10 0.13
D302 0.32
Cc
d.= = = 0.769 ≈ 0.77
( D60 )( D10 ) (0.9)(0.13)
2.5 a.
Mass retained Percent retained
Sieve no. (g) on each sieve Percent finer
4 28 4.54 95.46
10 42 6.81 88.65
20 48 7.78 80.87
40 128 20.75 60.12
60 221 35.82 24.3
100 86 13.94 10.36
200 40 6.48 3.88
Pan 24 3.88 0
Σ 617 g
The graph for percent finer versus grain size is shown.
3
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, b. From the graph, D10 = 0.14 mm, D30 = 0.27 mm, D60 = 0.42 mm
D60 0.42
Cu
c. = = = 3
D10 0.14
( D30 ) 2 (0.27) 2
d. Cc
= = = 1.24
( D60 )( D10 ) (0.42)(0.14)
2.6 a.
Mass of soil retained Percent retained
Sieve no. on each sieve (g) on each sieve Percent finer
The grain-size distribution is shown in the figure.
4 0 0.0 100
6 0 0.0 100
10 0 0.0 100
20 9.1 1.82 98.18
40 249.4 49.88 48.3
60 179.8 35.96 12.34
100 22.7 4.54 7.8
200 15.5 3.1 4.7
Pan 23.5 4.7 0
Σ 500 g
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, b. D10 = 0.21 mm; D30 = 0.39 mm; D60 = 0.45 mm
D60 0.45
Cu
c. = = = 2.142 ≈ 2.14
D10 0.21
D302 0.392
Cc
d.= = = 1.609 ≈ 1.61
( D60 )( D10 ) (0.45)(0.21)
2.7 a. The grain-size distribution curve is shown in the figure.
5
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, b. Percent passing 2 mm = 100 GRAVEL: 100 − 100 = 0%
Percent passing 0.06 mm = 84 SAND: 100 − 84 = 16%
Percent passing 0.002 mm = 11 SILT: 84 − 11 = 73%
CLAY: 11 − 0 = 11%
c. Percent passing 2 mm = 100 GRAVEL: 100 − 100 = 0%
Percent passing 0.05 mm = 80 SAND: 100 − 80 = 20%
Percent passing 0.002 mm = 11 SILT: 80 − 11 = 69%
CLAY: 11 − 0 = 11%
d. Percent passing 2 mm = 100 GRAVEL: 100 − 100 = 0%
Percent passing 0.075 mm = 90 SAND: 100 − 90 = 10%
Percent passing 0.002 mm = 11 SILT: 90 − 11 = 79%
CLAY: 11 − 0 = 11%
2.8 The grain-size distribution curve is shown in the figure.
a. Percent passing 2 mm = 100 GRAVEL: 100 − 100 = 0%
Percent passing 0.05 mm = 94 SAND: 100 − 94·= 6%
Percent passing 0.002 mm = 42 SILT: 94 − 42 = 52%
CLAY: 42 − 0 = 42%
b. Percent passing 2 mm = 100 GRAVEL: 100 − 100 = 0%
Percent passing 0.075 mm = 97 SAND: 100 − 97 = 3%
Percent passing 0.002 mm = 42 SILT: 97 − 42 = 55%
CLAY: 42 − 0 = 42%
6
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, 2.9 a. The grain-size distribution curve is shown below.
b. Percent passing 2 mm = 100 GRAVEL: 100 − 100 = 0%
Percent passing 0.06 mm = 84 SAND: 100 − 84 = 16%
Percent passing 0.002 mm = 28 SILT: 84 − 28 = 56%
CLAY: 28 − 0 = 28%
c. Percent passing 2 mm = 100 GRAVEL: 100 − 100 = 0%
Percent passing 0.05 mm = 83 SAND: 100 − 83 = 17%
Percent passing 0.002 mm = 28 SILT: 83 − 28 = 55%
CLAY: 28 − 0 = 28%
d. Percent passing 2 mm = 100 GRAVEL: 100 − 100 = 0%
Percent passing 0.075 mm = 90 SAND: 100 − 90 = 10%
Percent passing 0.002 mm = 28 SILT: 90 − 28 = 62%
CLAY: 28 − 0 = 28%
2.10 Gs = 2.60; temperature = 24°; R = 43; time = 60 min. Referring to Table 2.10, L = 9.2.
L(cm)
Eq. (2.6): D (mm) = K
t (min)
From Table 2.9 for Gs = 2.60 and temperature = 24°, K = 0.01321.
9.2
D = 0.01321 = 0.0052 mm
60
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, 2.11 For Gs = 2.70 and temperature = 23°, K = 0.01297 (Table 2.9);
R = 25, L = 12.2 (Table 2.10).
L(cm) 12.2
D(mm) K= 0.01297
= = 0.0041 mm
t (min) 120
2.12 a. Total mass in the ternary mix = 8000 × 3 =24,000 kg
8000
Percent of each soil in the mix = ×100 =
33.33%
24, 000
Mass of each soil used in the sieve analysis, ΣmA =
ΣmB = 500 g
ΣmC =
If a sieve analysis is conducted on the ternary mix using the same set of sieves, the percent of mass
retained on each sieve, mM (%), can be computed as follows:
m m m
(%) 0.333 A ×100 + 0.333 B ×100 + 0.333 C ×100
mM=
500 500 500
The calculated values are shown in the following table.
Sieve size Mass retained Percent passing
(mm) for the mixture
mA (g) mB (g) mC (g) mM (%)
25.0 0.0 0 0 0.0 100
19.0 60 10 30 6.66 93.34
12.7 130 75 75 18.65 74.69
9.5 65 80 45 12.65 62.04
4.75 100 165 90 23.64 38.4
2.36 50 25 65 9.32 29.08
0.6 40 60 75 11.65 17.43
0.075 50 70 105 14.98 2.45
Pan 5 15 15 2.33 ≈0
b. The grain-size distribution curve for the mixture is drawn below.
8
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