Chapter 02 us - final solutions
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1. The document contains information about particle size analysis of soils, including procedures, calculations, and examples. Sieve analysis data is provided for different soil samples. 2. Grain size distribution parameters like D10, D30, D60, Cu, and Cc are calculated from the sieve analysis data for each sample. Soil classifications are determined based on percent gravel, sand, silt and clay. 3. Hydraulic conductivity values are calculated for two examples using grain size distribution data and the Kozeny-Carmen equation.
Chapter 02 us - final solutions
- 1. 1 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. Chapter 2 2.1 2.63≈=== 2.625 16.0 42.0 10 60 D D Cu ; 0.66≈=== 0.656 )16.0)(42.0( 21.0 ))(( 2 1060 2 30 DD D Cc 2.2 3.0=== 27.0 81.0 10 60 D D Cu ; 0.77≈=== 0.768 )27.0)(81.0( 41.0 ))(( 2 1060 2 30 DD D Cc 2.3 a. Sieve no. Mass of soil retained on each sieve (g) Percent retained on each sieve Percent finer 4 10 20 40 60 100 200 Pan 28 42 48 128 221 86 40 24 4.54 6.81 7.78 20.75 35.82 13.94 6.48 3.89 95.46 88.65 80.88 60.13 24.31 10.37 3.89 0.00 ∑ 617 g
- 2. 2 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. b. D10 = 0.16 mm; D30 = 0.29 mm; D60 = 0.45 mm c. 2.81≈=== 2.812 16.0 45.0 10 60 D D Cu d. 1.17≈=== 1.168 )16.0)(45.0( 29.0 ))(( 2 1060 2 30 DD D Cc 2.4 a. Sieve no. Mass of soil retained on each sieve (g) Percent retained on each sieve Percent Finer 4 6 10 20 40 60 100 200 Pan 0 30 48.7 127.3 96.8 76.6 55.2 43.4 22 0.0 6.0 9.74 25.46 19.36 15.32 11.04 8.68 4.40 100.00 94.0 84.26 58.80 39.44 24.12 13.08 4.40 0.00 ∑ 500 g b. D10 = 0.13 mm; D30 = 0.3 mm; D60 = 0.9 mm
- 3. 3 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. c. 6.92≈=== 6.923 13.0 9.0 10 60 D D Cu d. 0.77≈=== 0.769 )13.0)(9.0( 3.0 ))(( 2 1060 2 30 DD D Cc 2.5 a. Sieve no. Mass of soil retained on each sieve (g) Percent retained on each sieve Percent finer 4 10 20 40 60 80 100 200 Pan 0 40 60 89 140 122 210 56 12 0.0 5.49 8.23 12.21 19.20 16.74 28.81 7.68 1.65 100.00 94.51 86.28 74.07 54.87 38.13 9.33 1.65 0.00 ∑ 729 g b. D10 = 0.17 mm; D30 = 0.18 mm; D60 = 0.28 mm c. 1.65≈=== 1.647 17.0 28.0 10 60 D D Cu
- 4. 4 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. d. 0.68=== )17.0)(28.0( 18.0 ))(( 2 1060 2 30 DD D Cc 2.6 a. Sieve no. Mass of soil retained on each sieve (g) Percent retained on each sieve Percent finer 4 6 10 20 40 60 100 200 Pan 0 0 0 9.1 249.4 179.8 22.7 15.5 23.5 0.0 0.0 0.0 1.82 49.88 35.96 4.54 3.1 4.7 100.00 100.00 100.00 98.18 48.3 12.34 7.8 4.7 0.00 ∑ 500 g b. D10 = 0.21 mm; D30 = 0.39 mm; D60 = 0.45 mm c. 2.14≈=== 2.142 21.0 45.0 10 60 D D Cu d. 1.61≈=== 1.609 )21.0)(45.0( 39.0 ))(( 2 1060 2 30 DD D Cc
- 5. 5 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2.7 a. b. Percent passing 2 mm = 100 GRAVEL: 100 – 100 = 0% Percent passing 0.06 mm = 73 SAND: 100 – 73 = 27% Percent passing 0.002 mm = 9 SILT: 73 – 9 = 64% CLAY: 9 – 0 = 9% c. Percent passing 2 mm = 100 GRAVEL: 100 – 100 = 0% Percent passing 0.05 mm = 68 SAND: 100 – 68 = 32% Percent passing 0.002 mm = 9 SILT: 68 – 9 = 59% CLAY: 9 – 0 = 9% d. Percent passing 2 mm = 100 GRAVEL: 100 – 100 = 0% Percent passing 0.075 mm = 80 SAND: 100 – 80 = 20% Percent passing 0.002 mm = 9 SILT: 80 – 9 = 71% CLAY: 9 – 0 = 9%
- 6. 6 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2.8 a. b. Percent passing 2 mm = 100 GRAVEL: 100 – 100 = 0% Percent passing 0.06 mm = 30 SAND: 100 – 30 = 70% Percent passing 0.002 mm = 5 SILT: 70 – 5 = 65% CLAY: 5 – 0 = 5% c. Percent passing 2 mm = 100 GRAVEL: 100 – 100 = 0% Percent passing 0.05 mm = 28 SAND: 100 – 28 = 72% Percent passing 0.002 mm = 5 SILT: 72 – 5 = 67% CLAY: 5 – 0 = 5% d. Percent passing 2 mm = 100 GRAVEL: 100 – 100 = 0% Percent passing 0.075 mm = 34 SAND: 100 – 34 = 66% Percent passing 0.002 mm = 5 SILT: 66 – 5 = 61% CLAY: 5 – 0 = 5%
- 7. 7 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2.9 a. 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%
- 8. 8 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. 2.10 a. b. Percent passing 2 mm = 100 GRAVEL: 100 – 100 = 0% Percent passing 0.06 mm = 65 SAND: 100 – 65 = 35% Percent passing 0.002 mm = 35 SILT: 65 – 35 = 30% CLAY: 35 – 0 = 35% c. Percent passing 2 mm = 100 GRAVEL: 100 – 100 = 0% Percent passing 0.05 mm = 62 SAND: 100 – 62 = 38% Percent passing 0.002 mm = 35 SILT: 62 – 35 = 27% CLAY: 35 – 0 = 35% d. Percent passing 2 mm = 100 GRAVEL: 100 – 100 = 0% Percent passing 0.075 mm = 70 SAND: 100 – 70 = 30% Percent passing 0.002 mm = 35 SILT: 70 – 35 = 35% CLAY: 35 – 0 = 35% 2.11 Gs = 2.7; temperature = 24°; time = 60 min; L = 9.2 cm Eq. (2.5): (min) (cm) (mm) t L KD =
- 9. 9 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. From Table 2.6 for Gs = 2.7 and temperature = 24°, K = 0.01282 mm0.005 60 9.2 01282.0 ==D 2.12 Gs = 2.75; temperature = 23°C; time = 100 min; L = 12.8 cm Eq. (2.5): (min) (cm) (mm) t L KD = From Table 2.6 for Gs = 2.75 and temperature = 23°, K = 0.01279 mm0.0046 100 12.8 01279.0 ==D CRITICAL THINKING PROBLEM 2.C.1 a. Soil A: 18.33=== 6.0 11 10 60 D D Cu ; 3.78=== )6.0)(11( 5 ))(( 2 1060 2 30 DD D Cc Soil B: 35=== 2.0 7 10 60 D D Cu ; 3.15=== )2.0)(7( 1.2 ))(( 2 1060 2 30 DD D Cc Soil C: 30=== 15.0 5.4 10 60 D D Cu ; 1.48=== )15.0)(5.4( 1 ))(( 2 1060 2 30 DD D Cc b. Soil A is coarser than Soil C. A higher percentage of soil C is finer than any given size compared to Soil A. For example, about 15% is finer than 1 mm for Soil A, whereas almost 30% is finer than 1 mm in case of soil C. c. Particle segregation may take place in aggregate stockpiles such that there is a separation of coarser and finer particles. This makes representative sampling difficult. Therefore Soils A, B, and C demonstrate quite different particle size distribution.
- 10. 10 © 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part. d. Soil A: Percent passing 4.75 mm = 29 GRAVEL: 100 – 29 = 71% Percent passing 0.075 mm = 1 SAND: 29 – 1 = 28% FINES: 1 – 0 = 1% Soil B: Percent passing 4.75 mm = 45 GRAVEL: 100 – 45 = 55% Percent passing 0.075 mm = 2 SAND: 45 – 2 = 43% FINES: 2 – 0 = 2% Soil C: Percent passing 4.75 mm = 53 GRAVEL: 100 – 53 = 47% Percent passing 0.075 mm = 3 SAND: 47 – 3 = 44% FINES: 3 – 0 = 3%