164996943 an Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual

An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual Click here to download immediately!!!http://solutionsmanualtestbanks.blogspot.com/2011/10/introduction-togeotechnical.html Name: An Introduction to Geotechnical Engineering Author: Holtz Kovacs Edition: 2nd ISBN-10: 0132496348 Type: Solutions Manual - The file contains solutions and questions to all chapters and all questions. All the files are carefully checked and accuracy is ensured. This is a sample chapter CHAPTER 2 INDEX AND CLASSIFICATION PROPERTIES OF SOILS 2-1. From memory, draw a phase diagram (like Fig. 2.2, but don’t look first!). The “phases” have a Volume side and Mass side. Label all the parts. SOLUTION: Refer to Figure 2.2. 2-2. From memory, write out the definitions for water content, void ratio, dry density, wet or moist density, and saturated density. SOLUTION: Refer to Section 2.2. 2-3. Assuming a value of ρs = 2.7 Mg/m , take the range of saturated density in Table 2.1 for the six soil types and calculate/estimate the range in void ratios that one might expect for these soils. 3 SOLUTION: Create a spreadsheet using input values from Table 2.1 and Eq. 2.18. (Given) (see Eq. 2.18) ρ' - min (Mg/m ) 3 ρ' - max (Mg/m ) 3 emax 0.89 3.25 0.55 0.89 emin 0.21 0.55 0.21 0.42 16.00 1.13 0.9 0.4 1.1 0.9 0.0 0.3 1.4 1.1 1.4 1.2 0.1 0.8 ∞ 4.67 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2-4. Prepare a spreadsheet plot of dry density in 3 Mg/m as the ordinate versus water content in percent as the abscissa. Assume ρs = 2.65 Mg/m and vary the degree of saturation, S, from 100% to 40% in 10% increments. A maximum of 50% water content should be adequate. SOLUTION: Solve Eq. 2.12 and Eq. 2.15 for ρd = f(ρs, w, S, Gs), or use Eq. 5.1. ρ dry = ρwS w+ ρw ρs S 3 S= w (%) 100 ρdry (Mg/m3) 90 ρdry (Mg/m3) 80 ρdry (Mg/m3) 70 ρdry (Mg/m3) 60 ρdry (Mg/m3) 50 ρdry (Mg/m3) 40 ρdry (Mg/m3) 30 ρdry (Mg/m3) 20 ρdry (Mg/m3) 10 ρdry (Mg/m3) 0.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0 45.0 50.0 2.65 2.34 2.09 1.90 1.73 1.59 1.48 1.37 1.29 1.21 1.14 2.65 2.31 2.05 1.84 1.67 1.53 1.41 1.31 1.22 1.14 1.07 2.65 2.27 1.99 1.77 1.59 1.45 1.33 1.23 1.14 1.06 1.00 2.65 2.23 1.92 1.69 1.51 1.36 1.24 1.14 1.05 0.98 0.92 2.65 2.17 1.84 1.59 1.41 1.26 1.14 1.04 0.96 0.89 0.83 2.65 2.09 1.73 1.48 1.29 1.14 1.02 0.93 0.85 0.78 0.73 2.65 1.99 1.59 1.33 1.14 1.00 0.89 0.80 0.73 0.67 0.61 2.65 1.84 1.41 1.14 0.96 0.83 0.73 0.65 0.58 0.53 0.49 2.65 1.59 1.14 0.89 0.73 0.61 0.53 0.47 0.42 0.38 0.35 2.65 1.14 0.73 0.53 0.42 0.35 0.30 0.26 0.23 0.21 0.19 Problem 2-4 3.00 2.50 ρdry (Mg/m3) 2.00 1.50 S=100% S=90% S=80% S=70% S=60% S=50% S=40% S=30% S=20% S=10% 1.00 0.50 0.00 0 5 10 15 20 25 30 w (%) 35 40 45 50 55 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2-5a Prepare a graph like that in Problem 2.4, only use dry density units of kN/m and pounds per cubic feet. SOLUTION: From Eq. 2.12 and Eq. 2.15: ρ = dry ρwS ρ w+ w ρs S= w (%) 3 S 100 γdry (kN/m3) 90 γdry (kN/m3) 80 γdry (kN/m3) 70 γdry (kN/m3) 60 γdry (kN/m3) 50 γdry (kN/m3) 40 γdry (kN/m3) 30 γdry (kN/m3) 20 γdry (kN/m3) 10 γdry (kN/m3) 0.0 5.0 10.0 15.0 20.0 25.0 30.0 35.0 40.0 45.0 50.0 26.00 22.95 20.55 18.60 16.99 15.64 14.48 13.49 12.62 11.86 11.18 26.00 22.66 20.08 18.03 16.36 14.97 13.80 12.80 11.94 11.18 10.52 26.00 22.30 19.53 17.37 15.64 14.22 13.04 12.04 11.18 10.44 9.79 26.00 21.86 18.86 16.58 14.79 13.36 12.17 11.18 10.34 9.62 8.99 26.00 21.29 18.03 15.64 13.80 12.35 11.18 10.21 9.40 8.70 8.10 26.00 20.55 16.99 14.48 12.62 11.18 10.04 9.11 8.33 7.68 7.12 26.00 19.53 15.64 13.04 11.18 9.79 8.70 7.83 7.12 6.53 6.03 26.00 18.03 13.80 11.18 9.40 8.10 7.12 6.35 5.73 5.23 4.80 26.00 15.64 11.18 8.70 7.12 6.03 5.23 4.61 4.13 3.73 3.41 26.00 11.18 7.12 5.23 4.13 3.41 2.90 2.53 2.24 2.01 1.82 Problem 2-5a 30.00 S=100% S=90% S=80% S=70% S=60% S=50% S=40% S=30% S=20% S=10% 25.00 20.00 γ dry (kN/m3) 15.00 10.00 5.00 0.00 0 5 10 15 20 25 30 35 40 45 50 55 w (%) An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual only use dry density units of pounds per cubic feet.12 165.41 36.36 135.83 45.44 75.36 99.36 139.16 165.21 99.09 14.00 0.54 165.70 104.34 165.53 S=100% S=90% S=80% S=70% S=60% S=50% S=40% S=30% S=20% S=10% 165.12 63.07 95.0 50.70 99.27 75.00 160.36 146.27 71.95 105.93 71.95 59.54 45.30 38.26 12.00 60.00 40.12 64.77 55.0 15.46 87.46 71.30 33.00 20.77 61.12 66.24 30.24 29.0 30.46 82.30 40.12 80.0 5.36 144.72 108.42 71.53 38.36 124.08 92.0 35.12 85.16 57.12 65.94 76.59 71.25 23.12 59.92 53.25 21.12 45.89 165.46 90.25 55.47 94.33 26.43 71.00 γ dry (pcf) 80.60 Problem 2-5b 180.48 33.14 127.00 120.0 10.86 124.12 55.45 114.77 51.30 41.00 0 5 10 15 20 25 30 35 40 45 50 55 w (%) An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .80 81.4.35 49.94 71.45 82.85 45.34 33.24 26.70 87.0 40.75 114.0 20.79 80.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2-5b Prepare a graph like that in Problem 2.85 57.25 165.35 45.33 108.47 99.0 165.00 140.69 165.30 165.12 66.00 48.96 77.01 130.36 141.48 16. SOLUTION: γ dry = Gs γ w S S + Gs w S= w (%) 100 γdry (pcf) 90 γdry (pcf) 80 γdry (pcf) 70 γdry (pcf) 60 γdry (pcf) 50 γdry (pcf) 40 γdry (pcf) 30 γdry (pcf) 20 γdry (pcf) 10 γdry (pcf) 0.79 11.39 62.00 100.0 25.36 71.58 71.75 21.69 18.35 51.46 92.25 87.36 114.80 78.72 118.08 99.21 110.04 119.0 45.80 71.12 62.11 84.36 130. 2.6.52 1.0 10.38 2. SOLUTION: From Eq.0 5.73 1.0 30.00 0 5 10 15 20 25 30 w (%) 35 40 45 50 55 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .59 1.65 2.23 1.00 ρs (Mg/m3) 1.21 1.75 ρdry (Mg/m ) 3 2.63 1.79 1.92 1.77 1.41 1.29 1.00 2.50 2.0 3.0 35.30 2.51 1.4.71 1.90 1.95 1. Prepare a graph like that in Problem 2.70 2.19 1.16 2.49 1.0 20.50 0.60 to 2.16 1.12 and Eq.13 2.75 2.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.37 1.39 1.36 1.6 ρdry (Mg/m ) 3 2.0 25.0 45.42 2.65 ρdry (Mg/m ) 3 2. ρs = w (%) 2.48 1.61 1.34 2.15 2.13 1. You decide the size of the increments you need to “satisfactorily” evaluate the relationship as ρs varies.17 2. Prepare a concluding statement of your observations.30 1.31 1.14 2.60 2.27 1.8 ρdry (Mg/m ) 3 0.0 50.50 1.40 1.20 1.46 2.0 15.15: ρ = dry ρwS w+ ρw ρs S Note: The relationship between ρdry and w is not overly sensitive to ρs.80 2. 2.06 1.97 1.75 1.65 1.87 1.0 40.80 Mg/m .46 1.22 1.24 1.32 1.7 ρdry (Mg/m ) 3 2.09 1.58 1.00 0. only for S = 100% and vary the density of solids 3 from 2. 2 kg / m3 = dry 3 (1 + w) ⎛ ρ ⎞ = ⎜1 − w ⎟ ρ (1 + 0.0% Solving for w: w = ρ S w = (1Mg / m )(100%) ⎜ 1.87 Mg/m and the density of the solids is 2. Note: S = 100% S 3 ρs ⎛ 1 − 1 ⎞ = 16. What is the water content of the material when saturated? SOLUTION: From Eq. 2-15: ρdry = ρw S ρ w+ ⎛ 1 1⎞ − ⎟ ρ ⎜ ⎝ ρ ⎟ ⎠ ⎝ w 3 .19 = 2729.6 kg / m s An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .24 → M = 0.6 kg / m3 b) Solve using phase diagram relationships: assume Vt = 1.2 + 1000 − 2045) = 2729.67Mg / m3 ⎟ dry s ⎠ 2. What is the density of the solids? What is the dry density of the soil? SOLUTION: a) Solve using equations or phase diagrams: ρt ρ = 2045 = 1649.24M Ms w s Mt = Mw + Ms → 2045 = 0. The dry density of a compacted sand is 1.0 Mt = 2045 kg Mw = 0.24Ms M = 0. 2-12 and Eq.8.3958 m3 ρ = 1000 = w → V = w w 1000 Vw Ms 3 ρ = = 1649.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.2) (1649.87 Mg / m 3 ⎜ 2.67 3 Mg/m .7. A soil that is completely saturated has a total density of 2045 kg/m and a water content of 24%.19 kg 0.24Ms + Ms → Ms = 1649.24) +ρ dry w ρ sat ⎝ ρs ⎠ ρw ρdry ρ = s ρdry + ρ w − ρsat = (1000)(1649. An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual Vs 1 − 0.19 M = 1649.2 kg / m3 = s = dry ρ Vt 1 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .3958 1649. An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.9 What is the water content of a fully saturated soil with a dry density of 1.72 Mg/m ? Assume ρs = 2.72 Mg/m3. SOLUTION: From Eq. 2-12 and Eq. 2-15: ρdry = ρw S ρ w+ ⎛ 1 1⎞ − ⎟ ρ ⎜ ⎝ ρ ⎟ ⎠ ⎝ w 3 ; Note: S = 100% S 3 ρs ⎛ 1 − 1 ⎞ = 21.4% Solving for w: w = ρ S w = (1Mg / m )(100%) ⎜ 3 ⎜ 1.72 Mg / m 2.72 Mg / m3 ⎟ dry s ⎠ 2.10. A dry quartz sand has a density of 1.68 Mg/m . Determine its density when the degree of 3 saturation is 75%. The density of solids for quartz is 2.65 Mg/m . SOLUTION: Recognize that ρdry (initial) = ρdry (final); S( final) = 75% From Eq. 2-12 and Eq. 2-15: ρdry = ρw S ρ w+ ⎛ 1 1⎞ − ⎟ ρ ρ w 3 ρs S 3 ⎛ 1 − 1 ⎞ = 16.34% Solving for w: w = ρ S w = (1Mg / m )(75%) ⎜ 3 ⎜ 2.65 Mg / m3 ⎟ 1.68 Mg / m ⎠ ⎜ ⎟ s ⎠ ⎝ ⎝ dry 3 final ρ t = ρdry (1 + w) = 1.68(1 + 0.1634) = 1.95 Mg / m 2.11. The dry density of a soil is 1.60 Mg/m and the solids have a density of 2.65 Mg/m . Find the (a) water content, (b) void ratio, and (c) total density when the soil is saturated. SOLUTION: Given: S = 100% From Eq. 2-12 and Eq. 2-15: ρdry = ρw S ρ w+ ⎛ 1 w 3 3 ρs S (a) Solving for w: w =ρ S An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 1⎞ − ⎛ 1 − w = (1Mg / m )(100%) 3 1 ⎞ = 24.76% ⎜ ρ ρ ⎟ ⎜ 3 2.65 Mg / m3 ⎟ 1.60 Mg / m ⎝ ⎠ ⎜ ⎟ ⎝ dry s ⎠ wρ (24.76)(2.65) (b) From Eq. 2.15: e = s = (100)(1.0) = 0.656 Sρ w (c) ρ t = ρdry (1 + w) = 1.60(1 + 0.2476) = 1.996 = 2.00 Mg / m3 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.12. A natural deposit of soil is found to have a water content of 20% and to be 90% saturated. What is the void ratio of this soil? SOLUTION: w = 20% and S = 90%; assume Gs = 2.70 From Eq. 2.15: e = wρ s Sρ w = (20.0)(2.70) = 0.60 (90)(1.0) 2.13. A chunk of soil has a wet weight of 62 lb and a volume of 0.56 ft . When dried in an oven, the soil weighs 50 lb. If the specific gravity of solids Gs = 2.64, determine the water content, wet unit weight, dry unit weight, and void ratio of the soil. SOLUTION: Solve using phase diagram relationships. (a) Ww = Wt − Ws = 62 − 50 = 12 lb W 12(100) w = w × 100% = = 24.0% 50 Ws 62 = 110.7 pcf (b) γ = Wt = t 3 (c) γ dry Vt 0.56 Ws 50 = 89.29 pcf = = Vt 0.56 (d) Vw = Ww = 12 = 0.1923 ft 3 62.4 γw Vs = Ws 50 = 0.3035 ft 3 Gs γ w = (2.64)(62.4) Vv = Vt − Vv = 0.56 − 0.3035 = 0.2565 V 0.2565 e= v = = 0.8451 = 0.84 Vs 0.3035 2.14. In the lab, a container of saturated soil had a mass of 113.27 g before it was placed in the oven and 100.06 g after the soil had dried. The container alone had a mass of 49.31 g. The specific gravity of solids is 2.80. Determine the void ratio and water content of the original soil sample. SOLUTION: Solve using phase diagram relationships. Ms = 100.06 − 49.31 = 50.75 g Mw = 113.27 − 100.06 = 13.21 g An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 03) (b) From Eq.73 ρw S (1)(100) Ms Mw × 100% = 13.21(100) An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . 2.7288 = 0.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual (a) w = = 26.0% 50.03 = 26.80(26.15: e= s = = 0.75 ρ w 2. 286 = 89. C.857 g Mw = (0. Compute how many grams of water must be added to each 1000 g of soil (in its natural state) in order to increase the water content to 22.429 − 107.12 = w M s → M = (0. (Prof.12Ms = 1000 g Ms = 892.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.27 m of entrapped air.65)(1000 s s s s s t v kg m3 v t )(0.27 m = 370.857) = 107. C.143 g Target state (Note: Ms does not change between natural state and target state) Mw = w × Ms = (0.) SOLUTION: Vt = 1 m3 = 1. how much force is required to prevent it from sinking? Assume that a weightless membrane surrounds the specimen. SOLUTION: Natural state M w = 0.0%.60 m3 V = V − V = 1. Ladd.857) = 196.0 = ρt − ρoil = 1060 − 920 = 140 kg m 3 γ buoy = ρbuoy × g = (140)(9. It has been calculated that the maximum density for the soil will be obtained when the water content reaches 22.143 = 89. The natural water content of a sample taken from a soil deposit was found to be 12.92 g/cm .4 m3 N m3 × 0.0%.65) with a porosity of 60% is immersed in an oil 3 3 bath having a density of 0.40 m3 M = G × ρ × V = (2. A cubic meter of dry quartz sand (Gs = 2. If the sand contains 0.6)(1.15.12)M w s Mt = Ms + Mw = Ms + 0.429 g additional water necessary = 196.16.4 N Force = γ buoy × (entrapped air) = 1373.0) = 0.0 − 0.29 g 2.000.60 = 0.40 m3 ) = 1060 kg = M t 1060 M = 1060 kg ρ = t = t m 3 ρbuoy Vt 1.8 N 3 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .0%.22)(892.12)(892.000 cm3 V = n × V = (0.12Ms = 1.81) = 1373. 7718 (d) n = × 100 = 43.15. its compacted void ratio is 0.00822 = 0. SOLUTION: (a) ρt = 25. and degree of saturation for the moist soil.69 Find the density.15 + 1) = 124.803. A sample of moist soil was found to have the following characteristics: 3 Total volume: 0.10 = 0.73 + 1 Borrow Pit V s(borr ) =V s(emb) = Vt e +1 + 1) = (57. unit weight.006344 m3 0.000 m of soil is needed in its compacted state.00364 w 1000 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . porosity. The soil will be used 3 for a highway project where a total of 100.342.74 = 1767.64 3.34 kN m 3 3 (c) Vs = Ms 22.81) = 17.47)(1.18.000 m3 100.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.64 V = = 0.7718 = 0.00822 0.01456 m Total mass: 25.7718 (e) Mw = 25.857)(9.69)(1000) Vv = 0.10 kg Specific gravity of solids: 2.77 0.000 V = = 57.00822 m3 Gsρ w = (2. void ratio.6% 1 + 0.74 − 22.01456 − 0. A soil sample taken from a borrow pit has a natural void ratio of 1.56 = 43.68 N m = 17.01456 3 (b) γ t = ρ t × g = (1767. How much volume has to be excavated from the borrow pit to meet the job requirements? SOLUTION: V V = v = Vt − Vs → V = Vt s e e s e +1 Embankment Vt = 100.47 m3 s(emb) 0.17.803.006344 e= = 0.857 = 1768 kg m 0.73.277 m3 V t(borr ) = V × (e s borr 2.74 kg Mass after oven drying: 22.10 = 3. 00364 × 100 = 57.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 0.006344 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .377 = 57.4% S = 0. 4 lb ft ) (1 ft 0. t γ = t Mt V t = 3 453.45 kN m 3 2.288)(9.741 in ) (1 ft w= Mw Ms × 100% = 12 in 55.54 lb Oven temperature 110°C Drying time 24 hours Specific gravity of solids 2.95 lb Wt. SOLUTION: 777 g 1lb 2 ( ) ( ) (a) V = π(2.2 = 55.2 grams after drying.5 feet. (a) Calculate the wet density in pounds per cubic feet.6 g = 78. before drying in oven 2. 3 using the correct number of significant figures. Compute the water content.654) = 47. A cylindrical soil specimen is tested in the laboratory.542 in3 Gs γ w = (2.288 3 kg N m3 m3 = 0.65 What is the degree of saturation of this specimen? SOLUTION: Vt = Vs = π(3)2 4 × 6 = 42.02454 ft 3 Ws 2.81) = 7448.3048 m) ( 0.4 (1 + 0.54 = 0.4536 kg1 lb) = 759. A gray silty clay (CL) is sampled from a depth of 12.83 inches and weighed 777 grams.6 = 7.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.4% 85. The “moist” soil was extruded from a 6-inch-high brass liner with an inside diameter of 2.38 = 65. after drying in oven 2.65)(62. (b) A small chunk of the original sample had a wet weight of 140.2 (c) γ dry = γt (1 + w) 3 = 78.19.7 g.429 = 78. w t s = 65. (c) Compute the dry density in Mg/m and the dry 3 unit weight in kN/m .4) An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .4 pcf 4 ( 37.759 Mg m3 γ dry = (759.7 ) 3 (b) M = M − M = 140.9 grams and weighed 85.40 pcf ρdry = ( 47.83) × 6 = 37.9 − 85.741 in3 .4115 in3 = 0.20. The following properties were obtained: Sample diameter 3 inches Sample length 6 inches Wt.01536 ft 3 = 26. 00657 ft3 = 11.869 in = 0.4115 − 26.354 15.95 − 2.41 Vw = w = = 0.354 in3 62.5% An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 3 3 V = V − V = 42.542 = 15.54 = 0.41 lb W 0.869 × 100 = 71.009184 ft v t s Ww = Wt − Ws = 2.4 γw S= Vw Vv × 100% = 11. The sample is compressed to a 2-cm thickness without a change in diameter.527 cm3 v t e final = 73.797 cm3 = V M = ρ × V = (1 g w w w cm3 v t v s s s s s w )(112.594 × 100% = 50. Its void ratio in this state is 1.350 → V = 83. (b) Find the void ratio after compression and the change in water content that occurred from initial to final state.6 = 17. SOLUTION: (a) Vt = S = 1= π(10)2 4 × 2.6% 225. M = 225.350 cm3 Vw Vv →V =V w v Vv = e × Vs = 1. (a) Find the density of the silt sample.527 g.35.527 cm3 . in prior to being compressed.594 g cm3 M = G × V × ρ = (2.797 g ) = 225.594 = 338.527 s = 0.0% (c) w initial = final conditions V = V = 73.553)(1 g s s s w Mt = 112.797 + 225.527 × 100% = 32.553 cm3 V = (1.08 − 83.553) = 112.594 g (no change) w v w s w final = 73.35V = 196.0 − 32.35Vs V = V + V = 1.797 225.0 = 157.594 Δw = 50.35V + V = 2.08 cm3 4 Vs = 83.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.35)(83.5 = 196.70.553 = 73.553 112. and the specific gravity of solids is 2. M = 73.391 = 1.553 cm3 (no change) V = V − V = 157.4% An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .391 g M = 1723 kg ρ = t = 338.5 cm thick.70)(83.797) = 112.35 cm3 m3 (b) V t−2 = π(10)2 × 2.723 g t Vt 196.21 A sample of saturated silt is 10 cm in diameter and 2.88 83. total volume Vt =80 3 cm . water content w = 20%.667 g.70 = 49.667 cm3 w w Vv = Vw (when S = 100%) V = V + V = 49.053 = 3.383 = 30.053 cm3 t s v ΔV = 80 − 76.20)(133.95 cm3 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . A sample of sand has the following properties: total mass Mt = 160 g.383 + 26.383 cm .617 × 100% = 87. V = 26.10% Desired condition: S = 100% Vt changes.70. How much would the sample volume have to change to get 100% saturation.20Ms = 160 g Ms = 133.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2. but Vs and Ms remain the same M = 26. w V = (1 g w cm3 3 )M = 26.33) = 26.667 g.667 = 76.667 cm3 w V = s Ms Gs ρ w = 133.22.617 cm3 v S = i 26.33 2.20)Ms Mt = Ms + 0. specific gravity of solids G s =2. V = 80 − 49.33 g M = (0.667 30. assuming the sample mass Mt stayed the same? SOLUTION: Mw = w × Ms = (0. 2 32. The dry weight of the specimen is 128 g.2 lbm ft3 (see Appendix A) ⎜ ⎟ ⎝16. Give the answers to parts (e) and (f) in both SI and British engineering units. (c) porosity.281 = 13.8 mm3 e= 32.238. and the density of the 3 soil solids is 2.238. SOLUTION: (a) Mw 145 − 128 = 17 g M w = w × 100% = 17 × 100 = 13.13281) × m3 kg = 1600.9 lbm ft 3 (see Appendix A) ⎜ ⎟ ⎝16.2 = 32.0 (1 + 0.67 47.23.000 − 47. (d) degree of saturation.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.000 mm3 17.8 = 0.8125 g = 1812.68)(1) Vv = 80.761.5 kg t m3 × ⎛ 1 ⎞ = 113. 238.7612 cm3 = 47.000 mm and weighs 145 g.5 cm3 kg m3 ρ = 1812. (b) void ratio.238.2 mm3 Gs ρ w = (2.68 Mg/m .8 80.8 × 100% = 52.3% 128 Ms (b) Vs = Ms 128 = 47.018 ⎠ (f ) ρdry = ρ dry 1812.5 = 1600.761.7% (e) ρ = t (c) n = × 100% = 40.3% 145 80 = 1. Find the: (a) water content.0 kg ⎛ 1 ⎞ m3 = 99. and (f) dry density.018 ⎠ An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . Draw a phase diagram and begin to fill in the blanks: A soil specimen has total volume of 3 80.761.000 (d) Vw = Mw × ρ w = (17)(1) = 17 cm 3 = 17.000 S = 32. (e) wet density. 40 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . The container weighs 258. The weight of the wet soil plus container was 18.63 g.7 = 138.9 = Mw + Ms = 0. SOLUTION: Ms = 15.351 = 46.46 − 15.034)(130.7 g. Calculate the water content of the sample.03 g.6 g when the initial water content is 6. of Interior (1990). SOLUTION: Mt = 397.43 g M 3. Dept.03 − 7. How much water needs to be added to the original specimen if the water content is to be increased by 3.63 = 7.668 g ΔM w Δw = = 0.668) = 4.03 = 3.46 g. A sample of soil plus container weighs 397.4%? After U.034 Ms ΔMw = Δw × Ms = (0.3%.S.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2. and the weight of the dry soil plus container was 15.6 − 258.25.9 g Mw = 0. A water-content test was made on a sample of clayey silt.063Ms Ms = 130. Weight of the empty container was 7.40 g Mw = 18.3% Ms 7.24.063Ms Mt = 138.43(100) (a) w = w × 100% = = 46.44 g 2.063Ms + Ms = 1. 14 16.22 215.30 16.78 23. The oven-dried water content is 23.0 Water Content (%) 24. SOLUTION: Mt = 231. evaluate the water content and draw conclusions.19 215.32 = 0.48 215.05 23.72 215.0 19. GIVEN Time in Oven (min) 0 3 1 1 1 1 1 1 1 Total Oven Time (min) 0 3 4 5 6 7 8 9 10 Mass of Soil + Dish (g) 231.0 23.237)Ms 85. A soil sample is dried in a microwave oven to determine its water content.0 0 1 2 3 4 5 6 7 8 9 10 11 Time in Oven (min) An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . of Interior (1990).40 15.22 215. From the data below.63 23.32 215.82 25.40 16.87 15.33 23.9733 CONCLUSION: The loss of additional water in the soil sample becomes negligible after 8 to 10 minutes in the microwave oven used in the experiment.62 217.82 23.19 CALCULATED Mass of Water (g) -13.0 20.9733 g (this value is constant throughout drying period) column 5 = 231.30 grams.11 22.62 − 146. Dept. The mass of the dish is 146.3 = 85.43 Water Content (%) -20.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.40 23.32 g Mw = (0.237Ms + Ms Ms = 68.0 21.43 16.S.26. After U.90 16.7%.0 22.75 216.62 −column 3 × 100% 68. 96 = (2. find the water content. Compute the water content of the sample.30.76(100) (a) w = w × 100% = = 98.5% Ms 4. The volume of solids Vs is 0.84 = 4.28.29.3692) = 0.24) = 240 kg 240 w= × 100% = 36. compute (a) the void ratio and (b) the porosity. The volume of water in a sample of moist soil is 0. For the soil sample of Problem 2. What is the natural water content of the soil? SOLUTION: Mw = 1389 − 982 = 407 g M 407(100) (a) w = w × 100% = = 41. SOLUTION: Ms = 13.27.67 − 8.6)(0.4% Ms 982 2.67 g.9% 650 3 2. Given that the density of soil solids ρs is 2600 kg/m . The container weighs 8.84 g.76 g M 4. A specimen of fully saturated clay soil that weighs 1389 g in its natural state weighs 982 g after drying.83 g Mw = 18.0% 1 + 0.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2. SOLUTION: Assume S = 100% (a) e = Gs w S 0. SOLUTION: Ms = ρs × Vs = (2600)(0.29.25) = 650 kg Mw = ρ w × Vw = (1000)(0.24 m . The mass of a sample of silty clay soil plus container is 18.43 g and the weight of the dry soil plus container is 13.43 − 13.83 2.96 (b) n = × 100 = 50.67 = 4.96 (1) An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .25 3 3 m . 284 (b) n = 0. Compute the void ratio.284 = 248. Give your answers in Mg/m .68 Vv = 592 − 343. For the soil sample of Problem 2. and 3 total density in kg/m .716 (a) e = v = = 0.716 Vv ) ρ = (d t 1090 592 = 1.32. Its dry weight is 920 g and the density of 3 solids is 2680 kg/m . SOLUTION: ρs = 2680 kg m = 2.7245 (c) Mw = 1090 − 920 = 170 g 170 V = = 170 cm3 w ρw Vw S= × 100% = 170 × 100 = 68.25 + 0.284 cm3 2.53 kg = 1. compute (a) the total or wet density and (b) the dry 3 3 3 density.33 Mg = 82.49 3 3 3 2. degree of saturation.49 kg m3 = 1.98 0.49 m3 ρt = 710 = 1448.68 3 3 g cm3 Vs = 920 = 343.841 g cm3 = 1841 kg m3 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .24 = 0. porosity. A 592-cm volume of moist sand weighs 1090 g. and lbf/ft .8 lbm ft m m 0.31.0% 1 + 0.7245 = 0.29.45 Mg m3 = 90.7245 × 100% = 42. water content.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.72 Vs 343. kg/m .4% 248.4 lbm ft 3 (b) ρdry = 650 = 1326.716 cm3 V 248. SOLUTION: Assume S = 100% (a) Vt = 0. A sand is composed of solid constituents having a density of 2.58. 3 SOLUTION: γ ' = 137 − 62. v V = 1 + 0.70 Mg m3 For S = 100%.58.58 3 Mg ρ ' = 2.68 + 0.58 Mg Mt = 2. The saturated density γsat of a soil is 137 lbf/ft .6962 = 1.26 ρ = = 2.6 ' ρ = lbf ft 3 3 ( 74.06 m3 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .6 lbf ) ft3 ⎛ 16. Compute the density of the sand when dry and when saturated and compare it with the density when submerged.68 Mg/m3.68 1.58 = 1.58 m3 t ρ dry = 2.06 − 1. SOLUTION: Assume Vs = 1 m3 V = 0. V = V = 0.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.58 = 3.58 m3 w v Mw = (1)(0.58 = 1. Find the buoyant density of this soil in both 3 3 lbf/ft and kg/m .4 = 74.06 Mg sat m 1.018 kg m ⎞ ⎜ ⎝ 1lbm ft ⎟= ⎠ 1195 kg m3 3 2.33.58) = 0.34.0 = 1.26 Mg 3. The void ratio is 0. 7 (b) ρdry = = 1.67 Mass (kg) water solid Mw = 182 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . and void ratio.71 Mg m3 (d) n = (e) e = 1. P2. The water content is 10%.7) = 1.18 air Vs = 0.7 s s Ms = Gs × Vs × ρs = (2.4 2.404 m = V Vv w v w ρw v Vt = 1.35.404 1.404 + 1.673 = 0.10Ms = 1. Estimate the wet density.36. porosity.7)(1)(1) = 2. buoyant density.18 kg Mw = (0.0 × 100% = 58.673 m3 . Assume G = 2.404 m3 4. dry density.82 kg V = s Ms = 1818.404 = 4.4% = 1. Mw = w × Ms = (0.12 Mg m 2.404 Mg Mt = 2.181 = 0.0 Vv = 0.327 − 0.82 = 0.70 3 Mg/m .404 3 3 (c) ρ ' = 1.181 m3 ρs 2700 V = 1 − 0.404 1.404 2.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2. V = w 181. v 1000 V = 0.7 Mg.0 = 2. Assume ρs = 2. SOLUTION: Assume V = 1 m3 .7 + 1. The water content was found to be 52%. 3 SOLUTION: Mw = Ms × w = 0. Clearly state any necessary assumptions.18 = 0.146 m3 a Volume (m3) Va = 0.327 m3 .10Ms → Ms = 1818.18) = 181. A 1-m sample of moist soil weighs 2000 kg.0 = 0.104 Mg V 3 M S = w = 1→ V = V . A sample of natural glacial till was taken from below the groundwater table.104 = 1.33 Vw = 0. fill in all blanks in the phase diagram of Fig.15 Vt = 1.36.71 Mg (a) ρ = t m 2.71 − 1.10Ms Mt = 2000 = Ms + 0.404 2. V = w = 1.10)(1818. With this information.52)(2. An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual Ms = 1818 Mt = 2000 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . 6 and the degree of saturation is 75%.37. Assuming the density 3 of the solids is 2710 kg/m .0) = 2710 kg M An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . and (c) the dry density.49 × 100 = 32.6 m3 .6 + 1.0 m3 V = e × V = (0.6) = 0.0 3 2. SOLUTION: Assume Vs = 1. v s V = S × V = (0. compute (a) the water content and (b) dry and wet densities in both SI and British engineering units.0 1818.4859 = 0. SOLUTION: (a) e = (b) n = (c) ρ dry 0.45) = 450 kg.327 0. (b) the porosity.6 m3 Ms = ρs × Vs = (2710)(1. calculate (a) the void ratio.0 = 1.7% 1.36.18 = = 1818 kg m 1.0) = 0.327 = 0.38. The void ratio of clay soil is 0. For the information given in Problem 2.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.6)(1.75)(0.673 0. Mt = 450 + 2710 = 3160 kg V = 0.45 m3 w v Mtw = ρ w × Vw = (1000)(0. An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 450 Ms 2710 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual s dry An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual = 2710 = 1693.75 = 1694 kg ρ = 1694 kg ⎜ ⎛ An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 1 lbm An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual ft 3 kg An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . 8 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual ⎞ lbm ⎟ = 105. An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual ft 3 ρ = Mt An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . 018 3160 ⎝ = = 1975 kg An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 16. 018 3 ( ) 3 lbm ft3 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual m3 ⎠ Vt (a) w = w 1.3 16.6 (b) ρ t t m 3 = M V 1 ) ρdry = (1975) ( m m t = 123.6 × 100 = × 100 = 16.6 % 1. 0) = 1.8391 = 1.50) 1 + 0.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2. and saturated density.70) +(1. 2.6 % 1 + 1. What is the corresponding range in the saturated density in kg/m ? SOLUTION: Eq.0)(0.70) 1 + 0.13 Mg m3 minimum : ρ sat = (2.17: ρ sat = ρ +ρ e s w 1+ e = sat max imum : ρ (2.0)(0.70) +(1.17: ρ = s = = 1.70 Mg/m . S = 100%.84 sat Mg 3 1+ e 1 + 1. On the assumption that ρs 3 = 2. 2.026) (c) From Eq. porosity.026 m 2. A specimen of saturated glacial clay has a water content of 38%. respectively. The values of minimum e and maximum e for a pure silica sand were found to be 0.70) (a) From Eq.39.15: e = s = (100)(1.40.0)(2.70 = 2.026 × 100 = 50.0)(1.70.00 Mg m3 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .70 Mg wρ (38. compute the void ratio.50 and 3 0. SOLUTION: w = 38%.03 Sρ w 3 m (b) n = e 1+ e × 100 = 1.026 ρ + ρw e (2.026 = 1. ρs = 2.70) + (1.50 = 2. 2. An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.7405. thus.92 Vs 0.14 mm in diameter) and (b) tiny ball bearings 0.38 g W 2.0 − 0. Calculate the maximum possible porosity and void ratio for a collection of (a) tennis balls (assume they are 64.2595 1.7405 = 0.8 19. A plastic-limit test has the following results: Wet weight + container = 23. Ws = 20.38 Ws An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .46 g Compute the PL of the soil.52 = Vv Vt × 100 = × 100 = 48 % emax = Vv = 0.46 = 19. regardless of sphere size. A quick internet search on packing of equal spheres will reveal numerous mathematical theories and approaches for estimating the densest and loosest possible packing.28 PL = w × 100 = × 100 = 11.3 mm in diameter.48.84 = 2.52 Densest packing (not required in problem statement) Vs = 0. statisticians.48.84 g Container weight = 1.2595 = 0.0.12 g Dry weight + container = 20.48 = 0.41. 0.7405 = 0. As an aside.42.0 and Vs = 1 − 0.48 = 0. (These values are approximate – there is not a unified consensus in the Vs = 18 literature. thus.2595 0.35 2. the loosest arrangement of equal spheres yields a void fraction of about 0. In general.48 1. SOLUTION: Three-dimensional particle arrangement of equal spheres has been studied in depth by mathematicians. n max Vv = 0. Can the plastic limit be evaluated by a one-point method? SOLUTION: Ww = 23. and materials scientists since the 1600s. the densest possible packing of equal-size spheres yields a solids volume of about: π = 0.84 − 1.28 g.12 − 20.0 × 100 = 26% Vv = 1. nmin = emin 0.7405.) Loosest packing For Vt = 1. During a plastic-limit test. the following data was obtained for one of the samples: Wet weight + container = 23.43. 2.13 g An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual The plastic limit cannot be evaluated using a one-point method. Find the water content of the soil.13 − 19.0) t Volume (cm3) Va Vt = 937 Mass (g) air Vv = 613 Vw = Vs = 613 324 water solid Mw = 613 Ms = 876 Mt = 1489 2. Draw a phase diagram and properly label it.12 = 4. SOLUTION: S = 100%. V = 613 + 324 = 937 cm3 Gs ρ w = (2.70 thus.62 g W 4. compute the void ratio and the porosity.7)(1.0 Assume Gs = 2.4 cm3 .An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual Dry weight + container = 19. n= Vv × 100% = 613 × 100 = 65. Vs = Ms 876 = 324.12 g Container weight = 1.01 PL = w × 100 = × 100 = 22.50 g What is the PL of the soil? SOLUTION: Ww = 23.62 Ws 2. For the soil in the previous problem.01 g.89.8 17.50 = 17. The clay when wet weighs 1489 g and after drying weighs only 876 g.12 − 1.45. Mt = 1489 g. The degree of saturation of a cohesive soil is 100%. V V = = 613 cm3 = w w ρw 1.4% An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . Mt = 876 + 613 = 1489 g M 613 w = w × 100 = × 100 = 70.0 g cm v 3 S 1. Does your answer compare with what you would expect for a saturated cohesive soil? SOLUTION: e= Vv = 613 = 1. Ws = 19. Ms = 876 g Mw = 1489 − 876 = 613 g.44.0% Ms 876 M 613 V = w = 613 = 613 cm3 . and lbf/ft . compute (a) the total or wet density and (b) the 3 3 3 dry density. SOLUTION: An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . For the soil in the previous two problems.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual Vs 324 Vt 937 2. Provide your answers in units of Mg/m .46. kN/m . An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 1489 = 1. 3 γ t = γ sat = γ '+ γ w = 73 + 62.4)(16. SOLUTION: γ ' = 73 lb ft .3 lbf ft 2.2 lbf ft M = 876 = 0.47.59 kN m = 99. Calculate its wet 3 density in kg/m .4 lb ft kg m3 3 ρt = (135.018) = 2169 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .59 Mg cm3 m3 Vt 3 937 3 γ t = 15. A soil specimen had a buoyant density of 73 pounds per cubic foot.4 = 135.17 kN m = 58.5891 g (a) ρ = Mt = t = 1.9349 (b) = ρ s dry g = 0.93 Mg cm3 m3 Vt 3 937 3 γ dry = 9. 43 cm3 Vs = s = = = g g 2.79 × 100 = 8. (c) void ratio. determine the (a) wet density.47 Mg Vt m 3 m 3 120.28 − 4.8 g Mass of specimen + wax coating in air = 215.8 = 34.650 cm ρs ρ w 1.366 g → M = M − M = 181.0 = 157.8 = 177.72 Vv 53.0 g 3 t + wax = 157.43 53.28 cm3 0. Mass of specimen at natural water content in air = 181.1 = 36.0 − 36.025 177.9 g Natural water content = 2.93 = 49. and (d) degree of saturation of the sample.0 cm 3 3 M wax =M t + wax − M = 215.366 = s = = 1.9 = 157.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.43 = 53.72 cm3 (a) ρ = Mt = 181.50 Mg t (b) ρ dry Vt 120.366 g M = 66.43 − 66.2 % Volume (cm3) Mass (g) An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .9 − 58. (b) dry density.28 = 120.43 g = 4.43 g w t s Ms 1 + w 1 + .0 cm V air =V t + wax − V − V = 157.80 Vs 66.8 = 1.” From the information given below.93 (d) S = Vw Vv × 100% = 4.72 M 177.9 g Mass of specimen + wax in water = 58.0 − 36.36 + 4.8 − 177.93 cm3 → Vw Mw 4.9 − 181.53.5% 3 3 Soil solid density = 2650 kg/m Wax solid density = 940 kg/m SOLUTION: w= M −M t s → M = s Mt = 181.1 g t → V wax = M wax = 34.36 cm3 w s V = V + V = 49.940 ρ wax Mwater displaced = Mt + wax − Mwater displaced V = Mwater displaced ρw −V wax → Mwater displaced = 215. A specimen of cemented silty sand is treated in this way to obtain the “chunk density.79 (c) e = = = 0.79 cm3 v air w → V =V t t + wax −V wax = 157.0 1.4 = 4. The “chunk density” method is often used to determine the unit weight of a specimen of irregular shape. 28 49.1 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .93 Mw = 4.43 66.36 wax air water solid Mwax = 34.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual Vwax = Vt+wax = 157.72 53.0 36.4 Mt = 181.1 Va = Vv = 120.79 Vt = Vw = Vs = 4.43 Ms = 177. 0 air water solid Mass (Mg) Vv = 9. 3 3 In rechecking the above values.30 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .55 cm3 ρw (2) V = V + V = 9 + 1 = 10 cm3 ≠ 8.55 = 8.11) = (0.28 V = e × V = (9)(1) = 9 cm3 .75 Mg. Mw = w × Ms = (3. (b) e = 0.0? Mw = 8.90.30 M = 1.95)(9. v s V = Mw = w 8.13 Mg m 10 m 3 3 Volume (m3) Va Vw = 8.0 Vt = 8. Find the inconsistent value and report it correctly.55 1.55 Ms = 2. 11. (d) ρs = 2. Consequently.75 Mt = 11.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.83 cm3 (1) Vt = ρt 1.83 cm3 t v s check : Gs w = Se (2.11)(2. (c) S = 95%.75 Mg/m .54. one was found to be inconsistent with the rest.55 Vs = 1.30 Mt = = 8.28 Mg/m .0 = 8. Show all your computations and phase diagrams.55 = 11.75) = 8.55 Mg Mt = Ms + Mw = 2.75 + 8.65)(3.13 Mg Re-calculate ρ = t = t Vt Solution : ρ t = 1.83 or 10. SOLUTION: Assume Vs = 1 cm3 Ms = ρs × Vs = 2.55 → These values are in correct proportion. A sensitive volcanic clay soil was tested in the laboratory and found to have the following properties: (a) ρ = 1.30 Mg 11.0) 8. (e) w = 311%. the error must be in the ρt value. 54 − 279.54 cm3 t 1.95 cm3 t 1.75 = 0.95 − 279. 3 3 Mt = 740 g. Compute the void ratio.45 Mg m 3 (b) Sand under static load 510 V = = 504. n= Vv Vt × 100% = 230.65 Mg/m . dry sand: Mw = Vw = 0. n= Vv Vt × 100% = 225.initial condition Vt = 510 cm3 V = V − = 510 − 279.47 g = 1.47 Mg m (c) Vibrated and loaded sand 510 V = = 454.10 V = V − = 454. and then by vibration it is reduced 10% of the original volume.83.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.45 Mg m3 ρdry = ρt = 1.01 V = V − = 504.24 = 225.45 Vt 510 = 1.2% 279.24 = 175.24 cm .65 = 279. porosity.2% 279. dry density. (c) Vibrated and loaded sand. Under a static load of 200 kPa the volume is reduced 1%.0 (a) Loose sand .71 510 × 100 = 44.47 Mg cm3 m3 ρdry = ρ t = 1. (b) Sand under static load.71 = 0.30 cm3 V v t s An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .24 = 230.24 g cm3 ρdry = Ms = 740 = 1. A cylinder contains 510 cm of loose dry sand which weighs 740 g.95 3 ρ dry = Ms Vt = 1. SOLUTION: Vs = Ms ρs 3 740 = 2. and total density corresponding to each of the following cases: (a) Loose sand.55. Assume the solid density of the sand grains is 2.75 cm3 V v t s e= Vv Vs = 230.24 = 740 504.81.71 cm3 V v t s e= Vv Vs = 225.75 510 × 100 = 45. 6% 279. Vs ρ dry e= n= Vv × 100% = 175.30 × 100 = 38.63 g 454.24 = 740 454.30 = 0.63 Mg m An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .63.54 3 Vt = 1.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual Vv = 175.63 Mg cm3 m3 ρdry = ρ t = 1.54 = Ms Vt = 1. continued next page An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2. A through F.56. Determine also the percentages of gravel. For each soil determine the effective size as well as the uniformity coefficient and the coefficient of curvature. plot the grain-size distribution curves from the following mechanical analysis data on six soils. (b) AASHTO. and clay according to (a) ASTM. sand silt. (c) USCS. On five-cycle semilogarithmic paper. and (d) the British Standard. 6 3.1 0.09 1 0.0 1000.1 3.6 0.001 0.4 N/D continued next page An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .003 Cu 46.001 SILT or CLAY Percent finer by weight 90 80 70 60 50 40 30 20 10 0 100 LEGEND Soil A Soil B Soil C Soil D Soil E Soil F Grain size in millimeters Soil Type: Description: Depth: LL: PL: Cu = Cc = PI: Soil A B C D E F Effective Size D10 (mm) 0.56.0 0.22 0.06 0.01 0 0. continued.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.7 N/D Cc 2.16 0.7 18.015 N/D D60 (mm) 28 0.0 1.9 16.005 0. SOLUTION: C = u D60 D 10 C = c (D30 ) 2 D ×D 60 10 Particle Size Distribution GRAVEL Coarse 100 Fine Course SAND Medium Fine 100 90 80 70 60 50 40 30 20 10 10 1 0.6 1.04 0.006 N/D D30 (mm) 6 0.1 0.3 0. Soil A B C D E F Gravel (%) 80 18 30 0 11 0 Sand (%) 16 27 38 100 32 0 Fines (silt + clay) 4 55 32 0 57 100 Silt (%) 4 45 18 0 49 29 Clay (%) 0 10 14 0 8 71 (c) Percentages according to USCS.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2. Soil A B C D E F Gravel (%) 73 12 19 0 0 0 Sand (%) 23 33 49 100 43 0 Fines (silt + clay) 4 55 32 0 57 100 Silt (%) 4 45 18 0 49 29 Clay (%) 0 10 14 0 8 71 (b) Percentages according to AASHTO. continued. Soil A B C D E F Gravel (%) 80 18 30 0 11 0 Sand (%) 16 27 38 100 32 0 Fines (silt + clay) 4 55 32 0 57 100 Silt (%) 4 53 21 0 57 48 Clay (%) 0 2 11 0 0 52 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . (a) Percentages according to ASTM.56. Soil A B C D E F Gravel (%) 73 12 19 0 0 0 Sand (%) 23 33 49 100 43 0 Fines (silt + clay) 4 55 32 0 57 100 Silt (%) ------Clay (%) ------- (d) Percentages according to the British Standard. 8 Soil B 14 35 29 6 -2.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2. SOLUTION: Based on Atterberg’s definitions: LL > PL > SL. SOLUTION: PI = LL − PL. An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .56 have the following Atterberg limits and natural water contents. The soils in Problem 2. Comment on the validity of the results of Atterberg limits on soils G and H.58. Determine the PI and LI for each soil and comment on their general activity.24 Soil D 11 -NP 0 -Soil E 8 28 NP 0 -Soil F 72 60 28 32 1.38 Property wn (%) LL PL PI LI Soil A 27 13 8 5 3.5 Soil A: very sensitive.59. LI = w n − PL PI Soil C 14 35 18 17 -0. Soil H violates the definitions because the PL > LL (42 > 38). Soil G violates the definitions because the SL > PL (25 > 20). highly active Soil B: most likely a clay above the water table that has experienced a decrease in moisture Soil C: most likely a clay above the water table that has experienced a decrease in moisture Soil D: most likely a fine sand Soil E: most likely a silt Soil F: slightly sensitive and active 2. and the toughness index is the PI divided by the flow index. PI.6 Flow Index = slope = IF = w1 − w 2 = 10.6 − 23.60.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.4 PI = LL − PL = 42.1 + 23.2 PI 19. SOLUTION: From the plot below.6 (from linear regresion of best fit line) ⎛N ⎞ log ⎜ 2 ⎟ ⎝ N1 ⎠ use average value of PL tests.2 = = 1. PL = 23. Determine the LL. The following data were obtained from a liquid-limit test on a silty clay. Two plastic-limit determinations had water contents of 23. the flow index.4 = 19. LL = 42.6%. The flow index is the slope of the water content versus log of number of blows in the liquid-limit test.81 Toughness Index = IF 10.6 2 = 23.1% and 23.6 Number of Blows An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .6 50 48 46 Water Content (%) 44 42 40 38 36 34 32 30 10 15 20 30 40 LL = 42.6 Flow index = 10. and the toughness index. and boulders (or GP-GC) 2. The liquid limit is 33% and the plastic limit is 27%. and 2% fines. and is 21% when oven dried!) and the plastic index is 21% (not dried). The maximum size is 20 mm. 1990. (b) OL -. The maximum size of the particles is 75 mm. of the Interior. Classify the five soils in the preceding question according to the AASHTO method of soil classification. 25% fine to coarse subangular sand. wet. (b) A dark brown. 200 sieve. while the coefficient of curvature is only 0. and 16% fine to coarse subrounded to subangular sand. SOLUTION: (a) A-1-a (b) A-8 (c) A-2-4 (d) A-2-4 or A-8 (e) A-1-a An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . The uniformity coefficient is 40. The coefficient of curvature is 2.Organic clay (c) SM – Silty sand with gravel (d) SM – Silty sand with organic fines (e) GP-GM – Poorly graded gravel with silt.7. 23% silty fines. The liquid limit is 32% (not dried.7 and corresponding footnotes (a) GW – Well-graded gravel with sand. it has everything else! It has gravel content of 78% fine to coarse subrounded to subangular gravel. The liquid limit (not dried) is 37% while it is 26% when oven dried.) SOLUTION: Refer to Table 2. The maximum size of the subrounded boulders is 500 mm. (c) This sand has 61% predominately fine sand. The plastic index (not dried) is 6.61. while the uniformity coefficient is 12. Dept. Classify the following soils according to the USCS: (a) A sample of well-graded gravel with sand has 73% fine to coarse subangular gravel. cobbles.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2. (e) Although this soil has only 6% nonplastic silty fines.8. (d) This material has 74% fine to coarse subangular reddish sand and 26% organic and silty dark brown fines.4.62.S. (After U. and 16% fine subrounded gravel size. organic-odor soil has 100% passing the No. sand. C = c (D )2 30 (1. 20 No. Sieve ½” No.00 1. 4 No.39) = 0.01 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . (a) Using a spreadsheet.5)(0. 10 No.69 D 10 D ×D 60 10 Grain Size Distribution Plot 100 90 80 70 2" 1" 1/2" #4 #10 #20 #40 #100 #200 % Passing 60 50 40 30 20 10 0 100.39 = 24.10 0.00 2" 1" 1/2" #4 #10 #20 #40 #100 #200 10. 40 No.00 Grain Diameter (mm) 0. 60 No.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.5 = 0.6)2 = (9.63. The results of a sieve test below give the percentage passing through the sieve. (c) Calculate the coefficient of curvature. (b) Calculate the uniformity coefficient. plot the particle-size distribution. 100 Percent Finer by Weight 71 37 32 23 11 7 4 SOLUTION: D C = u 60 9. 4 sieve. A sample of soil was tested in the laboratory and the following grain size analysis results were obtained. give both the letter symbol and the narrative description. 4 sieve.00 0. SOLUTION: (a) GP – Poorly graded gravel with sand (b) CL-ML – Silty clay (c) GW – Well-graded gravel with sand 2. Sieve 1/2" 4 10 20 40 60 100 200 Sieve Opening (mm) 12. 200 sieve.25 0.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.64. Cu = 5.075 Percent Coarser by Weight 30 36 52 64 69 71 77 91 Percent Finer by Weight 70 64 48 36 31 29 23 9 SOLUTION: (a) PI = LL − PL = 26 − 23 = 3D Cu = 60 3. For each soil. For the data given below.9 = 0.00 2" 1" 1/2" 10 #4 48 36 31 29 23 10.15 0.65. 90% passed No. Classify this soil according to the USCS.5. 200 sieve.4)2 = (3.425 0. 32% retained on No. 4 sieve. LL = 23.08) = 0. 27% retained on No.75 2. providing the group symbol for it. (c) 70% material retained on No. Cc = (D )2 D ×D 60 10 30 (0. classify the soils according to the USCS. Cu = 3. Cc = 1.85 0.7 4.08 = 49. Cc = 1. (a) 65% material retained on No. 00 # #20 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . (b) 100% material passed No.51 →∴ well graded D 10 SW-SM (W ell-graded sand with silt) 100 90 2" 1" 1/2" #4 #10 #20 #40 #100 #200 80 70 70 64 % Passing 60 50 40 30 20 10 0 100. PL = 17.9)(0. 200 sieve. 00 9 #40 #100 #200 0.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 1. 0 1 Grain Diameter (mm) An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .10 0 . 075 Percent Coarser by Weight 37 52 64 69 71 77 90 Percent Finer by Weight 63 48 36 31 29 23 10 SOLUTION: (a) PI = LL − PL = 60 − 26 = 34 D 4.5 = 51.75 = 63. 4 10 20 40 60 100 200 Sieve Opening (mm) 4.7. 00 # #20 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . providing the group symbol for it.5%. and a plasticity index of 24.7 LL = PI + PL = 24.15 0.73 = PL 44.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.75)(0.075) SW-SC (W ell-graded sand with clay) 2" 1" 1/2" #4 #10 #20 #40 #100 #200 100 90 80 70 % Passing 63 60 50 40 36 48 30 20 10 0 100. provide the group symbol. Sieve No.73.67. CH (Fat clay) 24.2 2. 40 material had a liquidity index of 0.75 2.075 10 C = c (D 30 )2 D ×D 60 10 = (0. SOLUTION: LI = w − PL 0. Classify this soil according to the USCS.425)2 = 0. Classify this soil according to the USCS.5 − → PL = 26.25 0. A minus No. Cu = 60 = D 0. a natural water content of 44.5 PI PI = LL − PL.51 →∴ well graded (4.85 0.00 2" 1" 1/2" 31 29 23 10 #4 10.7 + 26.00 0.66.425 0. A sample of soil was tested in the laboratory and the following grain size analysis results were obtained. 10 0 . 0 1 Grain Diameter (mm) An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .00 #100 10 #40 #200 0.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 1. 425 0. 40 material were: LL = 36.01(45 − 15)(22 − 10) GI = 5. Extra credit .determine the full AASHTO classification for this soil (symbol plus group index).00 2" 1" 1/2" #4 #10 #20 #40 #100 #200 10.4 (a) USCS: SC (Clayey Sand) (b) AASHTO: A-6 (5) Grain Size Distribution Plot 100 90 80 2" 1" 1/2" #4 #10 #20 #40 #100 #200 % Passing 70 60 50 40 30 20 10 0 100.005(LL − 40)] + 0.25 0.00 1.15 0. A sample of soil was tested in the laboratory and the following grain size analysis results were obtained: Atterberg limits on minus No. Determine the USCS classification symbol for this soil.10 0.00 0.2 + 0. PL = 14.68.01(F − 15)(PI − 10) GI = (45 − 35)[0.01 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .075 -Percent Finer by Weight 100 100 100 94 82 66 45 0 SOLUTION: PI = LL − PL = 36 − 14 = 22 GI = (F − 35)[0. Sieve No. 4 10 20 40 60 100 200 Pan Sieve Opening (mm) 4.2 + 0.75 2.00 Grain Diameter (mm) 0.005(36 − 40)] + 0.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.85 0. 00 0.01(55 − 15)(20 − 10) = 9.425 0.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.85 0. 3 inch 1.Percent Passing 100 98 96 77 -55 -30 18 32 25 100 96 94 73 -55 52 32 B .15 0.044 = 29.5 inch 0. (a) Determine the USCS classification symbol for Sample A.075 A .Percent Passing SOLUTION: (a) PI = LL − PL = 32 − 25 = 7 D 1.2 + 0.5. Sieve No. GI = (F − 35)[0.1 19.2 AASHTO → A-7-5 (9) ∴A −7−5 100 2" 100 1" 98 1/2" #4 100 #10 #20 #40 #100 #200 96 96 94 Sample A Sample B 90 80 70 77 73 % Passing 60 55 55 50 40 30 20 30 18 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .3)(0. Laboratory testing was performed on two soil samples (A and B).2 + 0.393 →∴ poorly graded = 0.2 38. LL − 30 = 52 − 30 = 22 < 32. (b) Determine the AASHTO classification for Sample B.3 C = u 60 ( D )2 C = c 30 (0.75 2.15)2 = (1. D10 D60 × D10 USCS → SM (Silty sand with gravel) (b) PI = LL − PL = 52 − 32 = 20.044) = 0.005(52 − 40)] + 0.01(F − 15)(PI − 10) GI = (55 − 35)[0.69.005(LL − 40)] + 0.1 4.75 inch 4 10 20 40 100 200 Liquid Limit Plastic Limit Sieve Opening (mm) 76. 10 0.00 1.00 2" 1" 1/2" #4 #10 #20 #40 #100 #200 10.01 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .00 Grain Diameter (mm) 0.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 10 0 100. 1 10.9 90. Technically.2 10 0 100. 40 material were: LL = 62.5 100.8 90.8 15. PL = 20.2 9.9 90 90.0 94.43 C = u 60 (D )2 C = c 60 30 (0.00 0.0 Percent Finer by Weight 100.01 An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .00 0.3 29.0 80 70 % Passing 60 50 40 30 20 59.00 9.5 0. a well graded determination is not unreasonable. USCS → SP − SC (Poorly graded sand with clay) Grain Size Distribution Plot 2" 100 1" 1/2" #4 100. D 10 D ×D 10 The Cu value is close to 6.7 70. Sieve No.25 0.70.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.2 84.43)(0. however.0 59.00 1.9 = 0. A sample of soil was tested in the laboratory and the following grain size analysis results were obtained: Atterberg limits on minus No.75 2.5 2" 1" 1/2" #4 #10 #20 #40 #100 #200 10.7 ≅ 6.8 15.425 0.15 0.10 0.0 40.075) = 1.075 -Percent Coarser by Weight 0.3 29.85 0.0 #10 #20 #40 #100 #200 94. Determine the USCS letter symbol for this soil.075 = 5.0 5.0 SOLUTION: PI = LL − PL = 62 − 20 = 42 D 0. this soil classifies as poorly graded.25)2 = (0. 4 10 20 40 60 100 200 Pan Sieve Opening (mm) 4. An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual Grain Diameter (mm) An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual . 15 g.An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual 2.49 g.9 2 6.65. the water content was 47. the wet + dish = 11. A sample of a brown sandy clay was obtained to determine its Atterberg limits and then classify its soil type according to the Unified Soil Classification System.34 = 16.53 g and the dry weight + dish = 10. the soil fines classify as CL ∴ USCS → CL (Sandy lean clay) 60 58 56 Water Content (%) 54 52 50 48 44 42 40 10 15 20 30 40 LL = 48.0 Flow index = 10. 2.15 = 6. and for 36 blows.71. Evaluate the soil type.3%. The dish only weighed 4.49 = 1. Another plastic limit was 16. and give the Unified Soil Classification symbol.34 g M 1.04 w PL = PL = w × 100 = × 100 = 16. the water content was 49.5%. SOLUTION: Mw = 11.7 = 31 Based on Casagrande's plasticity chart (Fig.53 − 10.04 g Ms = 10.8%. From plot below: LL = 48 PI = 48 − 16. Three determinations of the liquid limit were made. For 17 blows. For one of the PL determinations.8 Number of Blows An Introduction to Geotechnical Engineering Holtz Kovacs 2nd Edition Solutions Manual .4 + 16.9%. for 26 blows. Compute the plastic limit.13). the water content was 46.4 Ms PL(avg) = 16. indicate the information on a plasticity chart.49 − 4.