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 ρ s = 2.7 Mg/m 3 , 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. SOLUTION: Create a spreadsheet using input values from Table 2.1 and Eq. 2.18. ρ ' - min ρ ' - max e max e min (Mg/m 3 ) (Mg/m 3 ) 0.9 1.4 0.89 0.21 0.4 1.1 3.25 0.55 1.1 1.4 0.55 0.21 0.9 1.2 0.89 0.42 0.0 0.1 ∞ 16.00 0.3 0.8 4.67 1.13 (Given) (see Eq. 2.18) Index and Classification Properties of Soils Chapter 2 2-4. Prepare a spreadsheet plot of dry density in Mg/m 3 as the ordinate versus water content in percent as the abscissa. Assume ρ s = 2.65 Mg/m 3 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 ρ d = f( ρ s , w, S, G s ), or use Eq. 5.1. ρ S w ρ = dry ρ w w+ S ρ s S = 100 90 80 70 60 50 40 30 20 10 w ρ dry ρ dry ρ dry ρ dry ρ dry ρ dry ρ dry ρ dry ρ dry ρ dry (%) (Mg/m 3 ) (Mg/m 3 ) (Mg/m 3 ) (Mg/m 3 ) (Mg/m 3 ) (Mg/m 3 ) (Mg/m 3 ) (Mg/m 3 ) (Mg/m 3 ) (Mg/m 3 ) 0.0 2.65 2.65 2.65 2.65 2.65 2.65 2.65 2.65 2.65 2.65 5.0 2.34 2.31 2.27 2.23 2.17 2.09 1.99 1.84 1.59 1.14 10.0 2.09 2.05 1.99 1.92 1.84 1.73 1.59 1.41 1.14 0.73 15.0 1.90 1.84 1.77 1.69 1.59 1.48 1.33 1.14 0.89 0.53 20.0 1.73 1.67 1.59 1.51 1.41 1.29 1.14 0.96 0.73 0.42 25.0 1.59 1.53 1.45 1.36 1.26 1.14 1.00 0.83 0.61 0.35 30.0 1.48 1.41 1.33 1.24 1.14 1.02 0.89 0.73 0.53 0.30 35.0 1.37 1.31 1.23 1.14 1.04 0.93 0.80 0.65 0.47 0.26 40.0 1.29 1.22 1.14 1.05 0.96 0.85 0.73 0.58 0.42 0.23 45.0 1.21 1.14 1.06 0.98 0.89 0.78 0.67 0.53 0.38 0.21 50.0 1.14 1.07 1.00 0.92 0.83 0.73 0.61 0.49 0.35 0.19 Problem 2-4 0.00 0.50 1.00 1.50 2.00 2.50 3.00 0 5 10 15 20 25 30 35 40 45 50 55 w (%) ρ dry (Mg/m3) S=100% S=90% S=80% S=70% S=60% S=50% S=40% S=30% S=20% S=10% Index and Classification Properties of Soils Chapter 2 2-5a Prepare a graph like that in Problem 2.4, only use dry density units of kN/m 3 and pounds per cubic feet. SOLUTION: ρ S w From Eq. 2.12 and Eq. 2.15: ρ = dry ρ w w+ S ρ s S = 100 90 80 70 60 50 40 30 20 10 w γ dry γ dry γ dry γ dry γ dry γ dry γ dry γ dry γ dry γ dry (%) (kN/m 3 ) (kN/m 3 ) (kN/m 3 ) (kN/m 3 ) (kN/m 3 ) (kN/m 3 ) (kN/m 3 ) (kN/m 3 ) (kN/m 3 ) (kN/m 3 ) 0.0 26.00 26.00 26.00 26.00 26.00 26.00 26.00 26.00 26.00 26.00 5.0 22.95 22.66 22.30 21.86 21.29 20.55 19.53 18.03 15.64 11.18 10.0 20.55 20.08 19.53 18.86 18.03 16.99 15.64 13.80 11.18 7.12 15.0 18.60 18.03 17.37 16.58 15.64 14.48 13.04 11.18 8.70 5.23 20.0 16.99 16.36 15.64 14.79 13.80 12.62 11.18 9.40 7.12 4.13 25.0 15.64 14.97 14.22 13.36 12.35 11.18 9.79 8.10 6.03 3.41 30.0 14.48 13.80 13.04 12.17 11.18 10.04 8.70 7.12 5.23 2.90 35.0 13.49 12.80 12.04 11.18 10.21 9.11 7.83 6.35 4.61 2.53 40.0 12.62 11.94 11.18 10.34 9.40 8.33 7.12 5.73 4.13 2.24 45.0 11.86 11.18 10.44 9.62 8.70 7.68 6.53 5.23 3.73 2.01 50.0 11.18 10.52 9.79 8.99 8.10 7.12 6.03 4.80 3.41 1.82 Problem 2-5a 0.00 5.00 10.00 15.00 20.00 25.00 30.00 0 5 10 15 20 25 30 35 40 45 50 55 w (%) γ dry (kN/m3) S=100% S=90% S=80% S=70% S=60% S=50% S=40% S=30% S=20% S=10% Index and Classification Properties of Soils Chapter 2 2-5b Prepare a graph like that in Problem 2.4, only use dry density units of pounds per cubic feet. SOLUTION: s w dry s G S S G w γ γ = + S = 100 90 80 70 60 50 40 30 20 10 w γ dry γ dry γ dry γ dry γ dry γ dry γ dry γ dry γ dry γ dry (%) (pcf) (pcf) (pcf) (pcf) (pcf) (pcf) (pcf) (pcf) (pcf) (pcf) 0.0 165.36 165.36 165.36 165.36 165.36 165.36 165.36 165.36 165.36 165.36 5.0 146.01 144.14 141.86 139.04 135.45 130.72 124.21 114.70 99.46 71.12 10.0 130.72 127.75 124.21 119.95 114.70 108.08 99.46 87.80 71.12 45.30 15.0 118.33 114.70 110.47 105.47 99.46 92.12 82.94 71.12 55.35 33.24 20.0 108.08 104.07 99.46 94.11 87.80 80.27 71.12 59.77 45.30 26.25 25.0 99.46 95.25 90.45 84.96 78.59 71.12 62.25 51.54 38.34 21.69 30.0 92.12 87.80 82.94 77.43 71.12 63.85 55.35 45.30 33.24 18.48 35.0 85.79 81.44 76.58 71.12 64.95 57.92 49.83 40.41 29.33 16.09 40.0 80.27 75.93 71.12 65.77 59.77 53.00 45.30 36.48 26.25 14.26 45.0 75.42 71.12 66.39 61.16 55.35 48.85 41.53 33.24 23.75 12.79 50.0 71.12 66.89 62.25 57.16 51.54 45.30 38.34 30.53 21.69 11.60 Problem 2-5b 0.00 20.00 40.00 60.00 80.00 100.00 120.00 140.00 160.00 180.00 0 5 10 15 20 25 30 35 40 45 50 55 w (%) γ dry (pcf) S=100% S=90% S=80% S=70% S=60% S=50% S=40% S=30% S=20% S=10% Index and Classification Properties of Soils Chapter 2 2.6. Prepare a graph like that in Problem 2.4, only for S = 100% and vary the density of solids from 2.60 to 2.80 Mg/m 3 . You decide the size of the increments you need to “satisfactorily” evaluate the relationship as ρ s varies. Prepare a concluding statement of your observations. SOLUTION: ρ S w From Eq. 2.12 and Eq. 2.15: ρ = dry ρ w w+ S ρ s Note: The relationship between ρ dry and w is not overly sensitive to ρ s . ρ s = 2.6 2.65 2.7 2.75 2.8 w ρ dry ρ dry ρ dry ρ dry ρ dry (%) (Mg/m 3 ) (Mg/m 3 ) (Mg/m 3 ) (Mg/m 3 ) (Mg/m 3 ) 0.0 2.60 2.65 2.70 2.75 2.80 5.0 2.30 2.34 2.38 2.42 2.46 10.0 2.06 2.09 2.13 2.16 2.19 15.0 1.87 1.90 1.92 1.95 1.97 20.0 1.71 1.73 1.75 1.77 1.79 25.0 1.58 1.59 1.61 1.63 1.65 30.0 1.46 1.48 1.49 1.51 1.52 35.0 1.36 1.37 1.39 1.40 1.41 40.0 1.27 1.29 1.30 1.31 1.32 45.0 1.20 1.21 1.22 1.23 1.24 50.0 1.13 1.14 1.15 1.16 1.17 0.00 0.50 1.00 1.50 2.00 2.50 3.00 0 5 10 15 20 25 30 35 40 45 50 55 w (%) ρ s (Mg/m3) Index and Classification Properties of Soils Chapter 2 2.7. The dry density of a compacted sand is 1.87 Mg/m 3 and the density of the solids is 2.67 Mg/m 3 . What is the water content of the material when saturated? SOLUTION: w dry w s 3 w 3 3 dry s S From Eq. 2-12 and Eq. 2-15: ; Note: S = 100% w S 1 1 1 1 Solving for w: w S (1Mg / m )(100%) 16.0% 1.87 Mg / m 2.67Mg / m ρ ρ = ρ + ρ ⎛ ⎞ ⎛ ⎞ = ρ − = − = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ρ ρ ⎝ ⎠ ⎝ ⎠ 2.8. A soil that is completely saturated has a total density of 2045 kg/m 3 and a water content of 24%. What is the density of the solids? What is the dry density of the soil? SOLUTION: a) Solve using equations or phase diagrams: 3 t dry w sat dry w s w dry 3 s dry w sat 2045 1649.2 kg / m (1 w) (1 0.24) 1 (1000)(1649.2) 2729.6 kg / m (1649.2 1000 2045) ρ ρ = = = + + ⎛ ⎞ ρ ρ = − ρ + ρ ⎜ ⎟ ρ ⎝ ⎠ ρ ρ ρ = = = ρ + ρ − ρ + − b) Solve using phase diagram relationships: t t w w s s t w s s s s 3 s w w w w 3 s s s 3 s dry t assume V 1.0 M 2045 kg M 0.24 M 0.24M M M M M 2045 0.24M M M 1649.19 kg 0.24M M 1000 V 0.3958 m V 1000 M 1649.19 2729.6 kg / m V 1 0.3958 M 1649.19 1649.2 kg / m V 1 = = = → = = + → = + → = ρ = = → = = ρ = = = − ρ = = = Index and Classification Properties of Soils Chapter 2 2.9 What is the water content of a fully saturated soil with a dry density of 1.72 Mg/m 3 ? Assume ρ s = 2.72 Mg/m 3 . SOLUTION: w dry w s 3 w 3 3 dry s S From Eq. 2-12 and Eq. 2-15: ; Note: S = 100% w S 1 1 1 1 Solving for w: w S (1Mg / m )(100%) 21.4% 1.72 Mg / m 2.72 Mg / m ρ ρ = ρ + ρ ⎛ ⎞ ⎛ ⎞ = ρ − = − = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ρ ρ ⎝ ⎠ ⎝ ⎠ 2.10. A dry quartz sand has a density of 1.68 Mg/m 3 . Determine its density when the degree of saturation is 75%. The density of solids for quartz is 2.65 Mg/m 3 . SOLUTION: dry dry (final) w dry w s 3 w 3 3 dry s t dry Recognize that (initial) (final); S = 75% S From Eq. 2-12 and Eq. 2-15: w S 1 1 1 1 Solving for w: w S (1Mg / m )(75%) 16.34% 1.68 Mg / m 2.65 Mg / m final (1 ρ = ρ ρ ρ = ρ + ρ ⎛ ⎞ ⎛ ⎞ = ρ − = − = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ρ ρ ⎝ ⎠ ⎝ ⎠ ρ = ρ + 3 w) 1.68(1 0.1634) 1.95 Mg / m = + = 2.11. The dry density of a soil is 1.60 Mg/m 3 and the solids have a density of 2.65 Mg/m 3 . Find the (a) water content, (b) void ratio, and (c) total density when the soil is saturated . SOLUTION: w dry w s 3 w 3 3 dry s s w Given: S = 100% S From Eq. 2-12 and Eq. 2-15: w S 1 1 1 1 (a) Solving for w: w S (1Mg / m )(100%) 24.76% 1.60 Mg / m 2.65 Mg / m w (24.76)(2.65) (b) From Eq. 2.15: e S (100)(1 ρ ρ = ρ + ρ ⎛ ⎞ ⎛ ⎞ = ρ − = − = ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ρ ρ ⎝ ⎠ ⎝ ⎠ ρ = = ρ 3 t dry 0.656 .0) (c) (1 w) 1.60(1 0.2476) 1.996 2.00 Mg / m = ρ = ρ + = + = = Index and Classification Properties of Soils Chapter 2 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: s s w w = 20% and S = 90%; assume G 2.70 w (20.0)(2.70) From Eq. 2.15: e 0.60 S (90)(1.0) = ρ = = = ρ 2.13. A chunk of soil has a wet weight of 62 lb and a volume of 0.56 ft 3 . When dried in an oven, the soil weighs 50 lb. If the specific gravity of solids G s = 2.64, determine the water content, wet unit weight, dry unit weight, and void ratio of the soil. SOLUTION: Solve using phase diagram relationships. w t s w s t t t s dry t 3 w w w 3 s s s w v t v (a) W W W 62 50 12 lb W 12(100) w 100% 24.0% W 50 W 62 (b) 110.7 pcf V 0.56 W 50 (c) 89.29 pcf V 0.56 W 12 (d) V 0.1923 ft 62.4 W 50 V 0.3035 ft G (2.64)(62.4) V V V 0.56 0.3035 0.2565 e = − = − = = × = = γ = = = γ = = = = = = γ = = = γ = − = − = v s V 0.2565 0.8451 0.84 V 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. s w w s s w M 100.06 49.31 50.75 g M 113.27 100.06 13.21 g M 13.21(100) (a) w 100% 26.03 26.0% M 50.75 w 2.80(26.03) (b) From Eq. 2.15: e= 0.7288 0.73 S (1)(100) = − = = − = = × = = = ρ = = = ρ Index and Classification Properties of Soils Chapter 2 2.15. The natural water content of a sample taken from a soil deposit was found to be 12.0%. It has been calculated that the maximum density for the soil will be obtained when the water content reaches 22.0%. 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 22.0%. SOLUTION: w w s s t s w s s s s w s w s Natural state M w 0.12 M (0.12)M M M M M M 0.12M 1.12M 1000 g M 892.857 g M (0.12)(892.857) 107.143 g Target state (Note: M does not change between natural state and target state) M w M (0.22)(89 = = → = = + = + = = = = = = × = 2.857) 196.429 g additional water necessary = 196.429 107.143 89.286 89.29 g = − = = 2.16. A cubic meter of dry quartz sand (G s = 2.65) with a porosity of 60% is immersed in an oil bath having a density of 0.92 g/cm 3 . If the sand contains 0.27 m 3 of entrapped air, how much force is required to prevent it from sinking? Assume that a weightless membrane surrounds the specimen. (Prof. C. C. Ladd.) SOLUTION: 3 3 3 3 3 t 3 v t 3 s t v 3 kg s s s s t m t kg t m t kg buoy t oil m buoy buoy V 1m 1 ,000,000 cm V n V (0.6)(1.0) 0.60 m V V V 1.0 0.60 0.40 m M G V (2.65)(1000 )(0.40 m ) 1060 kg M M 1060 1060 V 1.0 1060 920 140 g (140)(9.81) 1373.4 = = = × = = = − = − = = × ρ × = = = ρ = = = ρ = ρ − ρ = − = γ = ρ × = = 3 3 N m 3 N buoy m Force (entrapped air) 1373.4 0.27 m 370.8 N = γ × = × = Index and Classification Properties of Soils Chapter 2 2.17. A soil sample taken from a borrow pit has a natural void ratio of 1.15. The soil will be used for a highway project where a total of 100,000 m 3 of soil is needed in its compacted state; its compacted void ratio is 0.73. How much volume has to be excavated from the borrow pit to meet the job requirements? SOLUTION: t s t v s s 3 t 3 s(emb) t s(borr ) s(emb) 3 t(borr ) s borr V V V V V V e e e 1 Embankment V 100,000 m 100,000 V 57,803.47 m 0.73 1 Borrow Pit V V V e 1 V V (e 1) (57,803.47)(1.15 1) 124,277 m − = = → = + = = = + = = + = × + = + = 2.18. A sample of moist soil was found to have the following characteristics: Total volume: 0.01456 m 3 Total mass: 25.74 kg Mass after oven drying: 22.10 kg Specific gravity of solids: 2.69 Find the density, unit weight, void ratio, porosity, and degree of saturation for the moist soil. SOLUTION: 3 3 3 kg t m N kN t t m m 3 s s s w 3 v 25.74 (a) 1767.857 1768 0.01456 (b) g (1767.857)(9.81) 17,342.68 17.34 M 22.10 (c) V 0.00822 m G (2.69)(1000) V 0.01456 0.00822 0.006344 m 0.006344 e 0.7718 0.77 0.00822 0.7718 (d) n 1 0.7 ρ = = = γ = ρ × = = = = = = ρ = − = = = = = + w w 100 43.56 43.6% 718 (e) M 25.74 22.10 3.64 3.64 V 0.00364 1000 0.00364 S 100 57.377 57.4% 0.006344 × = = = − = = = = × = = Index and Classification Properties of Soils Chapter 2 2.19. A gray silty clay (CL) is sampled from a depth of 12.5 feet. The “moist” soil was extruded from a 6-inch-high brass liner with an inside diameter of 2.83 inches and weighed 777 grams. (a) Calculate the wet density in pounds per cubic feet. (b) A small chunk of the original sample had a wet weight of 140.9 grams and weighed 85.2 grams after drying. Compute the water content, using the correct number of significant figures. (c) Compute the dry density in Mg/m 3 and the dry unit weight in kN/m 3 . SOLUTION: ( )( ) ( )( ) ( ) 3 2 3 t t t 3 3 t w w t s s t dry lb dry ft 1lb 777 g 453.6 g M (2.83) (a) V 6 37.741in , 78.429 78.4 pcf 4 V 1 ft 37.741in 12 in M 55.7 (b) M M M 140.9 85.2 55.7 g, w 100% 65.38 65.4% M 85.2 78.4 (c) 47.40 pcf (1 w) (1 0.654) 1 ft 47.4 0.3 π = × = γ = = = = = − = − = = × = = = γ γ = = = + + ρ = ( ) ( ) 3 3 3 3 3 kg Mg m m N kN dry m m 0.4536 kg 759.288 0.759 048 m 1lb (759.288)(9.81) 7448.6 7.45 = = γ = = = 2.20. A cylindrical soil specimen is tested in the laboratory. The following properties were obtained: Sample diameter 3 inches Sample length 6 inches Wt. before drying in oven 2.95 lb Wt. after drying in oven 2.54 lb Oven temperature 110°C Drying time 24 hours Specific gravity of solids 2.65 What is the degree of saturation of this specimen? SOLUTION: 2 3 3 t 3 3 s s s w 3 3 v t s w t s 3 3 w w w w v (3) V 6 42.4115 in 0.02454 ft 4 W 2.54 V 0.01536 ft 26.542 in G (2.65)(62.4) V V V 42.4115 26.542 15.869 in 0.009184 ft W W W 2.95 2.54 0.41lb W 0.41 V 0.00657 ft 11.354 in 62.4 V 1 S 100% V π = × = = = = = = γ = − = − = = = − = − = = = = = γ = × = 1.354 100 71.5% 15.869 × = Index and Classification Properties of Soils Chapter 2 2.21 A sample of saturated silt is 10 cm in diameter and 2.5 cm thick. Its void ratio in this state is 1.35, and the specific gravity of solids is 2.70. The sample is compressed to a 2-cm thickness without a change in diameter. (a) Find the density of the silt sample, in prior to being compressed. (b) Find the void ratio after compression and the change in water content that occurred from initial to final state. SOLUTION: 3 3 2 3 t w w v v v s s 3 t v s s s s s 3 v w g w w w cm g s s s w cm (10) (a) V 2.5 196.350 cm 4 V S 1 V V V V e V 1.35V V V V 1.35V V 2.35V 196.350 V 83.553 cm V (1.35)(83.553) 112.797 cm V M V (1 )(112.797) 112.797 g M G V (2.70)(83.553)(1 ) π = × = = = → = = × = = + = + = = → = = = = = ρ × = = = × × ρ = 3 3 t t g kg t cm m t 2 3 t 2 3 s 3 v t s final initial 225.594 g M 112.797 225.594 338.391 g M 338.391 1.723 1723 V 196.35 (10) (b) V 2.0 157.08 cm 4 V 83.553 cm (no change) V V V 157.08 83.553 73.527 cm 73.527 e 0.88 83.553 112. (c) w − = = + = ρ = = = = π = × = = = − = − = = = = 3 w v w s final 797 100% 50.0% 225.594 final conditions V V 73.527 cm ; M 73.527 g; M 225.594 g (no change) 73.527 w 100% 32.6% 225.594 w 50.0 32.6 17.4% × = = = = = = × = Δ = − = Index and Classification Properties of Soils Chapter 2 2.22. A sample of sand has the following properties: total mass M t = 160 g; total volume V t =80 cm 3 ; water content w = 20%; specific gravity of solids G s =2.70. How much would the sample volume have to change to get 100% saturation, assuming the sample mass M t stayed the same? SOLUTION: 3 w s s t s s s 3 g w w w cm 3 3 s s v s w i t M w M (0.20)M M M 0.20M 160 g M 133.33 g M (0.20)(133.33) 26.667 g; V (1 )M 26.667 cm M 133.33 V 49.383 cm ; V 80 49.383 30.617 cm G 2.70 26.667 S 100% 87.10% 30.617 Desired condition: S 100% V ch = × = = + = = = = = = = = = = − = ρ = × = = s s 3 w w v w 3 t s v 3 anges, but V and M remain the same M 26.667 g; V 26.667 cm V V (when S = 100%) V V V 49.383 26.667 76.053 cm V 80 76.053 3.95 cm = = = = + = + = Δ = − = Index and Classification Properties of Soils Chapter 2 2.23. Draw a phase diagram and begin to fill in the blanks: A soil specimen has total volume of 80,000 mm 3 and weighs 145 g. The dry weight of the specimen is 128 g, and the density of the soil solids is 2.68 Mg/m 3 . Find the: (a) water content, (b) void ratio, (c) porosity, (d) degree of saturation, (e) wet density, and (f) dry density. Give the answers to parts (e) and (f) in both SI and British engineering units. SOLUTION: w w s 3 3 s s s w 3 v w w w (a) M 145 128 17 g M 17 w 100% 100 13.281 13.3% M 128 M 128 (b) V 47.7612 cm 47,761.2 mm G (2.68)(1) V 80,000 47,761.2 32,238.8 mm 32,238.8 e 0.67 47,761.2 32,238.8 (c) n 100% 40.3% 80,000 (d) V M (17)(1) − = = × = × = = = = = = ρ = − = = = = × = = × ρ = 3 3 3 3 3 3 3 3 g kg t cm m kg lbm t m ft kg dry m kg l dry m 17 cm 17,000 mm 17,000 S 100% 52.7% 32,238.8 145 (e) 1.8125 1812.5 80 1 1812.5 113.2 (see Appendix A) 16.018 1812.5 (f ) 1600.0 (1 0.13281) 1 1600.0 99.9 16.018 = = = × = ρ = = = ⎛ ⎞ ρ = × = ⎜ ⎟ ⎝ ⎠ ρ = = + ⎛ ⎞ ρ = × = ⎜ ⎟ ⎝ ⎠ 3 bm ft (see Appendix A) Index and Classification Properties of Soils Chapter 2 2.24. A sample of soil plus container weighs 397.6 g when the initial water content is 6.3%. The container weighs 258.7 g. How much water needs to be added to the original specimen if the water content is to be increased by 3.4%? After U.S. Dept. of Interior (1990). SOLUTION: t w s t w s s s s s w s w s M 397.6 258.7 138.9 g M 0.063M M 138.9 M M 0.063M M 1.063M M 130.668 g M w 0.034 M M w M (0.034)(130.668) 4.44 g = − = = = = + = + = = Δ Δ = = Δ = Δ × = = 2.25. A water-content test was made on a sample of clayey silt. The weight of the wet soil plus container was 18.46 g, and the weight of the dry soil plus container was 15.03 g. Weight of the empty container was 7.63 g. Calculate the water content of the sample. SOLUTION: s w w s M 15.03 7.63 7.40 g M 18.46 15.03 3.43 g M 3.43(100) (a) w 100% 46.351 46.3% M 7.40 = − = = − = = × = = =