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E. EXAMPLE PROBLEMS AND SOLUTIONS E.1. CHAPTER 2 2.1 A sampling tube has an outside diameter Do of 3.00 inches, a tip diameter De of 2.84 inches, and a wall thickness of 0.065 inches. Calculate the clearance ratio, area ratio, and indicate if the sampling tube meets the criteria for undisturbed soil sampling. Solution: Di = Do − 2 t (where t = wall thickness) Di = 3.00 − (2)(0.065) = 2.87 in. Do = 3.0 in. From Table 2.8, inside clearance ratio = 100 (Di −De)/De = 100 (2.87 − 2.84)/2.84 = 1.06% From Table 2.8, area ratio = 100 (Do 2 −De2)/De2 = 100 (32 − 2.842)/2.842 = 11.6% These values are close to meeting the criteria for undisturbed soil sampling. 2.2 A standard penetration test SPT was performed on a near surface deposit of clean sand where the number of blows to drive the sampler 45 cm was 5 for the first 15 cm, 8 for the second 15 cm, and 9 for the third 15 cm. Assume that Em = 60% (i.e., E m = 0.60), the borehole diameter is 100 mm, and the drill rod length is 5 m. Calculate the measured SPTN-value (blows per foot), N60, and (N1 )60 assuming that the vertical effective stress σ′vo = 50 kPa. Also indicate the density condition of the sand. Solution: N-value = 8 + 9 = 17 For 100 mm borehole diameter,Cb = 1.0 For drill rod length = 5 m,Cr = 0.85 From Table 2.11: N60 = 1.67 E m Cb Cr N = (1.67)(0.60)(1.0)(0.85)(17) = 14.5 From Table 14.22: (N1) 60 = CN N60 = (100/σ′vo)0.5N60 = (100/50)0.5 (14.5) = 20.4 Per Table 2.12, for N60 = 14.5, the sand is in a medium condition 2.3 A field vane shear test was performed on a clay, where the rectangular vane had a lengthH of 4.0 in. and a diameter D of 2.0 in. The maximum torque T max required to shear the soil was 8.5 ft-lb. Calculate the undrained shear strength s u of the soil. Solution: H = 4 in. = 0.333 ft D = 2 in. = 0.167 ft From Table 2.13: su = Tmax /[π (0.5 D2 H + 0.167 D3 )] = 8.5/[π (0.5 (0.167)2 0.333 + 0.167 (0.167)3 )] = 500 psf

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2.4 A construction site in New England requires excavation of rock. The geologist has determined that the rock is granite and from geophysical methods (i.e., seismic refraction), the seismic velocity of the in situ granite is 12,000 to 15,000 feet per second. A Caterpillar D11R tractor/ripper is available. Can the granite be ripped apart? Solution: D11R tractor/ripper, therefore use Fig. 2.17. For granite with a seismic velocity = 12,000 to 15,000 feet per second, it is nonrippable. 2.5 A plate load test is performed where a 1-foot wide square steel plate is subjected to a vertical stress of 2,000 psf and the recorded depth of penetration of the steel plate is 0.25 inch. If the width of the actual square footing will be 10 feet, calculate the settlement of the footing if it is subjected to a vertical stress of 2,000 psf. Solution: S 1 = 0.25 in., D1 = 12 in., D = (10)(12) = 120 in. From Table 2.13: S = 4 S1 /(1 +D1 /D)2 = (4)(0.25)/[1 + (12/120)]2 = 0.83 in. 2.6 Figure E.1 presents data on shallow seismic refraction surveys. Field testing data indicates that point A = 0.04 second and point d′ = 100 feet. At the junction of the V2 and V3 straight line segments, the travel time is 0.07 seconds at a distance of 300 feet. Determine the thickness of the upper stratum (i.e., H1 ). Can a Caterpillar D11R rip the upper and lower stratum? Figure E.1 Shallow seismic refraction survey for Problem 2.6. (From Hvorslev 1949.)

Solution:

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V2 = (300 ft)/(0.07 − 0.04) = 10,000 ft/sec (3,000 m/s) Determining the travel time at the junction of theV1 and V2 straight line segments: Travel time = 0.04 + (100)/10,000 ft/sec = 0.05 sec V1 = 100 ft/0.05 sec = 2,000 ft/sec (600 m/s) sin α = V1 /V2 = 2,000/10,000 = 0.2, or α = 11.5° H1 = [(T1 V1 )/(2 cos α)] = [(0.04)(2,000)]/(2 cos 11.5°) = 41 feet (12 m) Checking using the following equation, where d′ = 100 feet (30.5 m) H1 = ½ d′ [(V2 − V1 )/(V2 + V1 )]½ = ½ (100) [(10,000 − 2,000)/ (10,000 + 2,000)]½ = 41 feet (12 m) Since the upper stratum has a low seismic wave velocity of 2,000 ft/sec (600 m/sec), it will be easy for a Caterpillar D11R to remove this material. For the lower stratum, the seismic wave velocity is 10,000 ft/sec (3,000 m/sec) and it could be ripped if it is shale, but granite would probably be non-rippable (see Fig. 2.17).

E.2. CHAPTER 3 3.1 A water content test was performed on a specimen of soil. The following data was obtained: Mass of empty container = 105.6 g Mass of the container plus wet soil = 530.8 g Mass of the container plus dry soil = 483.7 g Calculate the water content of the soil. Solution: Mass of water = 530.8 − 483.7 = 47.1 g Mass of dry soil = 483.7 − 105.6 = 378.1 g From Table 3.3: w (%) = 100 (Mw/Ms ) = 100 (47.1/378.1) = 12.5% 3.2 Using the data from Problem No. 3.1 and assuming that the initial volume of the wet soil specimen is equal to 225 cm3 , calculate the total unit weight and the dry unit weight. Solution: From Table 3.4, total density ρt = M/V = (530.8 − 105.6)/225 = 1.89 g/cm3 = 1.89 Mg/m3 Total unit weight γ t = (ρt)(g) = (1.89 Mg/m3)(9.81 m/sec 2) = 18.5 kN/m3 Dry unit weight γd = γt/(1 + w) = 18.5/(1 + 0.125) = 16.5 kN/m3 3.3 A clay specimen has been obtained from below the groundwater table and the total unit weight γt of the soil specimen is 19.5 kN/m3 (124 pcf). Calculate the buoyant unit weight of the clay. Solution:

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From Table 3.4: γb = γsat − γw = 19.5 − 9.8 = 9.7 kN/m3 (61.6 pcf) 3.4 A specific gravity test was performed on a granular soil. The mass of dry soil used for the specific gravity test was equal to 102.2 g. The dry soil was placed in a pycnometer and it was determined that the volume of the soil particles was equal to 38.9 cm 3. Determine the specific gravity of solidsG. Solution: From Table 3.5: G = ρs /ρw = (Ms /Vs )/ρw = (102.2 g/38.9 cm3 )/1.0 g/cm3 = 2.63 3.5 In a constant head permeameter test, the outflow Q is equal to 782 mL in a measured time of 31 seconds. The sand specimen has a diameter of 6.35 cm and a length L of 2.54 cm. The total head loss Δh for the permeameter is 2.0 m. Calculate the hydraulic conductivity (also known as the coefficient of permeability). Solution: Q = 782 mL = 782 cm3, t = 31 seconds, L = 2.54 cm, Δh = 2.0 m = 200 cm D = 6.35 cm and therefore A = 31.67 cm2 From Table 3.14: k = Q L/(ΔhAt) = [(782)(2.54)]/[(200)(31.67)(31)] = 0.01 cm/s 3.6 In a falling head permeameter test, the time required for the water in a standpipe to fall fromho = 1.58 m to hf = 1.35 m is 11.0 hours. The clay specimen has a diameter of 6.35 cm and a length L of 2.54 cm. The diameter of the standpipe is 0.635 cm. Calculate the hydraulic conductivity. Solution: For the standpipe, diameter = 0.635 cm, therefore a = 0.317 cm2 For the specimen, diameter = 6.35 cm, therefore A = 31.7 cm2 h o = 1.58 m, hf = 1.35 m, t = 11 hr = 39,600 seconds, L = 2.54 cm From Table 3.14: k = 2.3 [(aL )/(At)] log (ho /hf) = 2.3 [(0.317)(2.54)/ (31.7)(39,600)] log (1.58 /1.35) = 1.0 × 10−7 cm/s 3.7 An in situ soil has a void ratio of 0.79. Laboratory tests indicate the sand has a maximum void ratio = 0.85 and a minimum void ratio = 0.30. Calculate the relative density and indicate the density state of the sand. Solution: From Table 3.16: D r = 100 (e max − e)/(e max − e min) = 100 (0.85 − 0.79)/(0.85 − 0.30) = 11% (very loose condition)

E.3. CHAPTER 4 4.1 A soil has the following particle size gradation based on dry mass:

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Gravel and sand size particles coarser than No. 40 sieve

=

65%

Sand size particles finer than No. 40 sieve

=

10%

Silt size particles

=

5%

Clay size particles (finer than 0.002 mm)

=

20%

Total =

100%

In accordance with ASTM test procedures, the Atterberg limits were performed on the soil finer than the No. 40 sieve and the results are LL = 93 and PL = 18. Calculate the activity (A) of this soil. Solution: 35% passes the No. 40 sieve, and therefore the clay size fraction = 20/35 = 57% From Table 4.1: Activity = PI/%clay fraction = (93 − 18)/57 = 1.3 4.2 A clay has a liquid limit LL = 60 and a plastic limit PL = 20. UsingFig. 4.2, determine the predominate clay mineral in the soil. Solution: Plasticity index = LL − PL = 60 − 20 = 40 Entering Fig. 4.2 with liquid limit = 60 and plasticity index = 40, the predominate clay mineral in the soil is montmorillonite. 4.3 Based on a particle size analysis, the following values were obtained: D60 = 15 mm, D50 = 12 mm, D30 = 2.5 mm, and D10 = 0.075 mm. Calculate C u and Cc. Solution: From Table 4.3: Cu = D60 /D10 = 15/0.075 = 200 Cc = D302 /(D10 D60 ) = (2.5)2 /[(0.075)(15)] = 5.6 4.4 For the soil in Problem No. 4.3, laboratory testing indicates that 18% of the soil passes the No. 40 sieve and the LL = 68 and PI = 34 for this soil fraction. What is the USCS group symbol? Solution: Using the data from Problem No. 4.3 and recognizing that 0.075 mm is the opening of the No. 200 sieve, therefore the percent passing No. 200 sieve = 10%. Since D50 = 12 mm, which is a larger size than the No. 4 sieve, the majority of the soil particles are gravel. From Table 4.4, because there are 10% fines, dual symbols must be used. The value of Cc does not meet the requirements for well-graded gravel. Also, the limits (LL = 68, PI = 34) plot below the A-line in Fig. 4.1. Hence, the dual classification is GP-GM. 4.5 A sand has a Cu = 7 and a Cc = 1.5. The sand contains 4% nonplastic fines (by dry mass). What is the USCS group symbol? Solution: From Table 4.4, based on the values of Cc and Cu , the sand is well-graded. Because of 4% nonplastic fines, the group symbol is SW.

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4.6 An inorganic soil has 100% passing (by dry mass) the No. 200 sieve. The LL = 43 and the PI = 15. What is the USCS group symbol? Solution: From Table 4.4, since all the soil particles pass the No. 200 sieve, the soil is fine-grained. The limits plot below the A-line in Fig. 4.1 and the LL is less than 50. Thus the group symbol is ML. 4.7 An inorganic clay has a LL = 60 and a PL = 20. What is the USCS group symbol? Solution: The plasticity index PI = LL − PL = 60 − 20 = 40. Based on a LL = 60 and a PI = 40, the limits plot above the A-line inFig. 4.1. Thus for the clay, the group symbol is CH. 4.8 A fine-grained soil has a black color and organic odor. The LL = 65 on non-oven dried soil and LL = 40 on oven-dried soil. What is the USCS group symbol? Solution: The LL (oven dry) divided by the LL (not dried) = 40/65 = 0.62 and therefore fromTable 4.4, the soil is an organic soil. Since the LL is greater than 50, the group symbol is OH. 4.9 Based on dry mass, a soil has 20% gravel size particles, 40% sand size particles, and 40% fines. Based on dry mass, 48% of the soil particles pass the No. 40 sieve and the LL = 85 and PL = 18 for soil passing the No. 40 sieve. What is the USCS group symbol? Solution: PI = LL − PL = 85 − 18 = 67 From Table 4.4, more than 50% of the soil particles are retained on the No. 200 sieve and therefore it is a coarse grained soil. The majority of the soil particles are of sand size, there are greater than 12% fines, and the limits (LL = 85, PI = 67) plot above the A-line in Fig. 4.1, and therefore the group symbol is SC. For Problem Nos. 4.10 through 4.14, use the laboratory testing data summarized in Fig. E.2. Consider all five soils to consist of inorganic soil particles with the Atterberg limits performed on soil passing the No. 40 sieve (per ASTM D 4318). Note in Fig. E.2 that LL is the liquid limit, PL is the plastic limit, PI is the plasticity index, and NP indicates that the soil is nonplastic.

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Figure E.2 Particle size distributions for Problems 4.10 to 4.14 (plot developed by gINT computer program).

4.10 For SB-14 at 8 to 12 feet in Fig. E.2, determine the soil classification per the USCS and AASHTO soil classification systems. Solution: Unified Soil Classification System (USCS): Since 28.4% of the soil particles pass the No. 200 sieve (i.e., 28.4% fines), the soil is coarse-grained. Of the soil particles retained on the No. 200 sieve, 50.6% are gravel-size particles (i.e., 36.2/0.716 = 50.6%) and 49.4% are sand-size particles (i.e., 35.4/0.716 = 49.4%). Since the larger fraction consists of gravel size particles, the primary soil classification is gravel (Table 4.4). Since there are more than 12% fines and the soil is nonplastic (PI = 0), the soil classification is silty gravel (GM). AASHTO Soil Classification System: Since there are less than 35% passing the No. 200 sieve, the soil is classified as a granular material. Using the particle size distribution (Fig. E.2), the percent passing is as follows: Percent passing No. 10 sieve = 55% Percent passing No. 40 sieve = 42% Percent passing No. 200 sieve = 28% For these percent passing values, and because the soil is nonplastic (PI = 0), all of the criteria are met for group A-2-4 (see Table 4.6).

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Summary: USCS: Coarse-grained soil, silty gravel (GM) AASHTO: Granular material, silty gravel and sand (A-2-4) 4.11 For SB-20 at 0 to 4 feet inFig. E.2, determine the soil classification per the USCS and AASHTO soil classification systems. Solution: Unified Soil Classification System (USCS): Since 27.8% of the soil particles pass the No. 200 sieve (i.e., 27.8% fines), the soil is coarse-grained. Of the soil particles retained on the No. 200 sieve, 10.8% are gravel-size particles (i.e., 7.8/0.722 = 10.8%) and 89.2% are sand-size particles (i.e., 64.4/0.722 = 89.2%). Since the larger fraction consists of sand size particles, the primary soil classification is sand (Table 4.4). Since there are more than 12% fines and the soil is nonplastic (PI = 0), the soil classification is silty sand (SM). AASHTO Soil Classification System: Since there are less than 35% passing the No. 200 sieve, the soil is classified as a granular material. Using the particle size distribution (Fig. E.2), the percent passing is as follows: Percent passing No. 10 sieve = 82% Percent passing No. 40 sieve = 53% Percent passing No. 200 sieve = 28% For these percent passing values, and because the soil is nonplastic (PI = 0), all of the criteria are meet for group A-2-4 (see Table 4.6). Summary: USCS: Coarse-grained soil, silty sand (SM) AASHTO: Granular material, silty sand (A-2-4) 4.12 For SB-25 at 4 to 8 feet inFig. E.2, determine the soil classification per the USCS and AASHTO soil classification systems. Solution: Unified Soil Classification System (USCS): Since 40.2% of the soil particles pass the No. 200 sieve (i.e., 40.2% fines), the soil is coarse-grained. Of the soil particles retained on the No. 200 sieve, 25.3% are gravel-size particles (i.e., 15.1/0.598 = 25.3%) and 74.7% are sand-size particles (i.e., 44.7/0.598 = 74.7%). Since the larger fraction consists of sand size particles, the primary soil classification is sand (Table 4.4). Since there are more than 12% fines and the soil is nonplastic (PI = 0), the soil classification is silty sand (SM). AASHTO Soil Classification System: Since there are more than 35% passing the No. 200 sieve, the soil is classified as a silt-clay material. Because the soil is nonplastic (PI = 0, LL = 0), all of the criteria are met for group A-4. By inserting values into the group index equation (see Table 4.6), a negative value is obtained and thus the group index = 0. Summary:

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USCS: Coarse-grained soil, silty sand (SM) AASHTO: Silt-clay material, silty soil A-4 (0) 4.13 For SB-29 at 4 to 8 feet inFig. E.2, determine the soil classification per the USCS and AASHTO soil classification systems. Solution: Unified Soil Classification System (USCS): Since 14.8% of the soil particles pass the No. 200 sieve (i.e., 14.8% fines), the soil is coarse-grained. Of the soil particles retained on the No. 200 sieve, 18.0% are gravel-size particles (i.e., 15.3/0.852 = 18.0%) and 82.0% are sand-size particles (i.e., 69.9/0.852 = 82.0%). Since the larger fraction consists of sand size particles, the primary soil classification is sand (Table 4.4). Since there are more than 12% fines and the soil is nonplastic (PI = 0), the soil classification is silty sand (SM). AASHTO Soil Classification System: Since there are less than 35% passing the No. 200 sieve, the soil is classified as a granular material. Using the particle size distribution (Fig. E.2), the percent passing is as follows: Percent passing No. 10 sieve = 77% Percent passing No. 40 sieve = 39% Percent passing No. 200 sieve = 15% For these percent passing values, and because the soil is nonplastic (PI = 0), all of the criteria are meet for group A-1-b (see Table 4.6). Summary: USCS: Coarse-grained soil, silty sand (SM) AASHTO: Granular material, gravel and sand mixture (A-1-b) 4.14 For SB-38 at 4 to 8 feet inFig. E.2, determine the soil classification per the USCS, AASHTO, and USDA soil classification systems. Solution: Unified Soil Classification System (USCS): Since 52.8% of the soil particles pass the No. 200 sieve (i.e., 52.8% fines), the soil is fine-grained. Since the Atterberg limits (i.e., liquid limit = 50 and plasticity index = 35) plot above the A-line in Fig, 4.1, the soil is classified (Table 4.4) as sandy clay of high plasticity (CH). AASHTO Soil Classification System: Since there are more than 35% passing the No. 200 sieve, the soil is classified as a silt-clay material. With a plasticity index = 35, a liquid limit = 50, and since the LL − 30 is less than the plasticity index, the soil meets all the classification requirements for group A-7-6 (clayey soils). Using the group index equation listed in Table 4.6, with F = 52.8, LL = 50, and PI = 35, the group index = 13.9. The soil classification is therefore an A-7-6 (14), clayey soil. USDA Textural Classification System: The percent sand, percent silt, and percent clay size particles are as follows:

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Sand (2 mm to 0.05 mm) = (99 − 50.5)/0.99 = 49% Silt (0.05 mm to 0.002 mm) = (50.5 − 31.5)/0.99 = 19% Clay (finer than 0.002 mm) = 31.5/0.99 = 32% From Fig. 4.4 with the percent sand = 49%, percent silt = 19%, and percent clay = 32%, the soil classification is sandy clay loam. Summary: USCS: Fine-grained soil, sandy clay of high plasticity (CH) AASHTO: Silt-clay material, clayey soil, A-7-6 (14) USDA: Sandy clay loam For Problem Nos. 4.15 through 4.18, use the laboratory testing data summarized in Fig. E.3. Consider all four soils to consist of inorganic soil particles with the Atterberg limits performed on soil passing the No. 40 sieve (per ASTM D 4318). Note in Fig. E.3 that LL is the liquid limit, PL is the plastic limit, PI is the plasticity index, and NP indicates that the soil is nonplastic. Figure E.3 Particle size distributions for Problems 4.15 to 4.18 (plot developed by gINT computer program).

4.15 For SB-39 at 4 to 8 feet inFig. E.3, determine the soil classification per the USCS, AASHTO, and USDA soil classification systems.

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Solution: Unified Soil Classification System (USCS): Since 39...