The soil profile at a site has been investigated and the typical soil profile in the 100 feet has been determined to be: Depth Shear Wave Soil Description "N" PI (feet) Velocity 0- 30 Clay 85 30 - 50 Silty Sand 35 50 - 100 Sand and Gravel 50 What is the soil profile type according to 1997 UBC.
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- Subject : Geotechnical Engineering Please show me the complete solution with detailed. A soil PSD is given below. The soil fraction passing 425 micron sieve was tested for Atterberg limits and returned a plastic limit of 23% and a liquid limit of 63%. Name the soil according to AS1726-2017.The soil profile at a site has been investigated and the typical soil profile in the 100 feet has been determined to be: Depth Shear Wave Soil Description "N" PI (feet) Velocity 0- 30 Clay 85 30 - 50 Silty Sand 35 50 - 100 Sand and Gravel 50 What is the soil profile type according to 1997 UBC. Format of answer: Sx Sb O Sa O sd O Se O Sc OSf asap please!!! thank you o oFrom your Atterberg limits tests on the -40 portion of these samples, you have obtained the following data Percentage finer by weight 100% 80% 60% 40% 20% 0% Soil Sample Identification 0.001 H I J L FINES Liquid Limit 60 H 0.010 46 37 Plastic Limit 26 nonplastic #200 34 18 Fine HOT SAND Medium 0.100 Particle size (mm) 1.000 GRAVEL Course Fine Coarse #4 10.000 Classify the soils H, I, J and L using 1) USCS Classification 2) AASHTO Classification 3" 100.000 I LLII. I
- Answer ALL Questions Time: 2 Hours Q3: Classify the following soil by using the Unified soil classification system. Give the group symbols and the group names. Table 1. Sample 1 Grain Size Results (ASTM D422). Table 2. Atterberg Limits Results (ASTM D4318). Sleve No. Dia. (mm) % Passing Sample PL LL % in 12.7 100 G0 40 14 4.75 97 #10 2 94 #20 0.85 84 # 40 0.425 57 #60 0.25 32 140 0.106 15 #200 0.075 100 90 Atterberg Limit Results: LL = 60%, PL = 40% E 80 * 70- 60 - 50 40 30- a 20 10 1 30 a40 100 0.1200 10 80 0.01 Particle Diameter (mm) Answer ALL Questions Time: 2 Hours 70 CH or OH 60 A-line PI = 0.73(LL - 20) Or 50 40 30 CL MH OL 20 or CL-ML OH ML 10 - or OL 10 16 20 30 40 50 60 70 80 90 100 Liquid limit Figure 5.3 Plasticity chart Percent (%) Finer by Weight Plasticity index U-line PI = 0.(LL - 8) O Cengage Learning 2014The soil tested at the laboratory was found that the particle passing sieve No.200 is 89%, retained on sieve No. 4 is 3%, Liquid limit of 70%, organic matter of 25%, and Plasticity Index of 40%. Determine the type of soil * PLASTICITY CHART 60 50 CH A LINE: Pl 0,73(LL-20)| 30 CL MH&OH 20 CL ML ML&OL 10 20 30 40 50 60 70 80 90 100 LIQUID LIMIT (LL) (%) O Organic clay O Clay with high plasticity O Organic Silt O Sandy Clay PLASTICITY INDEX (PI) (%)Determine the soil classification Moisture content = 5.75% Limit liquid = 53.5 Plastic limit = 31.6 Granulometry Sieve % Pass No. 4 100 # 40 68 # 200 54.3 As in the plasticity graph the coordinates are difficult to read use IP = (wL-20), to obtain IP and compare.
- For a soil specimen, given the following: % passing sieve #4-92 %, sieve #10 -81%, sieve # 40-78%, LL-48%, PL-16%, GI-18.2 classify the soil by AASHTO classification system. GI = (F200-35)[0.2 + 0.005 (LL - 40)] +0.01 (F20015)(PI – 10)A Classify the soil samples shown below using AASHTO and USCS systems: Sieve No. 4 10 20 40 60 100 200 LL PL A - 68.5 - 36.1 21.9 34.1 16.5 Soil, % passing B . 79.5 69.0 - 54.3 53.5 31.6 C 69.3 59.1 48.3 38.5 28.4 19.8 4.5 Non-plastic (NP)• Particle size analyses were carried out on two soils-Soil A and Soil B-and the particle size distribution curves are shown in Figure % Finer Clay 100 90 80 70 60 50 40 30 20 10 0 0.001 + Silt Soil A Sand Soil B 0.01 0.1 1 Particle size (mm) - log scale Gravel 10 Soil A B 1. Classify these soils according to USCS and ASTM-CS 2. Is either of the soils organic? LL 26 (oven-dried; assume same for not dried) Nonplastic PL 18
- Classify Soil C (group classification and group index) using AASHTO method. SOIL SAMPLE SIEVE NO. DIA. (mm) A В # 4 4.76 90 100 100 # 8 2.38 88 90 100 # 10 2 86 77 98 # 20 0.84 84 59 92 # 40 0.42 82 56 84 # 60 0.25 80 54 79 #100 0.15 78 51 75 #200 0.075 76 48 72 Characteristics of # 40 fraction LL 38 28 50 PL 30 45ZVo @ e l:11 LTE zain IQ Iı. elearning7.uokufa.edu.iq A Question 3 Not yet answered Marked out of 30 P Flag question For a soil sample in a natural condition, void ratio (e) is 0.67, degree of saturation (S) is 75%, Gs=2.7, calculate: a) dry unit weight b) total unit weight c) saturated unit weight d) initial water content and water content at saturation. f) if the void ratio was reduced to 0.5, what will be the final specific gravity. a. 15.86 kN/m3, 18.81 kN/m3,19.79 kN/m3,0.186,0.248, 2.7 respectively b. 15.86 kN/m3, 18.81 kN/m3,19.75 kN/m3,0.166,0.248, 2.7 respectively c. 15.84 kN/m3, 15.46 kN/m3,15.26 kN/m3,0.186,0.248, 2.7 respectively d. 15.86 kN/m3, 15.86 kN/m3,15.86 kN/m3,0.166,0.238, 2.67 respectively ->1. Classify the following soil to be used as a highway subgrade material using AASHTO method. Sieve Analysis % Finer No. 10 sieve = 88% No. 40 sieve = 75% No. 200 sieve = 34 % Plasticity for fraction passing No.4 Liquid Limit = 39% Plasticity Index = 12% %3D