LL=25% PI=15% a. D10, D30, and D60 b. Uniformity coefficient, Cu
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Following are the results of a sieve analysis from laboratory activity. Show the necessary calculation and draw the particle size distribution:
LL=25%
PI=15%
a. D10, D30, and D60
b. Uniformity coefficient, Cu
c. Coefficient of gradation, Cc
d.) USCS Classification
Step by step
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- Sieve # Maas retained (gr) 4 0 10 11,5 16 11,7 20 12 30 12,9 40 17,8 50 19,1 70 35,1 100 38,1 200 118 Pan 335 Draw grain size distribution curve,find D60, D30, D10 Classify the soilDo the following: a) Calculate the percent retained and percent passing each sieve, and graph the results on a semi-log graph (you may use the attached chart, or graph in Excel and include with your submission). Then, characterize the gradation of the material. b) Calculate the values of D10, D30, D60, Cu, and Cc c) If the LL is 48 and the PL is 26, determine the USCS classification of the soil, and state its suitability for the purpose stated in the scenario.SHOW ALL SOLUTIONS a. Effective size b. Uniformity Coefficient c. Coefficient of Gradation d. Sorting Coefficient e. Percent of gravel, sand, silt and clay of MIT f. Percent of gravel, sand, silt and clay of AASHTO g. Percent of gravel, sand, silt and clay of USDA h. Percent of gravel, sand, silt and clay of USCS
- Calculate the AASHTO group index number of the following soil Liquid Limit = 36% Plastic Index = 14% Sieve analysis data: US Sieve Size % Passing No. 10 72 No. 40 61 No. 200 33The coefficient of permeability (K value) of a soil describes how easily a liquid will move through a soil. True )False Permeability measurement is used to: calculation of seepage through earth dams calculation of seepage under sheet pile walls estimating the seepage rate from waste storage O all of the answers are correct Which state describes soil that does not show cracks when deformed (choose correct state that applies). A) Plastic state B) Liquid State C) Semisolid state D) Solid stateThe 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) (%)
- For a soil sample the liquid limit is 50, plasticity index is 10 and the Percent passing No. 4 sieve is 85 Percent passing No. 10 sieve is 77 Percent passing No. 40 sieve is 69 Percent passing No. 200 sieve is 41 Classify the soil sample using AASHTO classification system. A) A-7-6(1) B) A-7-5(1.25) C) A-7-6(2) D) A-7-5(1.5)Three groups of students from the Geotechnical Engineering class collected soil-aggregate samples for laboratory testing from a recycled aggregate processing plant in Palm Beach County, Florida. Three samples, denoted by Soil A, Soil B, and Soil C, were collected from three locations of the aggregate stockpile, and sieve analyses were conducted (see Figure 2.35). (a) (b) Figure 2.35 (a) Soil-aggregate stockpile; (b) sieve analysis (Courtesy of Khaled Sobhan, Florida Atlantic University, Boca Raton, Florida) a. Determine the coefficient of uniformity and the coefficient of gradation for Soils A, B, and C. b. Which one is coarser: Soil A or Soil C? Justify your answer. c. Although the soils are obtained from the same stockpile, why are the curves so different? (Hint: Comment on particle segregation and representative field sampling.) d. Determine the percentages of gravel, sand and fines according to Unified Soil Classification System.PROBLEM No. 1 Particle Size Distributior Weight of Container Weight of Container + Dry Soil Weight of Dry Soil Sample 198 g 722 g 524 g Mass of Mass of Mass of Soil Sieve Diameter Percent Percent Empty Sieve (g) 116.00 Sieve + Soil Retained No. (mm) Retained Passing Retained (g) (g) 4 4.750 2.000 148.00 10 99.00 213.00 20 0.840 0.425 98.00 175.00 40 97.00 182.00 60 0.250 94.00 157.00 140 0.106 93.00 184.00 200 0.075 91.00 117.00 Pan 70.00 106.00 Total Weight of Soil = Determine the following: Using USCS: % Gravel (Hint: use interpolation in lieu of graphing) D10 = D30 = D60 mm mm % Sand mm % Fines Cu = Cc =
- AASHTO SOIL CLASSIFICATION (by AASHTO) AASHTO = American Association of State Highway and Transport Officials Situation 2 Classification of Highway Subgrade Materials % FINER SOIL B SIEVE General classification Granular materials (35% or less of total sample passing No. 200) SOIL A SOIL C 10 83 100 90 A-1 А-2 Procedure: 40 48 92 76 Group classification A-1-a А-1-b А-3 A-2-4 A-2-5 A-2-6 A-2-7 O Group Classification with required test data in mind, proceed from left to right, in chart. The correct group will be found by process of elimination. The first group from the left consistent with the test data is the corect classification. 86 34 Sieve analysis (percentage passing) No. 10 200 20 LL 20 70 37 50 max. 30 max. No. 40 No. 200 Characteristics of fraction PI 32 18 50 max. 51 min. 15 max. 25 max. 10 max. 35 max. 35 max. 35 max. 35 max. Using AASHT0 method: 1. Determine the group symbol and index for soil A. 2. Determine the group symbol and index for soil B. 3. Determine the group symbol…5. Determine which isotherm equation (linear, Freundlich or Langmuir) best fits the GAC adsorption test data. Also determine the corresponding coefficients along with their units. The liquid volume for the tests was 1.0 liter. Show your trails for the three cases. Mass of Adsorbent Equilibrium concentration of adsorbate in solution mg/1 0 0.00175 0.0175 0.175 0.875 5.90 5.72 4.85 3.26 2.33Q#4: Provide three applications of each in light of soil stability. Hydraulic conductivity Liquid limit Plasticity index Moisture content Permeability SIR, PLS GIVE ME PROPER AND CORRECT ANSWER WITH FULL DETAILS. THANKS