Problem 3: Calculate the Dry Unit Weight for the five samples of soil below then graph them in the space provided and determine the maximum dry unit weight and its corresponding moisture content. BONUS: If a project requires at least 93% compaction in the soil under the footing, what range of water contents will be acceptable to achieve desired compaction/dry unit weight? Dry Unit Weight (lbs/ft³) 120 119 118 117 116 115 114 113 112 111 110 109 0 Weight (lbs) Compacted Samples of Soil Unit Dry Unit Weight Weight (lbs/ft³) (lbs/ft³) Volume (ft³) 3.94 0.033333 4.32 0.033333 4.44 0.033333 4.45 0.033333 4.41 0.033333 5 Water Content Compaction Results for Soil Samples 10 15 Moisture Content (%) 5% 10% 12.50% 15% 20% 20 25

Problem 3: Calculate the Dry Unit Weight for the five samples of soil below then graph them in the space provided and determine the maximum dry unit weight and its corresponding moisture content. BONUS: If a project requires at least 93% compaction in the soil under the footing, what range of water contents will be acceptable to achieve desired compaction/dry unit weight? Dry Unit Weight (lbs/ft³) 120 119 118 117 116 115 114 113 112 111 110 109 0 Weight (lbs) Compacted Samples of Soil Unit Dry Unit Weight Weight (lbs/ft³) (lbs/ft³) Volume (ft³) 3.94 0.033333 4.32 0.033333 4.44 0.033333 4.45 0.033333 4.41 0.033333 5 Water Content Compaction Results for Soil Samples 10 15 Moisture Content (%) 5% 10% 12.50% 15% 20% 20 25

Principles of Geotechnical Engineering (MindTap Course List)

9th Edition

ISBN:9781305970939

Author:Braja M. Das, Khaled Sobhan

Publisher:Braja M. Das, Khaled Sobhan

Chapter2: Origin Of Soil And Grain Size

Section: Chapter Questions

Problem 2.12P: Repeat Problem 2.11 with the following data. 2.11 The grain-size characteristics of a soil are given...

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Problem 1 The following grain size distribution test results were obtained from sieve analysis test. Percent Passing U.S. Sieve Number 4 100 10 90 20 64 40 38 08 200 18 13 Also, the Atterberg test results revealed the following results LL = 23 and PL = 19 Classify the soil by using a) AASHTO soil classification system b) USCS soil classification system Give group names and group symbols. Show the GSDC and sketch of finding necessary information from GSDC. Problem 2
Given the following Sieve Analysis and Atterberg limits results. Answer the following questions: 1- What is the effective diameter of this soil? 2- Calculate the uniformity coefficient and coefficient of gradation 3- Determine the percentage of Gravel, Sand, Fine soils according to USCS. 4- Classify the soil using AASHTO (with GI) & USCS (with group name). Diameter Sieve # % Passing 100 (тm) 90 4 4.75 96 80 3.35 90 70 10 78.8 60 20 0.85 56.8 50 40 0.425 37.8 40 60 0.25 23.6 30 100 0.15 13 20 200 0.075 4.6 10 Рan LL 10 10 0.1 0.01 Diameter PL 5 Percent Passing (%) 2.
Repeat Problem 2.11 with the following data. 2.11 The grain-size characteristics of a soil are given in the following table. a. Draw the grain-size distribution curve. b. Determine the percentages of gravel, sand, silt, and clay according to the MIT system. c. Repeat Part b using the USDA system. d. Repeat Part b using the AASHTO system.
Repeat Problem 2.8 using the following data. 2.8 The following are the results of a sieve and hydrometer analysis. a. Draw the grain-size distribution curve. b. Determine the percentages of gravel, sand, silt and clay according to the MIT system. c. Repeat Part b according to the USDA system. d. Repeat Part b according to the AASHTO system.
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)
The modified Proctor test (ASTM D 698, method C) was performed for a soil with G, = 2.70, and the results are as follows: Moist Weight in Mold, gf Water Content, % 6.5 3250 9.3 3826 12.6 4293 14.9 4362 17.2 4035 18.6 3685 (a) Plot the Ya versus w relation. (b) Determine Ya.max and wopt• (c) Calculate S and e at the maximum dry unit weight point. (d) What is Y at w. (e) What is the range of water content if the relative compaction (R.C.) is required to be 95% of the modified Proctor Ya.max? pt? opr?
Percent Passing (by weight) You are provided the mechanical sieve/hydrometer and Atterberg Limit test results for three soils, labeled as Soil #1, Soil #2, and Soil #3. The mechanical sieve results are shown in Figure 1, while Table 2 presents the Atterberg Limit test results. Classify the soils in accordance with the Unified Soil Classification System (USCS) and AASHTO. 100 90 80 60 50 40 30 20 10 0 0.0001 CLAY SIZE Soil #3 Soil #2 Soil #1 0.001 } Soil #1 #2 SILT SIZE #3 0.01 0.1 Grain Size (mm) Figure 1. Mechanical Sieve and Hydrometer Results for Problem #2. 0.075 mm Table 2. Atterberg Limits (ASTM D4318) for Problem #2. LL 68 39 SAND SIZE ΝΑ 1 PL 19 31 ΝΑ GRAVEL 10
Classify the soil using AASHTO. The liquid limit and plastic limit are found to be 30 and 20, respectively. The following data are obtained: Percent passing No. 4 sieve: 100% Percent passing No. 10 sieve: 91% Percent passing No. 40 sieve: 60% Percent passing No. 200 sieve: 20%
SITUATION 9 The result of sieve analysis of three soils is given next. It is required to classify these soils according to the AASHTO Classification System. Use the attached AASHTO Classification of Highway Subgrade Materials Table. (a) Classification of Soil A. (b) Classification of Soil B. (c) Classification of Soil C. Sieve Analysis Results for Situation 9 Sieve Analysis Sieve Diameter, Percent Passing Number mm. 4.76 90 100 100 8. 2.38 84 90 100 10 2. 72 77 98 20 0.84 66 59 91 40 0.42 58 50 85 60 0.25 50 42 79 100 0.149 44 36 70 200 0.074 38 33 LL 42 46 47 PL 23 29 24 63
10 11 percent. Classify the using the textural classification chart: a) Compute the percentage of clay. b). Compute the percentage of silt. c) Classify the type of soil. Problem: soil to be used as highway Classify the given subgrade material using the AASHTO method by using the attached table for its classification. Sieve analysis % finer No. 10 sieve = 47% %3D No.40 sieve = 28% No.200 sieve = 9% Plasticity for the minus No. 40 fraction Liquid limit 0 Plasticity index = 2%
The result of sieve analysis of three soils is given next. It is required to classify these soils according to the AASHTO Classification System. Use the attached AASHTO Classification of Highway Subgrade Materials Table. (a) Classification of Soil A. (b) Classification of Soil B. (c) Classification of Soil C. Sieve Analysis Results for Situation 9 Sieve Analysis Sieve Diameter, Percent Passing Number mm. 4.76 90 100 100 8. 2.38 84 90 100 10 2. 72 77 98 20 0.84 66 59 91 40 0.42 58 50 85 60 0.25 50 42 79 100 0.149 44 36 70 200 0.074 38 33 LL 42 46 47 PL 23 29 24 63
Classify the given soils using the Unified Soil Classification System (USCS). Give the group symbols and group names. (Show the path of the classification.) Sieve Analysis (percent finer) Soil No. No. 4 No. 200 LL PL Comments A 94 3 NP Cy = 4.48 and C = 1.22 - В 100 77 63 25 B.