HW05-CE357-Compaction_Solution_SI

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.

Q1: Classify the below soil according to Unified Soil Classification System (USCS) and .the AASHTO System Sieve No. sieve Diameter 4.75 2 1.18 06 pass % 65 4. 10 52 40 16 30 40 26 0.425 0.25 0.075 21 60 14 200 10 And the Liquid Limit (LL) = 35. Plastic Limit (PL) = 18 100 90 80 70 60 50 40 30 20 10 0.01 0.1 10 Sieve Dia mm passing%

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

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?

Q1: Classify the below soil according to Unified Soil Classification System (USCS) and the AASHTO System. Sieve No. sieve pass % 65 Diameter 4 4.75 10 52 16 1.18 40 30 0.6 26 40 0.425 21 60 0.25 14 200 0.075 10 And the Liquid Limit (LL) = 35 , Plastic Limit (PL) = 18 100 90 80 70 60 50 40 30 20 10 0.01 0.1 10 Sieve Dia mm % Buyssed

1. Classify the following soils using USCS and give the group symbols. Sieve Analysis, % finer Soil Liquid Limit No. 4 No. 200 1 (SC) 70 30 33 2 (GC) 48 20 41 3 (CH) 95 70 52 4 (CL) 100 82 30 5 (CL) 100 74 35 6 (SC) 87 26 38 7 (CH) 88 78 69 8 (CH) 99 57 54 9 (SP-SC) 71 11 32 10 (SW) 100 2 11 (CL) 89 65 44 12 (SP-SC) 90 8 39 Plasticity Index 21 22 28 19 21 18 38 26 16 NP 21 31 Cu 4.8 7.2 3.9 Cc 2.9 2.2 2.1

Problem 1: For a gravel with D60 0.48mm, D30 0.25mm, and D10 coefficient and the coefficient of gradation. Is it a well-graded or poorly graded soil? - = = = 0.11mm, calculate the uniformity Problem 2: 0.3 mm, D30 The following values for a sand are given: D10 Determine Cu and Cc and state if it is a well-graded or a poorly graded soil. = 0.41 mm, and D60 = = 0.77 mm.

Question 18 Classify the following soils using the Unified soil classification system. Give the group symbol. No. 4 = 92 No. 200 = 48 LL = 30 PL = 8 A) SP В GC SM D SC

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

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 2014

L II. Table 02 Sieve analysis data of coarse-grained soil Sieve Opening (mm) 20 10 4.75 2.36 1.18 0.6 0.3 0.15 0.075 Pan Mass of Soil Retaining on each Sieve (g) 5.27 20.27 180.27 360.27 280.17 70.17 50.27 17.27 10.17 8 Percentage of clay, silt, sand and Gravel in the soil sample Uniformity coefficient and coefficient of curvature Classify the soil according to BS soil classification system

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%

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)

2. The sieve analysis data for a granular soil are given below. NICS LECTUREY % passing 19 mm sieve =100 % passing 4.75 mm (4 sieve) = 61 % passing 2 mm = 45 % passing 0.075 mm (200 sieve) = 16 %3D % passing 0.06 mm (4 sieve) = 13 % passing 0.05 mm ( sieve) = 8 % passing 0.002 mm (4 sieve) = 5 What are the percentages of gravels, sands and fines within this soil, based on the Unified Soil Classification System (USCS), AASHTO, USDA and MIT Standards? Mass Opening (mm) Sieve No. Retained (g) 4.75 4

The soil sample is 75% gravel with laboratory result shown below. Is it well graded or poorly graded? D10 = 0.08 mm D30 = 0.22 mm D60 = 0.41 mm   Select one: a. Poorly graded b. Well graded

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

Percentage finer by weight 100% 80% 60% 40% 20% 0% 0.001 (c) Soil J (d) Soil L H 0.010 Ma 1.000 Soil Sample ID H I J #200 L 0.100 Particle size (mm) 46 37 #4 Liquid Limit Plastic Limit 60 26 Non plastic 34 18 10.000 AASHTO Classification: Classify the following soils using AASHTO criteria. (a) Soil H (b) Soil I 3° 100.00

A sample soil is obtained and sieve analysis is performed to determined its particle distribution. Sieve Opening Mass Retained % Finer # 6 3.35 146 # 8 2.36 88 #16 1.18 125 # 40 0.425 56 #50 0.3 42 #140 0.106 127 # 200 0.075 126 Pan 29 Determine: 1. Effective Size (in mm and in 4 decimal places) 2. Uniformity Coefficient (in 2 decimal places) 3. Coefficient of Curvature (in 2 decimal places) 4. Sorting Coefficient (in 2 decimal places)

dys.mu MSKÜ-LMS 016 ŞB: 1 ÖRGÜN Soil Mechanics Mühendislik Fakültesi Mac Atterberg limits of a soil sample are given: LL:55% and PL:25%. Give the name of the soil typle considering the chart given below? Remember PI=LL-PL 60 U-line PI - 0.9 (LL -8) 50 Z 40 CH or OH A-line PI0.73 (LL - 20) 30 CL OH 20 or OL MH CL- ML 10 ML& OL 10 20 30 40 50 60 70 80 90 100 Liquid limit (%) O a. CL or OL O b. CH OR OH O c. OH or ML O d. GL or ML O e. M ail E AR .ill lenovo Plasticity index (%)

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.

soil mechanic. Classify the soil according to unified soil classification system USCS; if the following data is given : Passing sieve No. 200 =7% / Passing sieve No. 4 = 40% / D10= 0.18, D30= 0.69 and D60= 3/ LL-28%, Pl=19%

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)

Given the data of sieve analysis of the soil; 1. Determine the approximate effective size of the soil (choices: 0.115mm, 0.140mm, 0.093mm, 0.079mm) 2. Determine the approximate coefficient of curvature (choices: 0.84, 1.03, 0.51, 0.69) 3. Classify the soil using the AASHTO classification system

These equations can be applied for dry soil, except: a) Sre = wGs b) Gspw /1+ e c) Ms / V d) Pp/1 + w

Question 4 Classify the following soils using the Unified soil classification system. Give group symbols and group names. % Passing No. 4: 99 % Passing No. 200: 76 LL = 60 PL = 32 %3D Group Symbol: Blank 1 Group name: Blank 2 Blank 1 Add your answer Blank 2 Add your answer

Soil lab Particle size distribution (Sieve method) In addition to that, calculate the value of D25 and D75 and calculate it in curvature

Q.1. Select the right answer of the following: 1. The void ratio of soil can be: a. greater than void ratio b. less than void ratio c. equal to void ratio d. no one 2. When Gravel had coefficient of uniformity of 3, the soil could be according USCS: a. GW b. GP c. SP d. any one of them 3. In a wet soil mass, air occupies 1/6 of volume and water occupies 1/3 of volume. The void ratio is: a. 0.25 b. 0.50 4. For the soil with LL= 45%, PL = 25% and water content = 15% the liquidity index is: B.0 с. 1.50 d. 1.00 a.0.6 5. The type of unit weight of soil s depends on: с. 1 dinone of them а. е b. n c. YW d. S 6. Soil has LL of 30, the corresponding plasticity index given by A-line: a. 7.3 b. 7.5 c. 9.0 d. 9.5 7. The porosity in of soil is function of: c. Vv & n P.= L.L-PL 12:45-25=2 a. Vv&Vs b. Vv& Vt d. Vs & n 8. For a soil deposit having n= 33% and G 2.6, ysaturated= 15KN/m3, S is: %3D а. 1.0 b. 1.05 с. 1.07 d. 1.10 10. When saturation in soil sample occurs, the degree of saturation: a.…

5.5 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 Wopt? What is the range of water content if the relative compaction (R.C.) is required to be 95% of the modified Proctor Ya.max?

5.5 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 Content, % in Mold, gf Water 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.