Q3The results of a particle size analysis of a soil are given in the followingTable. No Atterberg limit tests were conducted.Sieve No.Percent finerNo.40 No.100 | No.20049.8No.4No.10 No.2070.262.59.53 mm10089.828.64.1a) Would you have conducted Atterberg limit tests on this soil? Justify youranswer.b) Classify the soil according to USCS and AASHTO.c) Is this soil a good foundation material? Justify your answer

Calculating the percentage retained: Sieve no. Sieve size (mm) % finner % retained 3/8 in 9.53…

Q3 The results of a particle size analysis of a soil are given in the following Table. No Atterberg limit tests were conducted. Sieve No. Percent finer 9.53 mm 100 No.20 No.40 | No.100 | No.200 No.4 No.10 70.2 4.1 89.8 62.5 49.8 28.6 a) Would you have conducted Atterberg limit tests on this soil? Justify your answer. b) Classify the soil according to USCS and AASHTO. c) Is this soil a good foundation material? Justify your answver YOL

Q3 The results of a particle size analysis of a soil are given in the following Table. No Atterberg limit tests were conducted. Sieve No. Percent finer 9.53 mm 100 No.20 No.40 | No.100 | No.200 No.4 No.10 70.2 4.1 89.8 62.5 49.8 28.6 a) Would you have conducted Atterberg limit tests on this soil? Justify your answer. b) Classify the soil according to USCS and AASHTO. c) Is this soil a good foundation material? Justify your answver YOL

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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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)
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
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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 (%)
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From 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
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The results of the sieve analysis done on a soil are shown below. Sieve No. Diameter Soil A 4 4.760 88 6 2.380 69 10 2.000 62 20 0.840 51 40 0.420 39 60 0.250 31 100 0.149 18 200 0.074 14 LL 60 PL 20 Determine its classification using USCS. SC SM ML None of the above. CL
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?
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