Determine : 1. Soil classification of soil A using AASHTO method. 2. Soil classification of soil B using USCS method. 3. Soil classification of soil C using USDA method.
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A: Solution: Given: Liquid Limit (LL)=52 Plastic Limit (PL)=40 Plasticity Index (PI)= LL-PL=52-40=12
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A: Given, Sieve analysis data To find - Classification of soil
Determine :
1. Soil classification of soil A using AASHTO method.
2. Soil classification of soil B using USCS method.
3. Soil classification of soil C using USDA method.
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- 5. The following information is obtained from a sieve analysis to determine the range of particle sizes in a granular soil sample. Sieve Size Sieve Opening (4.76) Percent Finer # 4 4.76 96 #10 2.00 80 #20 0.84 51 # 40 0.42 38 #60 0.25 25 #100 0.149 12 #200 0.074 Present the information as grain-size curve on semi-logarithmic coordinates of percent finer against particle size diameter. From the plot, determine the uniformity coefficient Cu.2. For the given data, shows a sieve analysis of soil samples A,B, and C. Diameter Soil A Soil B Soil C Sieve No. (mm) %Passing %Passing %Passing # 4 4.760 100 90 100 a. Classify soil sample A using AASHTO Method #8 2.380 97 64 100 #10 2.000 92 54 98 b. Classify soil sample B using AASHTO Method c. Classify soil sample C using #20 0.840 87 34 92 # 40 0.420 53 22 84 #60 0.250 42 17 79 AASHTO Method #100 0.149 26 9. 70 # 200 0.074 17 4 63 LL 35 47 PL 20 244. From the given data, shows a sieve analysis of soil samples A, B and C. SOIL SAMPLE Sieve No. Diam (mm) A В C PERCENT PASSING 4 4.460 90 100 100 8 2.380 64 90 100 10 2.00 54 77 98 20 0.840 34 59 92 40 0.420 22 51 84 60 0.250 17 42 79 100 0.149 9. 35 70 200 0.074 4 33 63 Characteristics of -40 Fraction LL 46 47 PL 29 24 a. Classify the soil sample A using USCS method (Group Symbol only), D.=2.3, D = 0.17, b. Classify the soil sample B using AASHTO method c. Classify the soil sample C using the AASHTO method D = 0.70 30
- 5. The following information is obtained from a sieve analysis to determine the range of particle sizes in a granular soil sample. Sieve Size Sieve Opening (4.76) Percent Finer#4 4.76 96#10 2.00 80#20 0.84 51#40 0.42 38#60 0.25 25#100 0.149 12#200 0.074 5 Present the information as grain-size curve on semi-logarithmic coordinates of percent finer against particle size diameter. From the plot, determine the uniformity coefficient Cu.The given data shows a sieve analysis of a soil sample. Classify the soil using AASHTO Method. Sieve No. Diameter Soil A 4 4.760 80 8 2.380 71 10 2.000 69 20 0.840 63 40 0.420 57 60 0.250 43 100 0.150 40 200 0.075 39 LL 36 PL 21The given data shows a sieve analysis of a soil sample. Classify the soil using AASHTO Method. Sieve No. Diameter Soil A 4 4.760 96 8 2.380 80 10 2.000 78 20 0.840 61 40 0.420 56 60 0.250 48 100 0.150 39 200 0.075 33 Characteristics of No. 40 Fraction LL 46 PL 31
- Question 46 The results of the sieve analysis done on a soil are shown below. Sieve No. Diameter Soil A 4. 4.760 88 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. ASUS43.) The results of the sieve analysis done on a soil are shown below. Sieve No. Diameter Soil A 4 4.760 100 6 2.380 98 10 2.000 91 20 0.840 85 40 0.420 84 60 0.250 79 100 0.149 61 200 0.074 55 LL 60 PL 40 ML None of the above. CL CH MHGiven below is the result of a sieve analysis done on a soil sample.Sieve No. Diameter Percent Passing %4 4.750 9510 2.000 8220 0.850 7340 0.425 5360 0.250 23100 0.150 14200 0.075 7Pan --- 0Determine the value of the coefficient of uniformity
- 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 45The given data shows a sieve analysis of a soil sample. Classify the soil using AASHTO Method. Sieve No. Diameter Percent Passing 4 4.760 100 6 2.380 91 10 2.000 88 20 0.840 76 40 0.420 55 60 0.250 50 100 0.150 47 200 0.075 46 Characteristics of No. 40 Fraction LL 50 PL 34Plasticity index o max. NP T0 max. 10 max. I1 min. 11 min. Fine Silty or clayey gravel and sand Usual types of significant Stone fragments, gravel, and sand sand constituent materials Excellent to good General subgrade rating General classification Silt-clay materials (more than 35% of total sample passing No. 200) A-7 A-7-50 A-7-6 Group classification A-4 A-5 A-6 Sieve analysis (percent passing) No. 10 No. 40 No. 200 36 min. 36 min. 36 min. 36 min. Characteristics of fraction passing No. 40 Liquid limit Plasticity index 40 max. 10 max. 41 min. 10 max. 40 max. 11 min. 41 min 11 min. Usual types of significant constituent materials Silty soils Clayey soils General subgrade rating Fair to poor *For A-7-5, PI LI. – 30 O A. A-2-4(0) O B.A-1-a(0) OC.A-2-5(0) O D.A-3(0) O E. A-1-b(0) F2 F3 F4 FS F6 F7 %23 $4 4 % & 5 7 因 LO