1. From the result of sieve analysis and the grain size curves for soils 1 and 2. a. Compute the value of uniformity coefficient Cu. b. Classify soil No. 1 using USCS Method. c. Classify soil No. 2 using USCS Method.

1. From the result of sieve analysis and the grain size curves for soils 1 and 2. a. Compute the value of uniformity coefficient Cu. b. Classify soil No. 1 using USCS Method. c. Classify soil No. 2 using USCS Method.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Mechanical analysis on four different samples designated as A, B, Cand D were carried out in a soil laboratory. The results of tests are given below. Hydrometer analysis was carried out on sample D. The soil is non-plastic. Sample D. liquid limit w 42, plastic limit = 24, plasticity index = 18 Classify the soils per the Unified Soil Classification System. Samples ASTM Sieve Percentage finer than Designation 63.0 mm 100 93 20.0 mm 64 76 6.3 39 100 65 20 mm 600 212 63A 20 24 98 59 12 90 54 5 9. 47 100 34 95 23 69 7. 46 2 31 100 HO 60 20 Sand Course lo medium Cotles l>762 mm Clay Silt Gravel Fine 0.001 001 0.075 0.1 10 100 Particle size (mum) Figure Ex. 3.21 Percentage finer
2. Prove n-e/(1+e) and e-n/(1-n) based on definitions of void ratio and porosity. 3. Classify the following soils according to the unified soil classification system: group symbols and typical names Soil A B Percent passing U.S. sieve No. 4 No. 10 No. 20 No. 40 No. 60 No. 100 No. 200 95 86 65 52 46 38 25 98 95 88 82 75 65 100 LL PL 38 65 22 58
46.) The results of the sieve analysis done on a soil are shown below. Sieve No. Diameter Soil A 4 4.760 89 6 2.380 70 10 2.000 63 20 0.840 55 40 0.420 40 60 0.250 31 100 0.149 20 200 0.074 4 LL 60 PL 40 Determine its classification using USCS SP GW SW GP None of the above
The given data shows a sieve analysis of a soil sample. Classify the soil using AASHTO Method. Tables are provided below. Sieve No. Diameter (mm) Percent Passing No. 4 4.760 100 No. 8 2.380 100 No. 10 2.000 95 No. 20 0.840 81 No. 40 0.420 74 No. 60 0.250 69 No. 100 0.150 55 No. 200 0.075 51 Characteristics of No. 40 Fraction LL 48 PL 40 A. A-5(3) B. A-7-5(4) C. None of these. D. A-5(4) E. A-7-6(3)
1. Sieve analysis data for two inorganic soils are provided in Table 1. Soil 1 has LL = 64 and PL = 38. Soil 2 has LL = 26 and PL =18. To plot the particle size distribution curves for Soils 1 and 2 using the attached sheet. Determine the effective grain size, D10 for Soil 2. Determine the USCS designation for each soil. List step by step how you make your choices. Table 1. Sieve Analysis Data. Sieve No. 14 10 20 40 60 100 200 Mesh opening (mm) 4.750 2.000 0.850 98 0.425 93 0.250 88 0.150 83 0.075 77 Soil 1 % finer 100 100 Soil 2 % finer 100 90 64 38 29 21 11
a.) Classify soil A using AASHTO method. b.) Classify soil B using AASHTO method. c.) Classify soil C using AASHTO method.
Within few minutes
C 1. Sieve analysis data for two inorganic soils are provided in Table 1. Soil 1 has LL = 60 and PL = 36. Soil 2 has LL = 23 and PL =15. To plot the particle size distribution curves for Soils 1 and 2 using the attached sheet. Determine the effective grain size, D10 for Soil 2. Determine the USCS designation for each soil. List step by step how you make your choices. Table 1. Sieve Analysis Data. Mesh Sieve No. 4 10 20 40 60 100 200 opening (mm) 4.750 2.000 0.850 0.425 0.250 0.150 0.075 Soil 1 % finer 100 100 98 93 88 83 77 Soil 2 % finer 100 90 64 38 29 21 11 % Finer 120 100 80 60 40 20 II 1 11 10 T+4- I I J FT T I 1 Bullie aller 1 1 1 TEE J 1 I 01: 1 H I I F + +4 F T LLU 1 L 1 1 4 II FT 7 LI 1 TT IT 1 1 31 11 1 LI J-L IT J { LIL II 1 1 rt II I T LI L 11 J J FIL I 31 I I 1 I B 1 1 T r T 1 CA J <-+ I 1 Particle Size Distribution 1 I THE SE DET L II LL t FT4 JII 1 1 1 1 Fr-r 7 1 J Lead L 1 I T I P E S I - 1 1 I E T I Particle size (mm) II LL 1 T E T TH 0.1 J J E IF LIL II F 1 1 T J II T…
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
Mechanical analysis on four different samples designated as A, B, Cand D were carried out in a soil laboratory. The results of tests are given below. Hydrometer analysis was carried out on sample D. The soil is non-plastic. Sample D. liquid limit w 42, plastic limit = 24, plasticity index = 18 Classify the soils per the Unified Soil Classification System. Samples A D ASTM Sieve Percentage finer than Designation 63.0 mm 100 93 20.0 mm 64 76 6.3 39 100 65 20 mm 600 212 63A 20 24 98 59 12 90 54 9. 47 100 2 34 95 23 69 46 2 31 Samples A,BandC 100 60 20 Sand Course le medium Cotles l>762 mm Clay Silt Gravel Fine 0.001 001 0.075 0.1 10 100 Particle size (mun) Figure Ex. 3.21 Percentage finer
Write in clean white paper...
Solutions plz