A soil sample is oven-dried from its liquid limit state to its dry state causing a volume change of 19.68% of the dry volume. Likewise, the same soil is oven-dried from its plastic limit state to its dry state causing a volume change of 8% of the dry volume. The liquid limit and plastic limit are 60.54% and 50.37%, respectively. Determine the specific gravity of soil solids, shrinkage limit and shrinkage ratio.

A soil sample is oven-dried from its liquid limit state to its dry state causing a volume change of 19.68% of the dry volume. Likewise, the same soil is oven-dried from its plastic limit state to its dry state causing a volume change of 8% of the dry volume. The liquid limit and plastic limit are 60.54% and 50.37%, respectively. Determine the specific gravity of soil solids, shrinkage limit and shrinkage ratio.

Principles of Geotechnical Engineering (MindTap Course List)

9th Edition

ISBN:9781305970939

Author:Braja M. Das, Khaled Sobhan

Publisher:Braja M. Das, Khaled Sobhan

Chapter5: Classification Of Soil

Section: Chapter Questions

Problem 5.10P

See similar textbooks

Similar questions

A soil sample is oven-dried from its liquid limit state to its dry state causing a volume change of 60% of the dry volume. Likewise, the same soil is oven-dried from its plastic limit state to its dry state causing a volume change of 20% of the dry volume. The liquid limit and plastic limit are 54% and 26%, respectively. Determine the specific gravity of soil solids, shrinkage limit and shrinkage ratio. 1. INPUT THE VALUE OF SPECIFIC GRAVITY OF SOIL SOLIDS IN 2 DECIMAL PLACES 2. INPUT THE VALUE OF SHRINKAGE LIMIT IN PERCENTAGE FORM AND IN 2 DECIMAL PLACES 3. INPUT THE VALUE OF SHRINKAGE RATIO IN 2 DECIMAL PLACES Answer with text and/or attachments:
Question 11 The results of the particle-size analysis of a soil are as follows: Percent passing the No. 10 sieve = 80 • Percent passing the No. 40 sieve = 40 Percent passing the No. 200 sieve = 20 %3D The liquid limit and plasticity index of the minus No. 40 fraction of the soil are 30 and 5, respectively. Classify the soil by the AASHTO system. A A-2-4 B А-2-6 C) A-3 D A-2-5
Give its consistency index. EXERCICE 3 Classify the following soil using the Unified Soil Classification (ASTM standard). Percentage passing N° 4 seive = 98 Percentage passing N° 10 seive = 86 Percentage passing N° 40 seive = 80 Percentage passing N° 200 seive = 71 The liquid limit = 102% The Plasticity index = 68% The water content is = 72%Give its consistency index. Indicate the activity of this soil and numerous some potential problems linked to its use in different geotechnical works. EXERCICE 4 Some tests were performed on the sandy silt according to Standard Proctor test. The volume of the mold is 944 cm. The specific gravity was measured as 2.67 1) Plot the Proctor compaction curve (see the page 2) 2) Find: 2-1 The maximum dry unit weight of sandy-silt fill, 2-2 Its OMC, 2-3 The moisture ranges for 85% of Standard Proctor Optimum, 2-4 The degree of saturation at the optimum Proctor state. 3) Plot the zero air voids curve (see the page 2) 4) Determine the values of degree of…
Classify the soil by IS classification Given:- Liquid limit=22% Plasticity limit=19% Percentage passing through 4.75mm IS sieve=98.5% Percentage passing through 75 micron IS sieve=35% D60=0.22 D30=0.19 D10=0.16
Using Unified Soil Classification System, determine the following: (a) Coefficient of Gradation (b) Group Symbol. Passing No. 4 = 47 % Passing No. 200 = 10 % D10 = 3.45 mm %3! D30 = 6.29 mm De0 = 13.72 mm Plasticity for the minus No. 40 fraction: Liquid Limit = 28 % Plastic Limit = 17 % O GC-GM GP-GC O GC O Coef. Of Gradation = 0.84 O GP O Coef. Of Gradation = 3.97
2. A sample of soil has 60% passing the No. 4 sieve and 42% passing the No. 200 sieve. The liquid limit (LL) for the soil is 28%, and the plasticity index (PI) is 4.5%. Classify this soil according to the USCS.
Q.1.A A soil specimen have void ratio (e) of 0.7, Gs 2.72. Calculate the total unit weight, dry unit weight, and water content at degree of saturation equals 75%. B. Classify the following soil using the USCS. Finer (%) 100 90 60 30 19 4 Sieve No. 4 No. 25 No. 50 No. 100 No. 200 12.5 mm (Diameter) (4.7 mm) (0.7 mm) (0.3 mm) (0.15 mm) (0.075 mm)
The results of the particle-size analysis of the soil are as follows: Percent passing No. 10 sieve = 100; Percent passing No. 40 sieve = 80; Percent passing No. 10 sieve = 58.The LL and PI of minus No. 40 fraction of the soil are 30 and 10 respectively. Classify the soil by USCS system a. CH b. SP-SC c. CL-ML d. CL-ML
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
.49 Plastic limit of a soil is 25% and plasticity index is 8%. When soil is dry from plastic limit, the volume change is 25% of its volume at plastic limit. Similarly when soil is dry from liquid limit, the volume change is 34% of its volume at liquid limit. Determine shrinkage limit and shrinkage ratto.
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?
Using AASHTO Classification System, determine the following: (a) Group Index (b) Soil Classification Passing No. 10 = 55 % Passing No. 40 = 47 % Passing No. 200 = 23 % Plasticity for the minus No. 40 fraction: Liquid Limit = 22 % Plastic Limit = 17 % O GI = - 0.40 say 0 O GI = 0.60 say 1 O GI = 0 O A-2-4 (1) O A-2-7 (0) O A-1-b (0) O A-2-6 (1)