(%) 32.1 |(mm) 14.4 41.6 16.4 51.8 18.2 64.2 21.1 72.6 22.3 Plastic Limit Test Weight of wet soil + container = 22.12 g Weight of dry soil + container = 21.42 g Weight of container = 13.07 g Natural Water Content Test Weight of wet soil + container = 17.53 g Weight of dry soil + container = 14.84 g Weight of container = 7.84 g Compute the liquid limit Compute the plastic limit Compute the natural water content Compute the plasticity index Compute the liquidity index

(%) 32.1 |(mm) 14.4 41.6 16.4 51.8 18.2 64.2 21.1 72.6 22.3 Plastic Limit Test Weight of wet soil + container = 22.12 g Weight of dry soil + container = 21.42 g Weight of container = 13.07 g Natural Water Content Test Weight of wet soil + container = 17.53 g Weight of dry soil + container = 14.84 g Weight of container = 7.84 g Compute the liquid limit Compute the plastic limit Compute the natural water content Compute the plasticity index Compute the liquidity index

Principles of Geotechnical Engineering (MindTap Course List)

9th Edition

ISBN:9781305970939

Author:Braja M. Das, Khaled Sobhan

Publisher:Braja M. Das, Khaled Sobhan

Chapter4: Plasticity And Structure Of Soil

Section: Chapter Questions

Problem 4.8P

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Problem F In a liquid limit test, using a cone penetrometer, the following readings were recorded. Moisture content (%) Cone Penetration 31.8 15.4 40.6 17.4 52.8 19.2 63.2 22.1 71.6 24.3 PLASTIC LIMIT TEST Weight of wet soil + container = 23.22 g Weight of dry soil + container = Weight of container 21.52 g 15.07 g NATURAL WATER CONTENT TEST Weight of wet soil + container = Weight of dry soil + container = Weight of container 18.33 g 15.64 g 8.05 g a. Compute the liquid limit. b. Compute the plastic limit. c. Compute the natural water content. d. Compute the plasticity index.
76 .ll Q3.png The data from standard and modified proctor test on a soil sample (density of solids: 2680 kg/m3 ) are provided below: Standard proctor test Modified proctor test Dry density (kg/m) 1691 Dry density (kg/m') 1873 Water content (%) Water content (%) 9.3 9.3 1715 11,8 1910 12.8 1755 14.3 1803 15.5 1747 17.6 1699 18.7 1685 20.8 1641 21.1 1619 23 a) Plot the compaction curves (both standard and modified in one graph). b) Establish the maximum dry density and optimum water content for each test.
Problem C: For Questions 12 to 15. Following are the results from the liquid and plasticlimit tests for a soil. Liquid Limit Test Number of Moisture Blows,N Content (%) 16 42.150 20 40.775 28 39.115 12) From the flow curve what is the liquid Limit ? a) 39.683% b) 40.555 % c) 38.992 % d) 39.875 % 13) If the Plasticlimit is 18.750 % what is its plasticity index? a) 22.431 % c) 21.344 % b) 20.918 % 14) If in-situ moisture content is 31% determine its liquidity index ? c) 0.585 d) 20.819 % a) 0.665 b) 1.59 15) If the percentage of clay-size fraction is 4.831. Based on its Activity,this mineral is more likely to be a) Illites b) kaolinite d) 20.588 c) Attapulgite d) Smectites
Field and oven-dry weights of a soil sample taken with a 12 cm long by 6 cm diameter cylindrical tube are Field mass 440 g, and Oven dry mass 420 g, and density, pm of soil solids is 2.75 g/cm3, and density, pw of water is 1 g/cm3. Calculate Degree of Saturation (S). Select one: a. 45.04% O b. 12.67% O c. 22.82% O d. 76.91% O e. 31.58% O f. 13.09%
A liquid limit test conducted on a soil sample in the cup device gave the following results: N 10 19 23 27 40 w (%) 60.0 45.3 39.8 36.5 25.2 The plastic limit test and natural water content. Natural Water Content Test Test No. Weight of Container + Lid (g) Weight of Container + Lid + Wet Soil (g) Weight of Container + Lid + Dry Soil (g) Plastic Limit 2 2 13.07 13.18 7.84 7.5 22.12 21.84 17.53 16.97 20.59 20.35 15.45 14.93 The shrinkage limit test. Weight of Dish with petroleum jelly (g) Weight of Dish with petroleum jelly + Wet Soil (g) Weight of Dish with petroleum jelly + Dry Soil (g) Volume of Shrinkage Dish (cm³) Volume of Oven-Dried Soil Pat (cm³) 11.3 38.51 32.81 15.26 12.83 27. Determine the liquid limit. b. 39% 28. Determine the plastic limit. а. 38% С. 37% d. 40% a. 20.56% b. 21.20% C. 21.85% d. 20.20% 29. Determine the shrinkage limit. b. 22.40% а. 15.20% C. 18.59% d. 16.45%
The coefficient of permeability of a soil sample was found out in a soil mechanics laboratory by making use of a falling head permeameter. The data used and the test results obtained were as follows: Determine the coefficient of permeability in m/day. a. 10.4b. 2.25c. 5.5d. 6.2
Problem A liquid limit test conducted on a soil sample in the cup device gave the following results: N 10 19 23 27 40 w (%) 60.0 45.3 39.8 36.5 25.2 The plastic limit test and natural water content. Plastic Limit Natural Water Content Test Test No. Weight of Container + Lid (g) 2 1 2 13.07 13.18 7.84 7.5 Weight of Container + Lid + Wet Soil (g) 22.12 21.84 17.53 16.97 Weight of Container + Lid + Dry Soil (g) 20.59 20.35 15.45 14.93 The shrinkage limit test. Weight of Dish with petroleum jelly (g) Weight of Dish with petroleum jelly+ Wet Soil (g) Weight of Dish with petroleum jely + Dry Soil (g) Volume of Shrinkage Dish (cm³) Volume of Oven-Dried Soil Pat (cm?) 11.3 38.51 32.81 15.26 12.83 8. Determine the liquid limit. b. 39% 9. Determine the plastic limit. a. 38% C. 37% d. 40% b. 21.20% 10. Determine the shrinkage limit. 20.55% C. 21.85% d. 20.20% a. a. 15.20% b. 22.40% C. 18.59% d. 16.45%
A permeability test was run on a soil sample with 8cm in radius and 10cm long. The level of water at point A and B from the datum is 18cm and 10cm, respectively. Determine the flow of water in cubic centimeter per hour that passes through the soil sample if the coefficient of permeability is 0.1388cm/sec. Point A Point B Sol Sample
Constant head test was conducted to determine the permeability constant of a certain soil sample. The table below shows the data gathered: Constant Head (cm) 29.14 Elapsed Time (s) Add your answer 141 Outflow Volume (cm³) 939 KT (cm/sec) The inner dimensions of the ring are: d = 6.2cm and h = 1.92cm where d is the diameter of the ring and h is the height of the ring. Determine the coefficient of permeability (cm/sec) of the soil sample. Round your answer to 4 decimal places.
A constant head permeability test was performed on a 40 cm² (cross sectional area) soil sample. The length of the sample was 5 cm, and a 20 cm head difference was applied during testing. If 10 cm³ of water was collected over 60 sec, calculate the coefficient of permeability of the soil. (b) In a falling head permeability test, the head fell from 52 to 25 cm, in 3 minutes. The cross-sectional area of the soil sample is 80 cm²,and the length was 8 cm. If the area of the standpipe was 0.5 cm², compute the coefficient of permeability of the soil.
- Find the moisture content, using dry weight per trial based on the table. - Solve the mean, sd, and rsd of the measurements. Description Trial 1 Trial 2 Trial 3 Trial 4 Weight of empty watch glass (g) 38.1151 38.1474 38.9473 38.7400 Weight of watch glass and soil sample (g) 90.1440 85.7554 86.6980 85.1235 Weight of watch glass and dry soil 67.0170 64.1915 65.1110 63.8859
a. The hydraulic head in the 1.1 cm diameter standpipe through which test water passed dropped from 87.5 cm to 587 mm in half a minute of observation. in that duration, the water was collected in the graduated flask. Evaluate the diameter (mm) of soil specimen, if the perimeter in a falling head permeability test set up involves a cylindrical soil sample having a height of 18.4 cmn. Coefficient of permeability of soil sample evaluated is 0.23 m hr.