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 a.) Compute the liquid limit. (Tip: Use normal scale) b.) Compute the plastic limit. c.) Compute the natural water content.

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 a.) Compute the liquid limit. (Tip: Use normal scale) b.) Compute the plastic limit. c.) Compute the natural water content.

Fundamentals of Geotechnical Engineering (MindTap Course List)

5th Edition

ISBN:9781305635180

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter12: Subsurface Exploration

Section: Chapter Questions

Problem 12.2P

See similar textbooks

Similar questions

In a liquid limit test, using a cone penetrometer, the following readings were recorded. Moisture Content (%) Cone Penetration (mm) 32.26 14.4 41.92 16.4 51.96 18.2 64.52 21.2 72.76 22.3 Plastic Limit Test: Weight of wet soil + container 22.28 g Weight of dry soil + container 20.42 g Weight of container 13.39 g Natural Water Content: Weight of wet soil + container 17.69 g Weight of dry soil + container 14.84 g Weight of container 8.16 g Determine: A. Liquid Limit B. Plastic Limit C. Natural Water Content D. Plastic Index E. Liquidity Index
Fill out the table and determine the ELEVATION of BM2. Show your solution. IGNORE "REMARK" COLUMN. THAT WAS SUPPOSED TO BE WHERE YOU PUT YOUR COMMENTS ON THE CONDITION IN THE FIELDWORK.
A fall cone test was carried out using cones of masses 80 grams and 240 grams. The following results were obtained. Parameter 80 gram cone Penetration (mm) 5.9 8.4 15.6 21.3 30.7 Water content (%) 41.7 48 53.5 64.9 69.8 Parameter 240 gram cone Penetration (mm) 12.7 16.78 25.78 35.7 Water content (%) 28 36.4 43.56 52.4 If the natural water content is 35%, compute 1. PI ( 2 LI( 3. If (
Problem 1. A liquid limit test conducted on a soil sample in the cup device gave the following results: 1 2 3 4 Test No. Number of blows 40 27 23 19 35.62 36.92 41.37 46.80 Weight of Container + Lid + Wet Soil (g) Weight of Container + Lid + Dry Soil (g) Weight of Container + Lid (g) 31.08 30.19 33.11 35.78 13.05 11.74 12.36 11.45 The plastic limit test and natural water content. Natural Water Content Test Test No. 1 1 2 22.12 17.53 16.97 Weight of Container + Lid + Wet Soil (g) Weight of Container + Lid + Dry Soil (g) Weight of Container + Lid (g) 20.59 15.45 14.93 13.07 7.84 7.5 The shrinkage limit test. Weight of Dish with petroleum jelly (g) 11.3 38.51 Weight of Dish with petroleum jelly + Wet Soil (g) Weight of Dish with petroleum jelly + Dry Soil (g) 32.81 Volume of Shrinkage Dish (cm³) 15.26 Volume of Oven-Dried Soil Pat (cm³) 12.83 Create a spreadsheet for the result of the Atterberg Limit Test above which will determine and show the following: a. Flow curve b. Liquid Limit c.…
A pre-engineering baseline was run down a very steep hill. Rather than measure horizontally downhill with the steel tape, the surveyor measures the vertical angle with a theodolite and the slope distance with steel tape. The vertical angle is -21° 26’ turned to a point on a plumbed range pole that is 4.88 ft above the ground. The slope distance from the theodolite to the point on the range pole is 148.61 ft. The theodolite’s optical center is 4.66 ft above the upper baseline station at 110 + 71.25. What is the stationing chainage of the lower station? Recall 110+71.25 just means it is 11,071.25 feet from the origin point of the project. The plus sign is just a separator between the hundreds of feet and the tens of feet
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.
3. A constant head permeameter has an internal diameter of 62.5 mm and is fitted in the side with three manometer tappings at points A, B and C. During tests on a specimen of sand the following data were recorded: Test no. 3 Quantity of water collected in 5 min. 136.2 ml 184.5 ml 309.4 ml Manometer level (mm) above datum A 62 112 B 90 122 175 Length between tapping points: A-B 120 mm B-C-125 C 117 164 244
Q: I an oedometer test on a specimen of saturated clay (Gs = 2.72), the applied pressure was increased from 107 1o 214 kPa and the following compression readings recorded: TaBLE| Tmen 0 025 05 1 225 4 &5 9 16 Guugelmm) 78 742 732 721 6% 678 &6 643 637 Teme(mi) 25 % 4 s om0 X0 a0 Guge(nm) 629 626 621 618 616 615 610 602 After 1440 min, the thickness of the specimen was 15.30 mm and the water content was 23.2%. Determine the values of the coefficient of consolidation and the compression atios from (a) the root time plot and (b) the log time plot. Determine also the values of the coefTicient of volume compressibility and the coefficient of permeability.
for B, it says moisture range. so I am expecting an interval. Can I ask how did 8.2 came?
A seismic refraction test yielded the following data:v1 = 610 m/s; v2 = 4268 m/s; xc = 91.5 mThe profile was a two-layer system and no v3 was detected.Determine the thickness of the top layer.
A soil sampler has inner and outer radii of 23 mm and 35mm, respectively. The area ratio of the sampler is
In a two peg test using model Wild NA2 dumpy level, the following observations were taken: Instrument at C Instrument at D Rod reading on A 1.103 m 0.568 m Rod reading on B 0.991 m 0.289 m Point C is equidistant from A and B and D is 12m from A and 72m from B. What is the true difference in elevation between A and B?