A Standard Proctor Test was completed according to Method A described in Chapter 12. The following laboratory data was obtained: Mass of mold = Mass of moist specimen and mold = Volume of compaction mold = Mass of container= 51.5g Mass of container plus moist soil= 261.85g Mass of container plus oven-dried soil- 231.83g Determine the Moisture content, w. (expressed as a percentage, 1 decimal place) Your Answer: 1.98 kg 3.718 kg 0.0009438m3 4

A Standard Proctor Test was completed according to Method A described in Chapter 12. The following laboratory data was obtained: Mass of mold = Mass of moist specimen and mold = Volume of compaction mold = Mass of container= 51.5g Mass of container plus moist soil= 261.85g Mass of container plus oven-dried soil- 231.83g Determine the Moisture content, w. (expressed as a percentage, 1 decimal place) Your Answer: 1.98 kg 3.718 kg 0.0009438m3 4

Principles of Geotechnical Engineering (MindTap Course List)

9th Edition

ISBN:9781305970939

Author:Braja M. Das, Khaled Sobhan

Publisher:Braja M. Das, Khaled Sobhan

Chapter6: Soil Compaction

Section: Chapter Questions

Problem 6.9P

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QUESTION 2 a. The results given in Table 1 were obtained in a particle size distribution test for soil sample received in the laboratory. Table 1 US Sieve No. Size Opening(mm) Weight Retained(g) 4 4.75 0 10 2.0 18.5 20 0.85 53.2 40 0.425 90.5 60 0.25 81.8 100 0.150 92.2 200 0.075 58.5 Pan - 26.5i. Determine the percentage passing each sieve and plot the particle size distribution curve. ii. How much fine-grained soils are in the sample ? iii. Determine the amount of gravel and sand in the sample ? iv. Determine D10, D30, and D60 from the grain-size distribution curve. v. Calculate the uniformity coefficient, Cu. vi. Calculate the coefficient of gradation, Cc. vii. Describe the gradation curve (eg poorly graded)
Problem (6) The laboratory test data for a standard Proctor test are given below. Volume of Proctor mold Mass of wet soil in the mold (kg) Moisture conient (%) (ст3) 943.3 943.3 1.65 10 1.75 12 943.3 1.83 1.81 14 943.3 16 943.3 1.76 18 943.3 1.70 20 a) Find the maximum dry unit weight and optimum moisture content. b) Calculate the zero-air-void unit weights for a soil with G, =2.68 at moisture content of 5, 10, 15, 20, and 25%. Plot a graph of Yrav versus moisture content.
In a liquid limit test using cup apparatus, the following data were taken: Water No. of Sample content blows A 41.8% 39 В 43.5% 23 C 44.3% 20 45.5% 13 While in a plastic limit test, the following values were obtained: Weight of wet soil + container = 21.19 grams Weight of dry soil + container = 18.78 grams Weight of container = 8.20 grams Find the liquid limit.
Draw moisture density curve and determine the ddman- CALIBRATION OF MOLD: Weight of Mold = 4.30kg Height of Mold=11.6cm Diameter of Mold=10.16cm TRIAL WEIGHT OF MOLD + MOIST SOIL DRY SOIL WEIGHT SOIL SAMPLE, kg SAMPLE + CAN, g SAMPLE + CAN, g OF CAN, g 5.52 32.60 32.00 16.07 2 5.74 34.64 32.75 16.31 5.95 36.65 33.16 16.12 4 5.92 36.11 31.31 16.41 5 5.68 48.96 39.95 16.57 3. 1.
A Standard Proctor Test was completed according to Method A described in Chapter 12. The following laboratory data was obtained: Mass of mold = Mass of moist specimen and mold = Volume of compaction mold = Mass of container= 51.5g Mass of container plus moist soil= 261.85g Mass of container plus oven-dried soil= 231.83g 1.98 kg 3.718 kg 0.0009438m3 Your Answer: 4 Determine the Moist density of compacted specimen. (expressed in Mg/m³, 3 decimal places)
If air dry soil sample weighing 25 kg was sieved and results are given below. Calculate the percentage of the particles finer than 0.6 mm. IS sieve 2.0 1.0 0.6 0.425 0.212 0.150 0.075 Pan (mm) Mass retained 2.02 3.51 7.53 8.15 2.81 0.90 0.08 (kg) A) 72.19% B) 69.88% C) 96.68% D) 77.88%
A laboratory test was conducted according to Method A of the ASTM D4318 to find the liquid limit of a soil sample. The following data were obtained: Sample # Number of drops Can # 1 30 A-1 34.06 27.15 11.8 2 23 A-2 32.47 25.8 11.61 3 18 A-3 37.46 29 11.69 Mass of can +moist soil, Mcws (grams) Mass of can+dry soil, Mcs (grams) Mass of the can, Mc (grams) 1- Briefly describe the liquid limit and plastic limit test procedure its applications in practice 2- The water content in sample 1 is --,-- % 3- The water content in sample 2 is --,-- % 4- The water content in sample 3 is --,-- % 5- The liquid limit is --,-- %
2a. The following data was obtained from Liquid Limit Test conducted at a Soil Lab. Using the data given in Table 01, determine the following; Liquid Limit Determination Specimen No. No. of Blows Wt. of Container Wt. of Container + Wet Soil Wt. of Container + Dry Soil i. ii. Unit 1 14 (g) 7.17 (g) 27.97 (9) 21.37 2 20 7.37 27.37 21.27 3 28 7.77 28.47 22.37 4 32 10.97 30.17 24.67 I Determine the Liquid Limit and Flow Index If Plastic Limit of this soil is 22. Classify the soil based on BS Soil Classification system. 5 39 7.27 26.57 21.17
The results of a standard Proctor Test are tabulated as follows. Volume of Proctor Mold(ft³) Weight of Wet Soil in Mold(lb) Moisture Content(w%) 1/30 3.69 12 1/30 3.62 14 1/30 3.88 16 1/30 3.87 18 1/30 3.87 20 1/30 3.77 21 . a. Determine the moist unit weight (ymoist) and dry unit weight (yd). Plot the value of dry unit unit weight (yd) versus the value of moisture content (w%) and estimate the maximum dry unit weight (ydmax) and the corresponding optimum moisture content (w%) b. Determine the void ratio at the optimum moisture content if G s = 2.68 c. Determine the degree of saturation at the optimum moisture content. Use the graph form below for plotting the compaction curve. COMPACTION CURVE D 20.5 R 20 Y 19.5 U 189 N 18 17.5 T 1613 W 16 E 15.5 15 7 9 11 13 151719 212325 27293133 35 37 39 Moisture Content (w%) H T (KN/m^3)
The results of a Standard Proctor test are given in the following table. Moist Weight of Wet Sol Moisture Content in the mold (Ib) Trial (%) 26 4.08 28 4.12 30 4 4,18 32 Determine: 1. Maximum dry unit weight in Ib/ft^3 (4 decimal places) 2. Optimum water content in % (4 decimal places)
mpt=541762&cmid%3D103601 Marks 25 Question 1 A sample of AL Batina soil tested in the laboratory. The gradation analysis results are as shown in table Q1. The Atterberg limit tests indicate that the liquid limit is LL = 30% and the plastic limit is PL=27%. Table Q1 Sieve No.# Sieve size (mm) Finer (%) 4 4.75 90 8. 2.36 84 16 1.3 60 30 0.6 30 50 0.2 10 200 0.075 8 i. Determine the uniformity coefficient Marks] ii. Determine curvature coefficient [5 [5 Marks] iii. Classify AL Batina soil according to Unified Soil Classification System (USCS). [10 Marks] iv. Determine the group symbol and group name Marks) [5
In a liquid limit test on a fine-grained soil, using the Casagrande apparatus, the following results were recorded: Test # 1 2 3 4 Number of blows 50 36 24 15 Container number J11 F12 C20 C2 Wet soil + container 90 92 91 91 Dry soil + container 79 82 83 87 Container mass 82.00 81.00 83.00 82.00 dry soil moisture loss Moisture content, w% If the plastic limit on the same soil was found to be 25%, determine the liquid limit and plasticity index of the soil.