Sand cone equipment is used to perform a field density test on a compacted earth fill.• Soil sample dug from the test hole = 20.60 N• Dry weight of soil sample= 17.92 NOttawa sand used to fill the hole weighs 16.05 N and is known to have a density of 15.74kN/m3.Determine the in-place dry density of the tested soil.Answer:

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Fundamentals of Geotechnical Engineering (MindTap Course List)

5th Edition

ISBN:9781305635180

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter5: Soil Compaction

Section: Chapter Questions

Problem 5.7P

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1. Sand cone equipment is used to determine an in-place unit weight (field density test) on a compacted earth fill. Ottowa sand is used in the cone and is known to have a bulk density of 100 pcf. a. From the information given below, determine the in-place dry unit weight of the tested soil, and the water content. Soil sample dug from test hole wet weight = 4.62 lb Dried Weight of soil sample = 4.02 lb Weight of Ottowa Sand (Sand Cone) to fill test hole = 3.60 lb b. Determine the percentage of compaction of the tested soil if the laboratory moisture unit weight curve indicates a dry unit weight of 115 pcf and optimum moisture content of 13 percent.
4.2 For a field test the technician is using a sand with a density of 96.7 pcf in his sand cone test. Calculate the dry density, degree of saturation and moisture content of a compacted fill using the following data that the technician collected: Weight of sand cone bottle before test (full of sand): Final weight of sand cone (some sand remains): Weight of sand needed to fill cone: Weight of fill soil excavated from hole + bucket: Weight of bucket used to collect fill soil: 15.14 Ibs 3.97 Ibs 2.67 Ibs 12.37 Ibs 0.76 Ibs Moisture content of fill soil: Empty can: 48.1 g Can + moist soil: 253.7 g Can + dry soil: 237.1 g
Sand cone equipment is used to perform a field density test on a compacted earth fill. • Sil sample dug from the test hole = 20.60 N • Dry weight of soil sample = 17.92 N Ottawa sand used to fill the hole weighs 16.05 N and is known to have a density of 15.74 kN/m³. Which of the following gives the water content of the tested soil? O a. 15.45% O b. 12.08% O . 11.12% d. 14.96% e. 10.94% Clear my choice
Sand-cone equipment is used to determine an in-place unit weight (field density test) on a compacted earth fill. The sand used in the cone is known to have a bulk density of 15.73 kN/m3 Wet weight of soil sample dug from test hole = 2100 g Dried weight of soil sample = 1827 g Weight of sand (sand core) to fill the test hole = 1636 g a) Compute the water content. b) Compute the in-place dry unit weight of tested soil. c) Compute the percentage of compaction of the tested soil if the laboratory moisture-unti weight curve indicates a dry unit weight of 18.09 kN/m3 and a optimum moisture content of 13%.
A liquid limit test on a clay was performed with the following results. The natural water content of the clay is 38% and plastic limit is 21%. Number of Blows 6 12 20 28 32 Water content (%) 52.5 47.1 42.3 38.6 37.5 What is the liquidity index of this clay? (Use Interpolation) Group of answer choices 0.895 0.96 1.08 0.90 Please answer this asap. For upvote. Thanks
A sand cone test has been performed in a recently compacted fill. The test results obtained are as follows: Initial weight of sand cone + sand = 13.51 lb Final weight of sand cone + sand = 4.26 lb Weight of sand to fill cone = 2.12 lb Weight of soil from hole + bucket = 12.42 lb Weight of bucket = 1.21 lb Moisture content test: Mass of empty moisture content can = 23.11 g Mass of moist soil + can = = 273.93 g Mass of oven-dried soil + can = 250.10 g The sand used in the sand cone had a unit weight of 81.0 lb/ft³, and the fill had a maximum dry unit weight of 121 lb/ft³ and an optimum moisture content of 11.7%, based on the modified Proctor test. Compute the relative compaction based on the modified Proctor test.
Sand cone equipment is used to perform Field density test on a compacted arth fil. Soil sample dug from the test hole = 19.88 N Dry weight of soil sample = 17.92 N Ottawa sand used to fill the hole weighs 16.05 N and is known to have a density of 15.74 HN/m. Which of the following gives the water content of the tested soil? O a. 15.45% O b. 12.08% O c. 14.96% O d. 10.94% e. 11.12%
Maximum and minimum void ratios of a sand sample were determined in the laboratory to be 0.9 and 0.3, respectively. If the void ratio of another sample of the same sand is found to be 0.53, calculate the relative density of the second sand sample. (Type your answer as a percentage %)
Sand cone equipment is used to perform a field density test on a compacted earth fill. • Soil sample dug from the test hole = 20.60 N • Dry weight of soil sample = 17.92 N Ottawa sand used to fill the hole weighs 16.05 N and is known to have a density of 15.74 kN/m³. Determine the volume of the test hole. Answer:
An in-place density determination (in-place unit weight) is made for the sand in a borrow pit using a balloon-type density apparatus. The damp sample dug from a test hole is found to weigh 37.90 N. The volume of the test hole is 0.00184m2. A. Compute the wet unit weight. B. This soil is to have a water content of 15%. Compute the dry unit weight of the soil. C. Compute the specific gravity of the soil if it has a void ratio of 0.423
3. A dry sand is placed in a container 9 cm3. The dry weight of the sample is 15 g. Water is carefully added to the container so as not to disturbed the condition of the Sample. When the container is filled the combined weight of soil and water is 18 g. Determine the following: a) Void ratio of sample in the container b) Porosity c) Water constant d) Specific gravity of the soil particle e) Saturated unit weight f) Dry unit weight
Consider that a 58.7 gram sample of dry soil is used in making a mechanical analysis by using the hydrometer method. The laboratory was held at a constant temperature of 20 C (68 F). The 40 s hydrometer reading was found to be 32.5 and the 8 hr hydrometer reading was 15.5. The percentage of sand in this soil sample is (note, a 58.7 gram sample is used here): Your Answer: Answer units

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