The following compaction curve represents the compaction data from the Lab for soil. Following are the results of a field unit-weight determination test performed on the same soil by means of the sand cone method: Volume of hole is 0.0019 m3 and mass of moist soil 2.85 kg. Assume the moisture content of moist soil in the field is the same as optimum moisture content of lab proctor test compaction curve. 19.0 18.5 18.0 17.5 17.0 16.5 16.0 15.5 10 15 20 Moisture content (%) Moist density of compacted soil in the field= Moist unit weight of compacted soil in the field= Dry unit weight of compacted soil in the field= Degree of compaction (insert your answer as a fraction for example 0.82) = Dry unit weight (kN/m³) id Cytve

The following compaction curve represents the compaction data from the Lab for soil. Following are the results of a field unit-weight determination test performed on the same soil by means of the sand cone method: Volume of hole is 0.0019 m3 and mass of moist soil 2.85 kg. Assume the moisture content of moist soil in the field is the same as optimum moisture content of lab proctor test compaction curve. 19.0 18.5 18.0 17.5 17.0 16.5 16.0 15.5 10 15 20 Moisture content (%) Moist density of compacted soil in the field= Moist unit weight of compacted soil in the field= Dry unit weight of compacted soil in the field= Degree of compaction (insert your answer as a fraction for example 0.82) = Dry unit weight (kN/m³) id Cytve

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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Since laboratory or field experiments are generally expensive and time consuming, geotechnical engineers often have to rely on empirical relationships to predict design parameters. Section 6.6 presents such relationships for predicting optimum moisture content and maximum dry unit weight. Let us use some of these equations and compare our results with known experimental data. The following table presents the results from laboratory compaction tests conducted on a wide range of fine-grained soils using various compactive efforts (E). Based on the soil data given in the table, determine the optimum moisture content and maximum dry unit weight using the empirical relationships presented in Section 6.6. a. Use the Osman et al. (2008) method [Eqs. (6.15) through (6.18)]. b. Use the Gurtug and Sridharan (2004) method [Eqs. (6.13) and (6.14)]. c. Use the Matteo et al. (2009) method [Eqs. (6.19) and (6.20)]. d. Plot the calculated wopt against the experimental wopt, and the calculated d(max) with the experimental d(max). Draw a 45 line of equality on each plot. e. Comment on the predictive capabilities of various methods. What can you say about the inherent nature of empirical models?
The following are results of a field unit weight determination test using sand cone method: Volume of hole = 0.0019 m3 Mass of moist soil from hole = 3.48 kg Water content = 12.43% Max. dry unit weight from a laboratory compaction test = 19.56 kN/m3 Determine the relative compaction, in percent. Hint: Relative compaction = Field dry unit weight / Max dry unit weight (lab)
Explain in detail Maximum Dry Density and Optimum Moisture Content of soil? Explain in 5 sentences how you are going to interpret the results of Proctor’s Compaction test?
The laboratory compaction test of a certain type of soil gives a maximum dry density of 1486 g/m3 with an optimum moisture content of 12.5%. The following are the results of a field unit weight determination test using sand cone method. Volume of soil excavated from the hole = 0.001337 m3 Weight of soil from the hole when wet = 2220 g Weight of soil from the hole when dried = 1890 g Determine the field unit weight of soil in g/cm3. Determine the relative compaction, then tell whether it is within the allowable range for relative compaction of 95 ± 0.20?
arrow_forward Question A fine-grained soil has 60% clay with LL 220%, PL 545%, and a natural water content of 6%. A standard Proctor test was carried out in the laboratory and the following data were recorded Diameter of mold = 101.40 mmHeight of mold = 116.70 mmMass of mold = 4196.50 gramSpecific gravity, Gs = 2.69 a) Determine the maximum dry unit weight and optimum water content.b) If the desired compaction in the field is 95% of the standard Proctor test results, what values of dry unit weight and water content would you specify?c) Explain why you select these values.d) What field equipment would you specify to compact the soil in the field, and why?e) How would you check that the specified dry unit weight and water content are achieved in the field? I Need The Answer of part (d) and (e)...
It has been anticipated that the soil under the proposed dam is contaminated, and needs to be removed and backfilled to a depth of 2.5 feet. A Standard Proctor compaction test was conducted on soil from a borrow pit, with the results shown below right. Based on the results: a) Plot the Proctor curve for the borrow soil. b) Find the optimum water content and max. dry unit weight for the borrow soil. c) Determine whether the unit weight of the top layer can be met by compacting to 95% of maximum, and what the range of moisture content for that unit weight is. d) If we borrow soil from a pit with a void ratio of 0.71, determine how much soil we need to borrow to backfill below the dam to match the in-situ void ratio of the top layer of soil.
The laboratory compaction test of a certain type of soil gives a maximum dry density of 1.486 g/cc3 with an optimum moisture content of 12.5%. Determine the field unit weight of the soil in g/cc3 using the following results of a filed unit weight determination test using sand cone method. Volume of soil excavated from the hole = 0.001337 m3 Weight of moist soil from the hole = 2220 g Weight of oven dried soil = 1890 g
How does the dry unit weight, differ from the unit weight ? Why is the Proctor method of assessing soil compaction based on the dry unit weight and not the unit weight?
Calculate the coefficient of gradation, Cc. ? Classify the soil by AASHTO and USCS system. (LL = 30, PL = 10) ? Use this given M=Mass of soil retained in pan +∑Mass reatianed in each sieve⇒M=617 gM=Mass of soil retained in pan +∑Mass reatianed in each sieve⇒M=617 g Sieve No. Opening Size (mm) Mass retained (g) %Mass retained=MiM×100=MiM×100 %Cumulative mass retained % Finer 4 4.750 28 =28617×100%=4.54%=28617×100%=4.54% 4.54 95.46 10 2.0 42 =42617×100%=6.81%=42617×100%=6.81% 11.35 88.65 20 0.850 48 =48617×100%=7.78%=48617×100%=7.78% 19.13 80.87 40 0.425 128 =128617×100%=20.74%=128617×100%=20.74% 39.87 60.13 60 0.250 221 =221617×100%=35.82%=221617×100%=35.82% 75.69 24.31 100 0.150 86 =86617×100%=13.94%=86617×100%=13.94% 89.63 10.37 200 0.075 40 =40617×100%=6.48%=40617×100%=6.48% 96.11 3.89 Pan 24 =24617×100%=3.89%=24617×100%=3.89% 100 0 Now, to plot the GSD curve, the % finer is kept on the vertical axis and the opening size are kept on the horizontal axis -…
Laboratory compaction test results and fielddetermined in-place density (unit weight) results are reported in terms of dry density or unit weight and not wet density or unit weight. Outline the reasons for this procedure of reporting. (Hint: Consider the soil properties directly related to density or unit weight.)
10. The results of a Standard Proctoretest is tabulated as follows: (SEE PICTURE) Determine the maximum dry unit weight of compaction Determine the optimum moisture content Determine the void ratio at the optimum moisture content Gs = 2.70 Determine the degree of saturation at the optimum moisture content A field unit determination test for the soil, yielded the following data. Moisture content = 12% and moist unit weight = 106 pcf. Determine the relative compaction.
One cubic foot soil test pit , weigh 100 pounds net weight. The dry weight of the sample is 80 pounds. Calculate the effective unit weight of the soil if it is submerge below the ground water level when the specific gravity is 2.6