Laboratory compaction test results on a clayey silt are given in the following table:-Moisture content (%)Dry unit w3eight (kNm³)14.80817.4518.521118.91218.61416.9Following are the results of a field unit weight determination test on the samesoil with the sand cone method:-• Calibrated dry density of Ottawa sand = 1667 kg/m³• Calibrated mass of Ottawa sand to fill the cone = 0.117 kg• Mass of jar + cone + sand (before use) = 5.99 kg• Mass of jar + cone + sand (after use) = 2.81 kg• Mass of moist soil from hole = 3.331 kg• Moisture content of moist soil = 11.6%Determinea. Dry unit weight of compaction in the field

Soil compaction is a process in which the soil retains mechanical pressure and is compacted. The…

Answered: Following are the results of a field…

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 6.7P

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6. The relative compaction of a sand in the field is 93.5%. Given that dry unit weight max = 16.98 kN/m3 and dry unit weight min = 14.46 kN/m3. What is the dry unit weights in the field? (Ans: 15.88 kN/m3) What is the relative density compaction? (Ans: 60.1%) What is the water content of the sand if the moist unit weight is 18.42 kn/m3. (Ans: 16%)
7. The relative compaction of a sand in the field is 94%. Given that dry unit weight max = 16.2 kN/m3 and dry unit weight min = 14.9 kN/m3 . For the field conditions. What is the relative density compaction? (Ans: 27%) What is the dry unit weight? (Ans: 15.23 kN/m3) What is the moist unit weight at a moisture content of 10%? (Ans: 16.45 kN/m3 )
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)
The relative compaction of a sand in the field is 93.5%. Given the dry unit weight maximum= 16.98 kN/m3 and the dry unit weight minimum = 14.46 kN/m3. What is the water content of the sand in percent if the moist unit weight is 18.42 kN/m3? a.17.1 b.16.0 c.15.2 d.14.3
8. The max and min dry unit weights of the sand were determined in the laboratory to be 16.5 kN/m3 and 14.6 kN/m3 respectively. The relative density of compaction in the field of the same sand is 70%. What is the relative compaction? (Ans: 96%) What is the dry unit weight? (Ans: 15.84 kN/m3 ) What is the moist unit weight at a moisture content of 10%. (Ans: 17.42 kN/m3 )
The following results are obtained from standard compaction test for a soil, if Gs = 2.5 Water Content (%) 6.2 8.1 9.8 11.5 12.3 13.2 Unit weight (kN/m3) 16.9 18.7 19.5 20.5 20.4 20.1 Required: 1- The optimum water content (Wc %) 2- The maximum dry unit weight ( dry ) 3- The void ratio (e) 4- Degree of saturation (S %) 5- Find the weight of water needed to be added to 1 m3 to reach 100% saturation.
Refer to Problem 3.5. Using Eq. (3.28), determine the averagerelative density of the sand. Assume it is a fine sand.Use Eq. (3.13) to obtain (N1)60.
Results of a standard proctor ompaction test on a silty sand are shown. PLEASE COMPLETE SOLUTION. 1) Find the maximum dry unit weight and optimum moisture content. 2) What is the moist unit weight at optimum moisture content? 3) What is the degree of saturation at optimum moisture content? Given: ?? = 2.69 4) If the required field dry unit weight is 18.5 ??/?3, what is the relative compaction? 5) What should be the range of compaction moisture contents in the field to achieve the above relative compaction? 6) If the minimum and maximum void ratios are 0.31 and 0.82, respectively, what is the relative density of compaction in the field?
9. The results of a Standard Proctoretest is tabulated as follows: (CHECK PICTURE) Determine the maximum dry unit weight of compaction Determine the void ratio at the optimum moisture content Gs = 2.68 Determine the optimum moisture content Determine the degree of saturation at the optimum moisture content
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)...
Data for field density by sand cone method: Weight of bottle + cone + sand before going to the field = 8990.5 g Weight of soil excavated from hole = 7550 g Weight of bottle + cone + sand after use = 2931 g Weight of sand required to fill the cone, Wc = 1578 g Dry unit weight of sand = 1.533 g/cm3 Moisture content = 6% 1- Calculate dry unit weight of compaction in the field.
A sand-cone in place unit test was conducted on the field during the construction of a proposed embankment. The following data were obtained. moisture content=12.8% weight of sand to fill the funnel=320 g weight of wet soil from test hole=750 g unit weight of sand=15.4 kN/m3 weight of sand used to fill test hole and funnel of sand-cone device=870 g a) compute the wet unit weight of soil (Ans. 21.14 kN/m3) b) compute the dry unit weight of the compacted soil (Ans. 18.74 kN/m3) c) compute the relative compaction in the field if the maximum dry unit weight of the soil is 21.3 kN/m3. (Ans. 87.98%)

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