Compaction Test of Soil, please check attached photos and put all decmal places if applicable. Thank you. Results and ComputationA. Moisture Content DeterminationDetermination NumberMoisture Can NumberMass of wet soil and can, gMass of dry soil and can, gMass of water, gMass of can, gMass of dry soil12101102(a) | 166(b) | 153193180(a-b)(c) | 101(b-с)128a-bWater Content,%х 100%b-cB. Determination of Dry Unit WeightDetermination NumberMass of wet soil & cylinder, gMass of cylinder,gMass of wet soil,gVolume of Cylinder, cm^3Specific Gravity_Wet Unit Weight, KN/cm^312(A)(B)(A-B)52003600360036001943.861943.862.72.7YwYwYs =(w +G?YsYary = .Ys =(w +GYsYary =.Dry Unit Weight, KN/ cm^31+w.1+w Data:Data collected during this test consist of the following:Determination of Dry Unit Weight of the Soil"Mass of the compaction mold and the base plateMass of the compaction mold, bdse plate and the sollInside diameter of the compaction moldMoisture Content DeterminationMass of canMass of moist soil plus canMass of dry soil plus canCalculations:The zero-air voids unit weight for the compaction curve can bedetermined using the formulayz = yw /[w+ (1/Gs) !where:YZ = zero-air voids unit weightyw = unit weight of waterW.= moisture contentGs = specific gravity

In table B, The formula of wet unit wt is wrong. The correct formula is: ywet= (G+Se)yw/(1+e) Since…

Results and Computation A. Moisture Content Determination Determination Number Moisture Can Number Mass of wet soil and can, g Mass of dry soil and can, g Mass of water, g Mass of can, g Mass of dry soil 1 2 101 102 (a) | 166 193 (b) | 153 180 (a-b) (c) | 101 128 (b-с) a-b x 100% Water Content,% b-c B. Determination of Dry Unit Weight Determination Number Mass of wet soil & cylinder, g Mass of cylinder,g Mass of wet soil,g Volume of Cylinder, cm^3 Specific Gravity Wet Unit Weight, KN/cm^3 1 2 3600 (A) (B) (A-B) 5200 3600 3600 1943.86 1943.86 2.7 2.7 Yw Yw Ys = (w + Ys = (w +) Dry Unit Weight, KN/ cm^3 Ys Ys Ydry Yary 1+w 1+w

Results and Computation A. Moisture Content Determination Determination Number Moisture Can Number Mass of wet soil and can, g Mass of dry soil and can, g Mass of water, g Mass of can, g Mass of dry soil 1 2 101 102 (a) | 166 193 (b) | 153 180 (a-b) (c) | 101 128 (b-с) a-b x 100% Water Content,% b-c B. Determination of Dry Unit Weight Determination Number Mass of wet soil & cylinder, g Mass of cylinder,g Mass of wet soil,g Volume of Cylinder, cm^3 Specific Gravity Wet Unit Weight, KN/cm^3 1 2 3600 (A) (B) (A-B) 5200 3600 3600 1943.86 1943.86 2.7 2.7 Yw Yw Ys = (w + Ys = (w +) Dry Unit Weight, KN/ cm^3 Ys Ys Ydry Yary 1+w 1+w

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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Similar questions

You are checking a field-compacted layer of soil. The laboratory control curve has the following values: gd (kN/m3) w (%) 16.34 14 16.57 16 16.65 18 16.50 20 16.26 22 15.87 24 The specification for compaction states that the field-compacted soil must be at least 95% of the maximum control density and within of the optimum moisture for the control curve. You dig a hole 1m3 in the compacted layer and extract a sample that weighs 17.2 kN wet and 14.1 kN dry. (a) What is the compacted gd? The compaction w? The percent compaction? Does the sample meet the specifications? (b) If the specific gravity is 2.68, what is the compacted degree of saturation? If the sample were saturated at constant density, what would be the water content?
Discuss your results and compare your results with any published data. For the bulk density test, obtain the value of dry density. How does the values of (i) void ratio (ii) porosity influenced your results? How will these values differ if the soil is fully saturated, partially saturated or submerged? Support your assertions using relevant phase relationship equations. Whatwill happen to the values of particle density (specific gravity) obtained if the soil sample contain large quantities of peat or organic material? How will the results be used in specific practical conditions in the construction industry? Explain any source(s) of error in your experiment and state how it affected your result.
No wrong answers please. During the determination of the shrinkage limit of a sandy clay, the following laboratory data was obtained. Wet weight of soil dish- 87.85 g Dry weight of soil dish 76.91 gWeight of dish - 52.70 g Volumetric determination of soil--Weight of dish mercury-430.80 g -Weight of dish - 244.62 g- Density of solid soil 2650 kg/m Compute the shrinkage limit of the soil. Compute the shrinkage ratio. Compute the specific gravity of soil.
I need just the answer of (d) The following data were recorded during a shrinkage limit test on the sample from the soil: • Initial volume of soil in a saturated state = 20.6 cm3 • Final volume of soil in a dry state = 13.8 cm3 • Initial mass in a saturated state = 47.5 g • Final mass in a dry state = 34.6 g (i) Determine the shrinkage limit of the soil sample. (ii) Determine the shrinkage ratio.
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A dry sand is placed in a containerhaving a volume of 0.0089 cm3. The dry weight of the sample is 0.014 grams. Water is carefully added to the container so as not to be disturbthe condition of the sand. When the container is filled, the combined weight of the of soil plus water is 0.0174 grams. With this, find the void ratio, the porosity, the water content, the specific gravity thepertinent unit weights of the soil sample.
cylindrical soil specimen of 50 mm diameter and 100 mm long is to be prepared from a dry soil. If the specimen required 15% water of its own dry weight and the volume of air voids are 20% from its total volume, calculate the weight of the soil and water required for sample preparation, known that the specific gravity of the soil is 2.69.
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