Following are the results of a field unit weight determination test on a soil with the sand cone method. Calibrated density of Ottawa Sand%3D 1667 kg/m3 Calibrated mass of Ottawa sand to fill the cone%3D 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%

Following are the results of a field unit weight determination test on a soil with the sand cone method. Calibrated density of Ottawa Sand%3D 1667 kg/m3 Calibrated mass of Ottawa sand to fill the cone%3D 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%

Principles of Foundation Engineering (MindTap Course List)

9th Edition

ISBN:9781337705028

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter2: Geotechnical Properties Of Soil

Section: Chapter Questions

Problem 2.13P

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The following are the results of a field unit weight determination test performed on the soil by means of the sand cone method. Calibrated dry density of Ottawa sand = 1570 kg/m3 Calibrated mass of Ottawa sand to fill the cone = 0.545 kg Mass of jar + cone + sand (before use) = 7.59 kg Mass of jar + cone + sand (after use) = 4.78 kg Mass of moist soil dug from hole = 3.007 kg Water content of moist soil = 10.2% Max dry unit weight = 19 kN/m3 Find the dry unit weight in kN/m3.
Following are the results of a field unit-weight determination test performed on the same soil by means of the sand cone method: -Calibrated dry unit weight of Ottawa sand = 15.4019 kn/m3 -Calibrated weight of Ottawa sand to fill the cone = 5.3496 N -Weightof jar + cone + sand (before use) = 74.451 N -Weightof jar + cone + sand (after use) = 46.886 N -Weight of moist soil from hole = 29.509 N -Moisture content of moist soil = 10.21% DETERMINE AND SHOW COMPLETE SOLUTION 1. Dry density of compaction in the field in kg/cu.m.
Pressure (kN/m²)27 54 107 214 429 214 107 54 Void ratio1.243 1.217 1.144 1.068 0.994 1.001 1.012 1.024 The table above shows results obtained from an oedometer test on a specimen of saturated clay. A layer of this clay 2.5 m thick lies below a 10 m depth of sand, the water table being at the surface. The saturated weight for both soils is 19 kN/m³. A 10-m depth of fill of unit weight of 21 kN/m³ is placed on the sand over an extensive area. If the fill was to be removed some time after the completion of consolidation, what heave (mm) would eventually take place due to swelling of the clay? (0 d.p). Use unit weight of water of 9.8 kN/m3. Use log base 10.
Following are the results of a field unit weight determination test performed bymeans of the sand cone method:• Calibrated dry density of Ottawa sand = 104 lb/ft3• Mass of Ottawa sand to fill the cone = 0.258 lb• Mass of jar + cone + sand (before use) = 13.21 lb• Mass of jar + cone + sand (after use) = 6.2 lb• Mass of moist soil from hole = 7.3 lb• Moisture content of moist soil = 11.6%Determine: Dry unit weight of compaction in the field.
Following are the results of a field unit weight determination test performed by means of the sand cone method:• Calibrated dry unit weight of Ottawa sand = 104 lb/ft3• Weight of Ottawa sand to fill the cone = 0.258 lb• Weight of jar + cone + sand (before use) = 13.4 lb• Weight of jar + cone + sand (after use) = 6.39 lb• Weight of moist soil from hole = 7.3 lb• Moisture content of moist soil = 11.6%Determine: Dry unit weight of compaction in the field.
A sand cone test has been performed in a recently compacted fill. The test results were as follows: Initial weight of sand cone + sand = 6.14 kg Moisture Content Test: Final weight of sand cone + sand = 1.94 kg Mass of empty moisture can = 23.11 g Weight of sand to fill cone = 0.96 kg Mass of moist soil + can = 273.93 g Weight of soil from hole + bucket = 5.65 kg Mass of dry soil + can = 250.10 g Weight of bucket = 0.55 kg The sand used on the sand cone had a unit weight of 12.75 kN/m3 and the fill had a maximum dry unit weight of 19.04 kN/m3 and an optimum moisture content of 11.7 % a. Compute the volume of hole b. Compute the dry unit weight of soil. c. Compute the relative compaction.
A fully saturated clay sample has a mass of 1526 grams. After oven-drying, its mass was reduced to 1053 grams. The specific gravity of soil particles is 2.7. Calculate the porosity in percent (%). Select one: a. 25.17 b. 43.19 c. 54.81 d. 32.52
Undisturbed soil obtained from a test pit 0.0283 m3 (1 ft3) in volume is found to have a wet weight of 0.459 kN (103.2 lb). The dry weight of the sample is 0.376 kN (84.5 lb). What would be the effective unit weight of such a soil if it were submerged below the groundwater table? The specific gravity of the soil is determined to be 2.70.
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 8.5 lb. The volume of the test hole is 0.065 ft3. From these data, compute the wet unit weight of the soil.
During the determination of the shrinkage limit of a sandy clay, the following laboratory data was obtained: Wet wt of soil + dish = 87.85 g Dry wt. of soil + dish = 76.91 g Wt. of dish = 52.70 g The volumetric determination of the soil pat Wt. of dish+ mercury = 430.8 g Wt. of dish = 244.62 g Calculate the shrinkage limit, assuming Gs = 2.65 A. 12.54% B. 18.76 % C. 10.87% D. 15.23% OE 15.57 %
Following are the results of a field unit weight determination test on a soil using the Sand Cone Method. Determine the field unit weight of the soil. FEEDBACK GUARANTEED
A sample of soil compacted according to a standard Proctor test has a unit weight of 130.9 lb/ft^3 at 100% compaction and at optimum water content of 14%. What is the dry unit weight? If the voids become filled with water what would be the saturated unit weight? Assume Gs= 2.67.