gm/cu.cm with an optimum water content of 12.5%. The ff are the results of a field density test using sand code equipment: Volume of soil excavated from the hole= 0.001337 cu.m. Weight of wet soil from the hole- 2220 grams Weight of soil after oven drying= 1890 grams a. Determine the field unit weight of the soil in gm/cu.cm. b. Find the water content of soil. c. What is the relative compaction?

gm/cu.cm with an optimum water content of 12.5%. The ff are the results of a field density test using sand code equipment: Volume of soil excavated from the hole= 0.001337 cu.m. Weight of wet soil from the hole- 2220 grams Weight of soil after oven drying= 1890 grams a. Determine the field unit weight of the soil in gm/cu.cm. b. Find the water content of soil. c. What is the relative compaction?

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Problem 4: A balloon-type apparatus is used to determine an in-place unit weight for a soil (field density test). The volume of the test hole is 0.000708 cu.m. The weight of the soil obtained from the test hole is 1410 g. The soil sample was dried on a field stone and weighs 1235 g. 1. Compute the water content of the tested soil. 2. Compute the in-place dry unit weight of the tested soil. 3. Determine the $ of compaction for the test result if the soil is part of a compacted carth fill whose maximum unit weight (from laboratory compaction tests) is 18.67 KN/m³.
A fully saturated soil sample has a volume of 28 cc. The sample was dried in oven and the weight of the dry soil pat was found to be 48.76g. Given G= 2.68. Determine the dry unit weight of soil in kN/m3.
1. Undisturbed soil sample was collected from the field in steel Shelby tubes for laboratory evaluation. The tube sample has a diameter of 2.8 in., length of 22 in., and a moist weight a of 9.56 lb. If the oven-dried weight was 8.51 lb, and Gs = 2.69, calculate the following: a. Moist unit weight b. Field moisture content c. Dry unit weight d. Void ratio e. Degree of saturation
The following data were obtained from sand-cone experiment where a volume of a certain soil is replaced with a sand in order to find the density of the soil. Density of sand used in test = 1.3 g/cc Wt. Of soil excavated from the pit = 925 g Wt. of sand filling the pit = 675 g Water content of natural soil = 15% Specific gravity of Solid = 2.7 a) Determine the wet density of the soil in g/cc. b) Determine the dry density of the soil in g/cc. c) What is the porosity of the soil?
A sample of saturated clay was placed in a container and weighed. The weight was 6N. The clay in its container was placed in an oven for 24 hours at 105˚C. The weight reduced to a constant value of 5N. The weight of the container is 1N. If Gs = 2.7, determine: a) Water Content b) Void Ratio c) Bulk Unit Weight d) Dry Unit Weight e) Effective Unit Weight
8 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 = 3.24 kg Water content = 12.01% Max. dry unit weight from a laboratory compaction test = 19.70 kN/m3 Determine the relative compaction, in percent. Round off to two decimal places. Hint: Relative compaction = Field dry unit weight / Max dry unit weight (lab)
(detailed and clean solution please) large soil sample obtained from a borrow pit has a wet mass of 26.50kg. the in – place volume occupied by the sample is 0.013m3. A small portion of the sample is used to determine the water content; the wet mass is 135g, and after drying in an oven, the mass is 117g. a. Determine the soil’s water content. b. Determine the soil wet and dry density for conditions at the borrow pit.
9. The field unit weight of the soil sample is 1960 kg/m³, and the unit weight of the soil particle is 2700 kg/m³. If emax=0.69 and emin = 0.44. a) Compute the dry unit weight if the water content is 11% b) Compute the void ratio of the soil sample. c) Compute the relative density of the soil sample.
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.
1. A dry sand is placed in a container having a volume of 0.0092 cubic centimeter. The dry unit weightof the sample is 0.015 gram. Water is carefully added to the container so as not to disturbed thecondition of the sand. When the container is filled, the combined weight of soil plus water is 0.0182gram.a. Determine the specific gravity of soil particle.b. Determine the dry unit weight of soil.2. If the soil sample has a water content of 11% and a unit weight of 19.67 kN/m3 , determine thepercentage of air in voids and the degree of saturation if the specific gravity of soil solid is 2.75.3. If the soil sample has a specific gravity of 2.71 and a porosity of 0.345, determine the saturated unitweight and the bouyant unit weight.4. A soil sample having a liquidity index of 0.85, a plasticity index of 16.50 and a plastic limit of 11.45.The soil sample has a specicfic gravity of 2.68 and a void ratio of 0.77.a. Compute the degree of saturation.b. Compute the air void ratio.
A modified Proctor compaction test was carried out using a cylindrical moldwith a 1000 ml volume. Table 2 summarizes the results. a. Plot the compaction and ZAV curves (RD=2.75 and ρwater=1000 kg/m3)b. Determine the maximum dry density and the optimum moisture content Table 2 Moisture Content (%)Mass of WetSample (g)10.5 2033.211.8 2146.613.0 2203.514.9 2183.116.4 2147.617.6 2116.8