Problem 2: From the following data, solve for the degree of compaction. Field density test data: Weight of material taken from hole, kg. = Weight of sand to fill hole, kg. = Volume of hole, cu.m = 4.885 3.885 0.0028 Wet unit weight, kg/cu.m = Dry unit weight, kg/cu.m = 1745 1663 Laboratory Data: Maximum dry density, kg/cu.m. = Optimum Moisture Content, % = 2250 8.10 A BI - I

Problem 2: From the following data, solve for the degree of compaction. Field density test data: Weight of material taken from hole, kg. = Weight of sand to fill hole, kg. = Volume of hole, cu.m = 4.885 3.885 0.0028 Wet unit weight, kg/cu.m = Dry unit weight, kg/cu.m = 1745 1663 Laboratory Data: Maximum dry density, kg/cu.m. = Optimum Moisture Content, % = 2250 8.10 A BI - I

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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From the following data, solve for the degree of compaction. Field density test data: Weight of material taken from hole, kg. = Weight of sand to fill hole, kg. = Volume of hole, cu.m = 4.885 3.885 0.0028 Wet unit weight, kg/cu.m = Dry unit weight, kg/cu.m = 1745 1663 Laboratory Data: Maximum dry density, kg/cu.m. = 2250 Optimum Moisture Content, % = 8.10
Given in the following table is the result of the standard proctor test: Dry Unit Weight (KN/m³) 16.79 Test no. Water Content (%) 5.75 1 2 8.60 19.20 3 13.7 19.28 4 16.8 16.99 5 21.3 15.50 Then followed by a sand cone test has been performed in a compacted fill which has the following data: Initial mass of sand cone apparatus with sand = 5.915 kg Final mass of sand cone apparatus with sand = 2.380 kg Mass of soil removed from hole = 2.883 kg Moisture content of soil from hole = 7% Density of sand =1300 kg/m³ Volume of sand cone below valve = 0.001114 m³ a) Using compaction curve, plot the water content versus the dry unit weight to determine the maximum dry unit weight and the optimum moisture content b) Determine the field dry unit weight in KN/m³ c) Determine the relative compaction and rate its quality
Sand cone equipment is used to perform a field density test on a compacted earth fill. • Soil sample dug from the test hole = 20.60 N • Dry weight of soil sample = 17.92 N Ottawa sand used to fill the hole weighs 16.05 N and is known to have a density of 15.74 kN/m³. Determine the volume of the test hole. Answer:
Sand-cone equipment is used to determine an in-place unit weight (field density test) on a compacted earth fill. The sand used in the cone is known to have a bulk density of 15.73 kN/m3 Wet weight of soil sample dug from test hole = 2100 g Dried weight of soil sample = 1827 g Weight of sand (sand core) to fill the test hole = 1636 g a) Compute the water content. b) Compute the in-place dry unit weight of tested soil. c) Compute the percentage of compaction of the tested soil if the laboratory moisture-unti weight curve indicates a dry unit weight of 18.09 kN/m3 and a optimum moisture content of 13%.
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
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)
Question 40 The results of a constant head permeability test for a fine sand are as follows: Diameter of the sample = 25 cm Length of sample = 80 cm Constant head difference = 77 cm Time of collection = 388 secs Weight of water collected = 340 grams Find the seepage velocity in cm/min. if the void ratio is 0.49. Round off to four decimal places.
a completely dry sand sample fills a 200 cm 3 metal cylinder and weighs 260 g (Ws), having Ss = 2.6 Calculate e(void ratio), then complete the schematic. (see the picture). I WILL RATE IT.
Table 1 below is the result of compaction results. Please evaluate the correct combination of missing data. Table 1: Compaction test results Water content (%) 8.00 III IV Bulk Density II 2.35 2.09 (Mg/m³) Dry Density (Mg/m³) 1.89 2.08 1.76 || A B Answer A B с D I 5.00 1.87 I 1.68Mg/m³ 1780kg/m³ 1.80Mg/m³ 1.78Mg/m³ 2.04Mg/m³ 2.08Mg/m³ 2080kg/m³ 2040kg/m³ III 13% 11% 12% 13% 16.00 2.20 1.90 IV 19% 20% 19% 19%
The results of a constant head permeability test for a fine sand are as follows: Diameter of the sample = 31 cm Length of sample = 98 cm Constant head difference = 52 cm Time of collection = 364 secs Weight of water collected = 448 grams Find the seepage velocity in cm/min. if the void ratio is 0.33. Round off to four decimal places. Answer: 0.3943
The results of a constant head permeability test for a fine sand are as follows: Diameter of the sample = 35 cm Length of sample = 83 cm Constant head difference = 59 cm Time of collection = 310 secs Weight of water collected = 372 grams Find the seepage velocity in cm/min. if the void ratio is 0.48. Round off to four decimal places.
3. 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 a. Determine the moist unit weight of compacted soil in kN/m³. b. Find the dry unit weight in kN/m³. c. Compute the relative compaction in the field.