A laboratory modified Proctor compaction test gave: optimum water content 16 % and maximum dry density Pdmax 1.74 gm/cm°. If the specifications require that at least a relative compaction of 95% and a water content should be: optimum water content +2%. A field density test gave the following values: the volume of the hole was 1165 cm. The mass of wet soil removed from the hole is 2230 gm and the water content is 20.41%. Does the test meet specifications? Why?

A laboratory modified Proctor compaction test gave: optimum water content 16 % and maximum dry density Pdmax 1.74 gm/cm°. If the specifications require that at least a relative compaction of 95% and a water content should be: optimum water content +2%. A field density test gave the following values: the volume of the hole was 1165 cm. The mass of wet soil removed from the hole is 2230 gm and the water content is 20.41%. Does the test meet specifications? Why?

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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the modified compaction method for the same soil sample gives 10 points . .moisture value than the standard compaction method and ..2. max. dry density value 1 (lower) 2(lower) O 1 (lower) 2(greater) O 1(greater) 2(lower) O 1(greater) 2(greater)
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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.
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)
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)
Question 4 In a standard compaction test on a soil (grain specific gravity of 2.67) the following results were obtained: Water content (%) 12.8 Mass (g) 1890 14.5 2130 15.6 2200 16.8 2210 19.2 2160 The volume of the mould is 1000 cm (i) Determine the maximum dry density obtainable under standard test conditions and the water content at which this occurs. (ii) Using the above results, specify a detailed quality control plan for field compaction of a highway project.
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The following are results of a field unit weight determination test using sand cone method: Volume of hole= 0.0015 m^3 Mass of moist soil from hole = 3.07 kg Water content = 12.24% Maximum dry unit weight from a laboratory compaction test = 19.28 kN/m^3 Determine the relative compaction, in percent.
Estimate coefficient of permeability for a sand deposit where field testing indicates that the in-situ e=0.50 and D10=0.55 mm and Cu = 8. Using the information from (1) determine the D60 particle size using the Kozeny-Carman equation.
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