The results of a standard Proctor test are given in Table Q2 below. Determine the maximum dry unit weight of compaction and the optimum moisture content. Table Q2 Volume of Proctor Mass of wet soil in the mold Moisture content mold (cm³) (kg) (%) 1000 1.78 5.0 1000 1.87 7.5 1000 1.95 10.0 1000 1.98 12.5 1000 2.02 15.0 1000 1.97 17.5 1000 1.91 20.0
Q: Q 1/ The following table shows the standard proctor test for clay soil that required: Water 6 8. 10…
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Q: 1. The results of a standard Proctor test are given in the following table. Weight of Moist Soil in…
A: Weight of moist Moisture Moist unit Dy unit Trial No Soil in mold Content Weight Weight 1…
Q: Table Q2(b)(i): Result of bulk density and water content from proctor compaction test Bulk density…
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Q: You are checking a field-compacted layer of soil. The laboratory control curve has the following…
A: Draw the curve between water content and dry density: From the above graph: Maximum dry density is…
Q: The results of a standard and a modified Proctor tests ate given in the table below. a) Generate a…
A: STANDARD PROCTOR TEST MODIFIED PROCTOR TEST VOLUME OF MOULD Mass of wet soil in the mould (kg)…
Q: A modified Proctor compaction test was carried out on a clayey sand in a cylindrical mold that has a…
A: Calculating the moist unit weight of soil: γ=WVm×9.81γ=18501000×9.81 =18.15 KN/m3
Q: 3. Given below the tabulated result in the field of compaction test. A. Unit Weight Determination…
A: Water content is defined as the ratio of weight of water to the weight of soil solids= (Wt of…
Q: 6. Following is the result of 8 trials of a standard proctor test. Determine the optimum water…
A: The dry density of the given samples of soil are calculated below:
Q: From the following data, solve for the degree of compaction. If this is for Item 201, Aggregate Base…
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Q: Determine the relative compaction in the field based on the Laboratory compaction test results are…
A: Relative compaction - It is the ratio of the field density and the maximum density of soil which was…
Q: 6. Following is the result of 8 trials of a standard proctor test. Determine the optimum water…
A: The water content corresponding to the maximum dry density of the soil is called optimum moisture…
Q: A sample of field soil is tested in the laboratory. The results of Standard Proctor test at optimum…
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Q: The results of a standard Proctor compaction test on a silty sand are given in the following table.…
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Q: Problem 1: The table below present the results of a standard proctor test of a borrow pit soil. FEB…
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Q: Q/ The laboratory test results of a standard Proctor test are given in the following table Moisture…
A: The given data is shown below:
Q: A standard compaction test was performed in the laboratory and the follows: Trial Number 2 3 4 Moist…
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Q: A sample of field soil is tested in the laboratory. The results of Standard and Modified Proctor…
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Q: TABLE 2 RESULTS OF ORI TEST Water Content (%) Weight (Kg) 14 1.45 16.5 1.71 18.5 1.87 20 1.88 21…
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Q: The results of a standard and a modified Proctor tests ate given in the table below. a) Generate a…
A: * zero air voids curve is mention in step 3 and 4 in soil sample, moisture content=weight of water…
Q: How do i solve problem 6.8 a & b
A: The given data is shown below:
Q: For a given soil, following After compaction of soil in the field, sand cone of tests (control…
A: Following specification required a. γd must be at least 0.95 (γdmax).b. Moisture content, w, should…
Q: Standard Proctor test was conducted on a silty clay soil collected from a proposed construction…
A: As per our guidelines we are supposed to answer only first three subparts. So kindly please repost…
Q: The following compaction curve represents the compaction data from the Lab for soil. Following are…
A: please find attached solution below.
Q: 6. Following is the result of 8 trials of a standard proctor test. Determine the optimum water…
A: Trial Proctor Mold Volume (cc) Mass (g) Moisture content Ybulk(g/cc) Ydry(g/cc) 1…
Q: A standard compaction test was performed in the laboratory and the results are tabulate follows:…
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Q: The results of a standard Proctor test are given in the following table. Given: Mold volume = 1/30…
A: Note: Since you have asked multiple questions, we will solve the first question for you. If you want…
Q: 6. Following is the result of 8 trials of a standard proctor test. Determine the optimum water…
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Q: A standard Proctor test was conducted on a silty clay soil collected from a proposed construction…
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Q: The detailed results of a standard compaction test are shown in the table below. Determine the…
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Q: 6.4 The results of a standard Proctor test are given below. Determine the maximum dry unit weight of…
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Q: Following are the results from the liquid and plastic limit tests for a soil. Liquid limit test:…
A: Given data PL = 18.7%
Q: plume of Proctor Mold (cm³) Moisture Content (w% mold (kg) 943.3 1.65 5.5 943.3 1.70 7.3 943.3 1.80…
A: Yd = dry unit weight w = moisture content Yb = wet unit weight / bulk density
Q: Standard Proctor test was conducted on a silty clay soil collected from a proposed construction…
A: we know that, from standard proctor testDry densityρd=ρ1+wGiven,volume of the mouldV=943.3…
Q: Problem 4: A balloon-type apparatus is used to determine an in-place unit weight for a soil (field…
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Q: The soil sample is being compacted. It has a Specific gravity of 2.70 Find the Following: a. Using…
A: Compaction curve : It is the graph between water content (w) on x- axis and dry unit weight on…
Q: Bulk Unit Weight (kN/m3) 20.0 20.7 21.7 21.4 21.1 Water Content (%) 8.0 9.5 11.5 12.5 13.5 1.…
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Q: 5.1 The results of a standard Proctor test are given below. Determine the may imum dry unit weight…
A: Note: As per our policy for multiple questions only one question can be solved for other question…
Q: 1972 kg/mᶟ wet unit weight and 5.5% average actual moisture content were obtained from a laboratory…
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Q: Question 4 In a standard compaction test on a soil (grain specific gravity of 2.67) the following…
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Q: QUESTION 19 From a compaction test, if you are given: Water content (%) = 13.2 Bulk unit weight…
A: Use formula of dry unit weight in terms of bulk unit weight and water content.
Q: Q3. (a) The following results are obtained from a Standard Proctor compaction test: Mass of soil…
A: Given data, Specific gravity = 2.70 Volume V = 0.945×10-3 m3
Q: 8. The results of a standard compaction test on a soil sample from Glasgow are shown in the graph…
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Q: Which of the following are standard procedures used for determining the field unit weight of…
A: Let us proceed with a brief description about the aim of each method listed in the question. Sand…
Q: Make necessary calculations to see if the control tests meets the specifications. For a given soil,…
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Q: PROBLEM 2: A 50 cu.cm sample of moist clay was obtained by pressing a sharpened hollow cylinder into…
A: Given data- Total Volume of the soil-VT=50 cm3Initial weight of the soil-Wbulk=85 gmDry weight of…
Q: . Following is the result of 8 trials of a standard proctor test. Determine the optimum water…
A: Compaction is the process in which soil particles are artificial and packed together into closer…
Q: The compaction of a soil obtained from a standard compaction test is shown in the attached figure.…
A: Given:
Q: The results of a standard Proctor test are given in the following table. Determine the maximum dry…
A: It is required to determine the bulk density of soil. ρ=MassVolume=1.6943.3 cm3×1 m3106 cm3=1696.17…
Q: A modified Proctor compaction test was carried out using a cylindrical mold with a 1000 ml volume.…
A: Formulaeρ=Wet massVolumeρdry=ρ1+wρzav=Gρw1+GwExample:at…
Q: A fine-grained soil has 60% clay with LL = 220%, PL = 45%, and a natural water content of 6%. A…
A: Given:- Diameter of mold = 4 inHeight of mold = 4.584 inMass of mold = 4.27…
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- Calculate the coefficient of gradation, Cc. ? Classify the soil by AASHTO and USCS system. (LL = 30, PL = 10) ? Use this given M=Mass of soil retained in pan +∑Mass reatianed in each sieve⇒M=617 gM=Mass of soil retained in pan +∑Mass reatianed in each sieve⇒M=617 g Sieve No. Opening Size (mm) Mass retained (g) %Mass retained=MiM×100=MiM×100 %Cumulative mass retained % Finer 4 4.750 28 =28617×100%=4.54%=28617×100%=4.54% 4.54 95.46 10 2.0 42 =42617×100%=6.81%=42617×100%=6.81% 11.35 88.65 20 0.850 48 =48617×100%=7.78%=48617×100%=7.78% 19.13 80.87 40 0.425 128 =128617×100%=20.74%=128617×100%=20.74% 39.87 60.13 60 0.250 221 =221617×100%=35.82%=221617×100%=35.82% 75.69 24.31 100 0.150 86 =86617×100%=13.94%=86617×100%=13.94% 89.63 10.37 200 0.075 40 =40617×100%=6.48%=40617×100%=6.48% 96.11 3.89 Pan 24 =24617×100%=3.89%=24617×100%=3.89% 100 0 Now, to plot the GSD curve, the % finer is kept on the vertical axis and the opening size are kept on the horizontal axis -…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?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 moist unit weights and degrees of saturation of a soil are given in the table. y (kN/m3) S(%) 16.62 50 17.71 75 Determine: a. e b. Gs Show handwritten solutions.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 2116.8The laboratory compaction test of a certain type of soil gives a maximum dry density of 1.486 g/cc3 with an optimum moisture content of 12.5%. Determine the relative compaction of the soil using the following results of a filed unit weight determination test using sand cone method. Volume of soil excavated from the hole = 0.001337 m3 Weight of moist soil from the hole = 2220 g Weight of oven dried soil = 1890 g Select one: a. 97.25 b. 95.10 c. 90.34 d. 88.41
- The ff are results of a field unit weight determination test Volume hole 0.0016 Mass of moist soil from 3.10kg Water content 12.20% Mas dry 19.50 Determine the relative compaction in %Calculate the CEC (m.e./100g soil) using the following given miliequivalent weights obtained from a 25g soil sample (a. Ca++= 2.23 m.e; b. Mg++=1.32 m.e.; c. Na+= 1.98 m.e.; d. K+= 1.87 m.e.; e. H+= 1.27 m.e.; f. Al+++= 0.11 m.e.; g. NH4+= 2.3 m.e.; h. NO3-= 1.5 m.e.arrow_forward Question A fine-grained soil has 60% clay with LL 220%, PL 545%, and a natural water content of 6%. A standard Proctor test was carried out in the laboratory and the following data were recorded Diameter of mold = 101.40 mmHeight of mold = 116.70 mmMass of mold = 4196.50 gramSpecific gravity, Gs = 2.69 a) Determine the maximum dry unit weight and optimum water content.b) If the desired compaction in the field is 95% of the standard Proctor test results, what values of dry unit weight and water content would you specify?c) Explain why you select these values.d) What field equipment would you specify to compact the soil in the field, and why?e) How would you check that the specified dry unit weight and water content are achieved in the field? I Need The Answer of part (d) and (e)...
- Question 30 The data from a falling head test is shown. Diameter of standpipe: 6 cm. Initial Head: 90.4 cm. Duration of test: 18 min. Length of soil sample: 30.35 cm. Diameter of permeameter: 44 cm. If the soil permeability is 0.0038 cm/min, determine the final head (in cm.) of the permeability test. Round off to two decimal places.Calculate the CEC of a 100g soil sample that was found to contain the following cations: Ca2+ = 0.12g Mg2+ = 0.024g K+ = 0.078g Al3+ = 0.027g NH4+ = 0.036g H+ = 0.002gThe 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)