Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN: 9781305635180
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
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Chapter 5, Problem 5.12P
To determine
Find the maximum dry density of compaction and the optimum moisture content based on modified proctor test.
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The soil with Gs = 2.7 ande = 0.8 has been air dried for 6 hours. After
drying, the moisture content (w) and void ratio (e) reduce to 20% and 0.7,
respectively. Determine the degree of saturation, Sr after drying the
sample. (provide the answer in % and the value must be round off to
whole number; e.g 20%)
The soil with Gs = 2.7 and e = 0.8 has been air dried for 6 hours. After
drying, the moisture content (w) and void ratio (e) reduce to 20% and 0.7,
respectively. Determine the degree of saturation, Sr after drying the
sample. (the answer in % ,value must round off to whole number)
Your answer
Laboratory compaction test results on a clayey silt are given in the following table:-
Moisture content (%)
Dry unit w3eight (kNm³)
14.80
8
17.45
18.52
11
18.9
12
18.6
14
16.9
Following are the results of a field unit weight determination test on the same
soil with the sand cone method:-
• Calibrated dry density of Ottawa sand = 1667 kg/m³
• Calibrated mass of Ottawa sand to fill the cone = 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%
Determine
a. Dry unit weight of compaction in the field
Chapter 5 Solutions
Fundamentals of Geotechnical Engineering (MindTap Course List)
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- b. A soil layer with a moisture content of 10 percent and a unit weight of 16.5 KN per cubic meter has a decrease in volume by 10 percent after compaction. Find the dry specific weight of the soil. Please solve question (b) I have the answer of Question (a)if you need information please use the question a. given data: a. The moisture content of the soil specimen is 20%, and the total weight is 300 gr. Calculate thevoid ratio and degree of saturation if the total volume was 180 cm3. (Gs = 2.5)arrow_forwardThe soil with Gs = 2.7 ande = 0.8 has been air dried for 6 hours. After drying, the moisture content (w) and void ratio (e) reduce to 20% and 0.7, respectively. Determine the degree of saturation, Sr after drying the sample. (the answer in % & value only no unit must round off to the nearest whole number) Your answerarrow_forwardQ1. Undisturbed soil sample was collected from the filed in Melbourne in steel tube for laboratory evaluation. The tube has the diameter of 85 mm, length of 170 mm and the moisture weight of 1600 g. If the oven-dried weight was 1200 g and Gs = 2.69, calculate the followings:a) Bulk unit weight (?) in kN/m3b) Field moisture content (w)c) Void ratio & porosity (e and ⴄ) d) Degree of saturation (Sr)arrow_forward
- Sand cone equipment is used to perform Field density test on a compacted arth fil. Soil sample dug from the test hole = 19.88 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 HN/m. Which of the following gives the water content of the tested soil? O a. 15.45% O b. 12.08% O c. 14.96% O d. 10.94% e. 11.12%arrow_forward= 2.75, calculate the zero-air-void unit weight for a soil in lb/ft³ Given G, w = 5%, 8%, 10%, 12%, and 15%. 6.1 6.4 The results of a standard Proctor test are given below. Determine the maximum dry unit weight of compaction and the optimum moisture content. Weight of wet soil in the mold (Ib) GsYw Vzav = 1+ wG5 Volume of Moisture Proctor mold (ft³) content (%) GSYW 1/30 3.26 8.4 Ya = 1/30 1+ GW 4.15 10.2 1/30 4.67 12.3 1/30 4.02 14.6 Se = Gsw 1/30 3.63 16.8 6.5 For the soil described in Problem 6.4, if G, = 2.72, determine the void ratio and the degree of saturation at optimum moisture content.arrow_forwardDetermine the maximum dry unit weight of compaction Determine the optimum moisture content Determine the void ratio at the optimum moisture content. Gs = 2.70 Determine the degree of saturation at the optimum moisture content A field unit determination test for the soil, yielded the following data. Moisture content = 12% and moist unit weight = 106 lb/cu.ft. Determine the relative compaction.arrow_forward
- The detailed results of a standard Proctor test on a soil classified as CL-ML and group name silty clay with sand are shown in the table below. Determine the maximum dry unit .weight and optimum water content .Diameter of mould = 4 in Height of mould = 4.584 in .Volume of mould 1/30 ft3 .Weight of mould, M = 4.45 lb Unit weight data Water content data Mass of can and Weight of wet soil and mold (Ibs) Mass of can (grams) Mass of can and wet soil (grams) dry soil (grams) M M.. M. M. 111.48 155.08 7.32 114.92 46.50 163.12 190.43 193.13 7.57 46.43 7.68 7.74 178.64 46.20 178.24 46.50 7.65 188.77 171.58 46.10arrow_forwardResults of a standard proctor ompaction test on a silty sand are shown. PLEASE COMPLETE SOLUTION. 1) Find the maximum dry unit weight and optimum moisture content. 2) What is the moist unit weight at optimum moisture content? 3) What is the degree of saturation at optimum moisture content? Given: ?? = 2.69 4) If the required field dry unit weight is 18.5 ??/?3, what is the relative compaction? 5) What should be the range of compaction moisture contents in the field to achieve the above relative compaction? 6) If the minimum and maximum void ratios are 0.31 and 0.82, respectively, what is the relative density of compaction in the field?arrow_forwardA sample of field soil is tested in the laboratory. The results of Standard Proctor test at optimum moisture content are given in the Table below: At the field, a sample of 552 cm' of this soil weighted 1104 g. The weight of water in this sample was 184 g. Find the degree of compaction (Relative compaction) at Standard Proctor test. Table 1: Standard Proctor test data Mass of wet soil in the mold(g) Volume of proctor mold(cm°) Optimum moisture content% 1656 828 12 CS Scanned with CamScannerarrow_forward
- Afalling-head permeability test is performed on a fine-grained soil. The soil sample has a length of 120 mm and a cross-sectional area of 600 mm2. The water in the standpipe flowing into the soil is 0.60 m above the top of the sample at the start of the test. It falls 50 mm in 30 minutes. The standpipe has a cross-sectional area of 200 mm2. (a) Make a sketch of the described conditions. (b) What is the coefficient of permeability in millimeters per second? (c) What is the coefficient of permeability in feet per minute? (d) On the basis of the calculated value for khyd, what is the probable soil type?arrow_forwardThe field wt of soil sample is 1900 kg/m³and the unit wt of the soil particle is 2660 kg/m³. A. Compute the dry unit weight if the moisture content is 11.5%. B. Compute the void ratio. C. Compute the degree of saturation.arrow_forwardSoil Compaction Homework: the results of standard compaction test are shown in the table below. Determine the maximum dry unit weight and the optimum moisture content. W% 6.2 8.1 9.8 11.5 12.3 13.2 ybkN/m³ 18.7 20.4 16.9 19.5 20.5 20.1 а. What is the dry unit weight and w at 95% standard compaction? b. Determine the degree of saturation at the maximum yd , Gs = 2.7? С. Plot the zero air voids line?arrow_forward
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