The detailed results of a standard compaction test are shown in the table below. Determine themaximum dry unit weight and optimum water content.1 Diameter of mold = 101.4 mmHeight of mold = 116.7 mmMass of mold, M = 4196.50 gramsUnit weight dataWater content dataMass of can(grams)M.Mass of wet soilMass of can andMass of can andand mold (grams)Mwmwet soil (grams)dry soil (grams)5906.00114.92111.4846.506056.00163.12155.0846.436124.00190.43178.6446.206156.00193.13178.2446.506103.00188.77171.5846.10

Answered: Image /qna-images/answer/9164f282-12df-48f8-8e57-c89e337464e5.jpg

The detailed results of a standard compaction test are shown in the table below. Determine the maximum dry unit weight and optimum water content. 1 Diameter of mold = 101.4 mm Height of mold = 116.7 mm Mass of mold, M = 4196.50 grams Unit weight data Water content data Mass of wet soil Mass of can and Mass of can (grams) M. Mass of can and and mold (grams) Mwm wet soil (grams) dry soil (grams) M. 5906.00 114.92 111.48 46.50 6056.00 163.12 155.08 46.43 6124.00 190.43 178.64 46.20 6156.00 193.13 178.24 46.50 6103.00 188.77 171.58 46.10

The detailed results of a standard compaction test are shown in the table below. Determine the maximum dry unit weight and optimum water content. 1 Diameter of mold = 101.4 mm Height of mold = 116.7 mm Mass of mold, M = 4196.50 grams Unit weight data Water content data Mass of wet soil Mass of can and Mass of can (grams) M. Mass of can and and mold (grams) Mwm wet soil (grams) dry soil (grams) M. 5906.00 114.92 111.48 46.50 6056.00 163.12 155.08 46.43 6124.00 190.43 178.64 46.20 6156.00 193.13 178.24 46.50 6103.00 188.77 171.58 46.10

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Related questions

Q: A sample of sand has the following properties: Moist mass = 625.2 g Dry mass = 589.9 g absorption =…

Q: Following the sieve analysis and the specific gravity of "SAND 1", a hydrometer test was performed…

A: The hydrometer test is used to determine the distribution of soil particle sizes that are smaller…

Q: - Soil Compaction '1. A series or Modilied Proctor Tlests have been perfomed on a soil that has a…

Q: Plot the Standard Proctor curve of dry unit weight versus water content (see below table) and; a)…

A: Plot the compaction curve by keeping water content along the horizontal axis and the dry unit weight…

Q: 11. The range of water content between the liquid and plastic limits is termed as: A. liquid limit…

Q: a. Determine the percent finer than each sie and plot a grain-size distribution curve. b. Determine…

A: Sieve analysis To find a) Percent finer and grain size distribution curve b) D10, D30, D60 C)…

Q: Consider the grain size gradation curve for the soil shown in the image below and classify the soil…

A: Look at the % passing through sieve no 200=20<50%⇒COARSE GRAINED SOIL

Q: Based on the compaction curve shown in Figure A (below), choose the appropriate combination for each…

A: Finding out the fill in the blanks of curve.

Q: effective particle size.

A: “Since you have posted a question with multiple sub-parts, we will solve first three subparts for…

Q: A soil sample following results: Standard sieve No: TOATE RE 4 10 20 40 100 200 was analyzed Pan…

A: IntroductionA sieve is a tool or equipment used in soil engineering to measure the particle size of…

Q: 1. An air dry (not oven dry) processed clay powder sample has the following measurements: - Mass of…

Q: A laboratory test was conducted according to Method A of the ASTM D4318 to find the liquid limit of…

A: The wet mass of the soil sample with the mass of the can is measured, again the dry mass of the…

Q: The weight of the dry soil per unit volume is known as Bulk Unit Weight Saturated Unit Weight O Dry…

A: 7. The weight of the dry soil per unit volume is known as Dry Unit Weight Bulk Unit Weight…

Q: A column analysis of a flocculating suspension is shown below. The initial solids concentration is…

A: C0 = Intital solids concentration = 200 mg/L(a) Calculate the removal fraction at each depth and…

Q: lable 5.2 Unified Soll Classification System (Based on Material Passing 76-2-mm Sieve) Criteria for…

A: Passing from sieve 200 =17% Passing from sieve 4=100% it means sieve belongs to sand fines are…

Q: Question 9 Soil placed on a can has been weighed before and after an oven-drying procedure. The…

Q: Standard Proctor test was conducted on a silty clay soil collected from a proposed construction…

Q: Soil Compaction 1. A series of Modified Proctor Tests have been perfomed on a soil that has a unit…

Q: Soil placed on a can has been weighed before and after an oven-drying procedure. The procedure…

Q: What is dry unit weight of the soil sample below (va) in Ib/ft ? The combined weight of a mold and…

A: Given data : Combined weight of mold and compacted soil sample (W1)=8.7 lbMold weight (W2)=5 lbmolds…

Q: The results of a Standard Compaction test are shown in the table below: Moisture Content Soil Unit…

Q: In the table below, find the missing values for pure sand sample and without fine materials ote…

Q: A Standard Proctor Test was completed according to Method A described in Chapter 12. The following…

Q: The laboratory test data for a standard Proctor test are given below. Volume of Moisture content…

Q: The particle destribution of soil are as follows. Silt = 24% Sand = 12% Clay = 44% %3D If the soil…

Q: 943.3 1.70 943.3 1.85 943.3 1.90 943.3 1.99 943.3 2.05 943.3 1.96 943.3 1.92

A: Standard proctor test Moist unit weight = wet soil / volume Dry unit weight = moist unit weight /…

Q: 1. An air dry (not oven dry) processed clay powder sample has the following measurements: - Mass of…

Q: Classify the soil by the USCS system.

Q: Question 4 (1 Listen The laboratory compaction tests requires several iterations of the test to be…

Q: A constant-head permeability test is performed in a laboratory where the soil sample is 25 cm in…

A: L=25cm V=200cm3 A=6cm2 t=2min To find permeability of soil sample

Q: 2. Following are the results of a sieve analysis a. Determine the percent finer of each sieve and…

Q: The results given in Table 1 were obtained in a particle size distribution test for soil sample…

A: The weight retained of soil is,To find: The gradation curve and its properties.

Q: a) Plot the grain size curve b) Calculate the coefficient of uniformity and the coefficient of…

A: For the given data, the particle size distribution curve is plotted below: a.

Q: A sample of soil obtained from a test pit 2 ft3 in volume has a weight of 210 lb. The entire sample…

A: Volume= 2 ft3 Total weight = 210lb Dry weight =185 lb

Q: Soil placed on a can has been weighed before and after an oven-drying procedure. The procedure…

Q: Question 18 Given below is the result of a sieve analysis done on a soll sample. Sleve No. Diameter…

A: To find out the coefficient of uniformity.

Concept explainers

Soil Compaction Test

Soil compaction is the major area of geotechnical engineering. Soil is composed of multiple pores present on its surface, while the soil is mainly composed of gravel, sand, rocks, etc. The arrangement of these rocks and gravels is such that they build up voids (spaces) between them, which allows spaces for the roots of the plants and crops to spread and absorb the required amount of nutrients. Soil compaction is a major problem primarily caused by human activities, micro-organisms, and other multiple factors.

Question

100%

The detailed results of a standard compaction test are shown in the table below. Determine the
maximum dry unit weight and optimum water content.
1 Diameter of mold = 101.4 mm
Height of mold = 116.7 mm
Mass of mold, M = 4196.50 grams
Unit weight data
Water content data
Mass of can
(grams)
M.
Mass of wet soil
Mass of can and
Mass of can and
and mold (grams)
Mwm
wet soil (grams)
dry soil (grams)
5906.00
114.92
111.48
46.50
6056.00
163.12
155.08
46.43
6124.00
190.43
178.64
46.20
6156.00
193.13
178.24
46.50
6103.00
188.77
171.58
46.10

Expert Solution

This question has been solved!

Explore an expertly crafted, step-by-step solution for a thorough understanding of key concepts.

This is a popular solution!

SEE SOLUTION Check out a sample Q&A here

Step 1

VIEW

Step 2

VIEW

Step 3

VIEW

Trending now

This is a popular solution!

Step by step

Solved in 3 steps with 3 images

SEE SOLUTION Check out a sample Q&A here

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, civil-engineering and related others by exploring similar questions and additional content below.

Similar questions

1. The results of a standard Proctor test are given in Table Q1 below. Determine the maximum dry density of soil compaction and the optimum moisture content. Table Q1 Volume of Proctor Mass of wet soil in the mold Moisture content mold (cm³) (kg) 3.39 (%) 11.5 987 987 3.87 13.5 987 4.22 15.5 987 5.11 17.5 987 4.75 19.5 987 4.31 21.5
Kindly give me right solution with clear calculations
Show complete solutions. Ang graph using Excel.
Subject soil Mechanics
In a liquid limit test using Fall cone apparatus, the following readings were recorded. While in the plastic limit test, the following data were obtained: Weight of soil + container = 22.12 grams Weight of dry soil + container = 20.42 grams Weight of container = 13.07 grams Determine the ff. (SHOW COMPLETE SOLUTION AND GRAPH) A. Find the liquid limit B. Compute the plastic limit. C. Find the liquid limit. D. Find the plastic limit. 40% 45% 50% 58% Water Content 12 16 21 35 Penetration of cone(mm) -PLS SHOW COMPLETE SOLUTION AND GRAPH. SHOW YOUR ANSWER IN 4 DECIMAL PLACES. I'LL RATE, THANKS!
The results of a Standard Proctor test are given in the following table. Moist Weight of Wet Sol Moisture Content in the mold (Ib) Trial (%) 26 4.08 28 4.12 30 4 4,18 32 Determine: 1. Maximum dry unit weight in Ib/ft^3 (4 decimal places) 2. Optimum water content in % (4 decimal places)
Someone can help me with this one. Please
The table shows the laboratory results of the sieve analysis of a sample. Plot the grain size curve of the soil. The soil has a liquid limit of 35% and plasticity index of 26%. SIEVE SIZE (mm) WEIGHT RETAINED 0.25 18.96 0.149 33.18 0.074 45.03 0.052 54.51 0.02 42.66 0.01 11.85 0.004 4.74 0.001 4.74 Pan 21.33 a.) Classify the soil using AASHTO method b.) Classify the soil using USCS method c.) Classify the soil using USDA method Please answer this asap. For upvote. Thank you in advance
SG is 2.66 and temperature is 18 degrees celcius
Problem 4: The following are the results of a sieve analysis. a. Determine the percent finer than each sieve and plot a grain-size distribution curve. b. Determine D10, D30, D60, for the soil. c. Calculate the uniformity coefficient Cu. d. Calculate the coefficient of gradation Cc. U.S. Sieve N 4 10 20 40 60 100 200 Pan U.S. Sieve N Mass of Soil Retained (g) Problem 5: Repeat problem 4 with the following data. 4 6 10 20 40 60 100 200 Pan 0 18.5 53.2 90.5 81.8 92.2 58.5 26.5 Mass of Soil Retained (g) 0 0 0 9.1 249.4 179.8 22.7 15.5 23.5
Read the question carefully and give me right solution. Provide detailed solution with clear calculations. The hydraulic conductivity of a soil sample, measured at 25C, was 0.005 cm/s using water as the test fluid (density = 997.2 [kg m–3 ], viscosity = 0.0008872 [kg m–1 s –1 ]. Determine the conductivity for ethanol at 10C, density = 798 [kg m–3 ], viscosity = 0.001394 [kg m– 1 s –1 ]
In a hydrometer test, the results are as follows: Gs=2.55 Temperature of water=25 degree Celsius, and R=41 at 2 hours after the start of sedimentation. What is the diameter, d, of the smallest-size particles that have settled beyond the zone of measurement at that time? Please answer this asap