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 Weight of wet soil and mold (lbs) Mass of can and wet soil (grams) Mass of can (grams) Mass of can and dry soil (grams) M M. M. M. 7.32 114.92 111.48 155.08 46.50 257 7.68 163.12 190.43 46.43 178.64 178.24 46.20 46.50 46.10 774 193.13 765 188.77 171.58

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 Weight of wet soil and mold (lbs) Mass of can and wet soil (grams) Mass of can (grams) Mass of can and dry soil (grams) M M. M. M. 7.32 114.92 111.48 155.08 46.50 257 7.68 163.12 190.43 46.43 178.64 178.24 46.20 46.50 46.10 774 193.13 765 188.77 171.58

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: atory apparatus for the following Label and give the use/uses. 1. Determination of moisture content…

A: Note- We are authorized to answer three subparts at a time since you have not mentioned which part…

Q: The test results of a soil sample are given below. Percentage finer than 4.75 mm = 55 Percentage…

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: Following

Q: Determine the liquid limit. LL = b. Determine the flow index. Flow index = Determine the liquidity…

A: Given Data: The plastic limit for the first trial is, PL1=16.2% The plastic limit for the second…

Q: Organic soils are typically characterized by high void ratio, low specific gravity, and high…

A: The given data is as follows: Initial height of the specimen=20.6 mmMass of dry specimen=12 gArea…

Q: 6.41 The results of a standard Proctor test are given in the following table. a. Determine the…

Q: A soil layer with a moisture content of 10 percent and a unit weight of 16.5 KN per cubic meter…

A: Question b) Objective: To determine the dry specific weight of the soil (γd,compaction). Given:…

Q: f dr

A: According to given data inorganic clay has a volume of 21.4 cm3 and weighs 36.7 g. After drying,…

Q: The total volume of a soil soecimen is 80,000 mm³ and it weighs 150 grams. The dry weight of the…

A: First we summarise the given data : Total volume of soil specimen = 80,000 mm3 Weight of specimen =…

Q: The variation of the total volume ( V) of a specimen of a fine soil with its water content ( w ) is…

A: Appreciate if there is detailed work with explanation and finds you helpful for ease of…

Q: Undisturbed soil sample was collected from the filed in Melbourne in steel tube for laboratory…

A: Diameter of the steel tube = 85 mmLength of steel tube= 170 mm (0.17 m)Moisture weight = 1600…

Q: A sample of field soil is tested in the laboratory. The results of Standard and Modified Proctor…

Q: a soil sample was determined to have a water content of 8% and a degree of saturation of 42%. After…

Q: Paragraph Liquid Limit Determination Specimen No. No. of Blows Wt. of Container Wt. of Container +…

Q: A soil sample is subjected to laboratory sieve analysis using a complete set of standard IS sieves.…

Q: A sample of field soil is tested in the laboratory. The results of Standard and Modified Proctor…

A: ASKED: To compute: degree of compaction/relative compaction (R%) at Standard Proctor Test degree of…

Q: Determine the degree of saturation in percentage in a bulk density determination procedure on a…

Q: a. The following data was obtained from Liquid Limit Test conducted at a Soil Lab. Using the data…

Q: Following are the results from the liquid and plastic limit tests for a soil. Liquid limit test: let…

A: Given data in question Liquid limit test values Number of blows viz N1 = 20 W1= 30% N2 =…

Q: A sample of field soil is tested in the laboratory. The results of Standard and Modified Proctor…

A: The ratio of field dry density to maximum dry density ( density corresponding to optimum moisture…

Q: The Void Ratio of a soil sample is 0.54 and the specific gravity of its particles is 2.69. Calculate…

Q: a soil sample was determined to have a water content of 8% and a degree of saturation of 42%. After…

Q: The results from standard proctor test shows that the weight of the soil with the mould is 3.85 kg…

Q: a soil sample was determined to have a water content of 8% and a degree of saturation of 42%. After…

A: Solution:-The given data ,Water content (w)=8%=0.08The degree of saturation(S)=42%=0.42Specific…

Q: a soil sample was determined to have a water content of 8% and a degree of saturation of 42%. After…

Q: In order to determine the relative density of a sand sample, its natural moisture content and bulk…

Q: Subj: civil engineering / soil mechanics The volume of soil sample measured before oven drying…

A: V = 120 cm3 Wd = 122 gmLI = 1.25PL = 7.6%Gs = 2.70

Q: The following results were obtained from a liquid limit test on a soil using a liquid limit device.…

Q: a soil sample was determined to have a water content of 8% and a degree of saturation of 42%. After…

A: Given : Water content = 8% Degree of saturation = 42 % Altered degree of saturation = 53 % the…

Q: Following are the results from the liquid and plastic limit tests for a soil. Liquid limit test: let…

A: liquid limit is taken at 25 number of blows plastic limit is average of test results plasticity…

Q: Sieve and Atterberg limits tests Wt. of sieve (gm) Wt. of sieve + Soil (gm) 450 518 378 420 L.L =…

A: The given data is shown below:

Q: In a sieve analysis test, 95.5% of the sample is passing the 5mm sieve and 5.13% is passing sieve…

A: 1)clay = less than 2 micron 2)silt = 2 micron to 75 micron 3)sand = 75 micron to 4.75mm 4)Gravel =…

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

a .l zain IQ
elearning7.uokufa.edu.iq
CHAPTER SEVEN
STERSSES WITHIN SOIL MASS
Н.W
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
Weight of wet soil
and mold (Ibs)
Mass of can and
wet soil (grams)
Mass of can and
dry soil (grams)
Mass of can
(grams)
M.
M.
M.
7.32
114.92
111.48
7,57
163.12
155.08
7.68
190.43
178.64
46.20
193.13
178.24
46.50
7.65
188.77
171.58
46.10
II

Expert Solution

This question has been solved!

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

SEE SOLUTION Check out a sample Q&A here

Step 1 Given

VIEW

Step 2 Tabular data

VIEW

Step 3 Formula used

VIEW

Step 4 Tabular Calculation

VIEW

Step 5 Tabular Calculation

VIEW

Step 6 Compaction Curve

VIEW

Step 7 Interpretation of maximum dry unit weight and water content

VIEW

Step by step

Solved in 7 steps with 4 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

I need the answer soon as possible, please
a soil sample was determined to have a water content of 8% and a degree of saturation of 42%. After adding some water, the degree of saturation was altered to 53%. take the specific gravity of the soil solids to be 2.7. assuming no change in the volume of the voids, determine the following A. void ratio B. water content% C. bulk specific gravity of the mass of the sample in the altered state.
U B. ( A Soil Specimen has the following characteristics: % passing No.4 sieve = 85 % passing No.200 sieve = 11 D60 = 2 mm D30 = 0.35 mm D₁0 = 70 μm L.L = 36% P.L = 31% D8 = 2 µm 1. Classify the specimen according to the Unified Soil Classification System (USCS). Assign the group name and the group symbol. 2. Determine the soil activity. 3. Determine the soil Liquidity index and consistency index if we = 26%.
Someone can help me with this one please. Following are the results of a shrinkage limit test:
liquid limit was 25.72%.
Hello can any one help me with this
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)
A soil specimen has a total volume of 70 cm3 and a total mass of 135.5g. After complete drying in an oven, the specimen has a mass of 122.0g. The Gs is 2.60. Determine the void ratio. Report your answer to three significant figures, and draw the full phase diagram.
A 59m3 soil sample having a degree of saturation of 96% at the liquid limit of 29%. At the plastic limit of the soil has a volume of 42m3. This soil sample was subjected to sieve analysis in the attached photo below. Find: a.Void ratio at Plastic limit b.Plastic limit c.Modified percentage of sand, silt and clay d.Classification of soil according to USDA
05: The Figure below represents the data that was obtained from a liquid-limit test on a soil. 50 percent of its particles is passing sieve #200. Two determinations for the plastic limit gave water contents of 20.3 % and 20.8 %. Determine 1- Liquid Limit (LL). 2- Plasticity index. 3- Liquidity index if the water content is 27.4% 4- Void ratio at the liquid limit, if G, 2.7 5- Classify the soil based on AASHTO system. 30 10 20 40 60 80 100 Penetration, d (mm) Moisture content, w ()