1. The results of a standard Proctor test are given in the following table.Trial No.Weight of Moist Soil in the Mold(kg)Moisture Content(%)11.689.91.7110.612.1231.7741.831.861.8813.815.117.4671.871.8519.421.28a) Determine the maximum dry unit weight of compaction and the optimum moisturecontent. Given: mold volume = 943.3 cm3b) Determine the void ratio and the degree of saturation at the optimum moisturecontent. Given: Gs = 2.7.

Weight of moist Moisture Moist unit Dy unit Trial No Soil in mold Content Weight Weight 1…

1. The results of a standard Proctor test are given in the following table. Weight of Moist Soil in the Mold (kg) Trial No. Moisture Content (%) 1.68 9.9 1.71 10.6 12.1 1.77 1.83 13.8 1.86 15.1 1.88 1.87 1.85 17.4 19.4 8 21.2 a) Determine the maximum dry unit weight of compaction and the optimum moisture content. Given: mold volume = 943.3 cm3 b) Determine the void ratio and the degree of saturation at the optimum moisture content. Given: Gs = 2.7. -23 4567 CO

1. The results of a standard Proctor test are given in the following table. Weight of Moist Soil in the Mold (kg) Trial No. Moisture Content (%) 1.68 9.9 1.71 10.6 12.1 1.77 1.83 13.8 1.86 15.1 1.88 1.87 1.85 17.4 19.4 8 21.2 a) Determine the maximum dry unit weight of compaction and the optimum moisture content. Given: mold volume = 943.3 cm3 b) Determine the void ratio and the degree of saturation at the optimum moisture content. Given: Gs = 2.7. -23 4567 CO

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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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%

1. The results of a standard Proctor test are given in the following table.
Trial No.
Weight of Moist Soil in the Mold
(kg)
Moisture Content
(%)
1
1.68
9.9
1.71
10.6
12.1
2
3
1.77
4
1.83
1.86
1.88
13.8
15.1
17.4
6
7
1.87
1.85
19.4
21.2
8
a) Determine the maximum dry unit weight of compaction and the optimum moisture
content. Given: mold volume = 943.3 cm3
b) Determine the void ratio and the degree of saturation at the optimum moisture
content. Given: Gs = 2.7.

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2. 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
The results of a standard Proctor test are given in the following table. Determine the maximum dry unit weight of compaction and the optimum moisture content. Also determine the moisture content required to achieve 90% of ꙋd (max).
2. Table 2 shows the results of the standard proctor test Water content (%) Weight (kg) 14 16,5 18,5 20 21 22,5 Table 2. Proctor Test Results 1,45 1,71 1,87 1,88 1,76 1,67 If this test uses a sample print with a size of 943 cm^3, answer these questions: a. interpret the results of this proctor test, and determine the optimum moisture content and maximum dry unit weight. (Gs = 2.7)
Determine dry unit weight , void ratio and water content
Soil mechanics سنذن
the following data have been obtained in a standard laboratory proctor compaction test : Trial No. moisture content. % 1 9 2 3 4 6 11.9 14.6 15-8 19.0 21.6 weight of container and compacted soil 137.43 38.16 38-67 38-8439-06 38.76 (N) The container is 943 cm³ ih volume and its weight is 201 A-Determine the optimum moisture content and the maximum dry unit weight B-what is the dry uhit weight and water content at 95% standard compaction in the field ?
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
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.
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)
6. Following is the result of 8 trials of a standard proctor test. Determine the optimum water content and maximum dry unit weight. Determine also the amount of energy expended in the compaction of one of the trials assuming that the test done is standard proctor test. Proctor Mold Volume Mass of wet soil in the Moisture Content Trial (cm³) mold (kG) (%) 1 943.3 1.68 9.9 943.3 1.71 10.6 3 943.3 1.77 12.1 4 943.3 1,83 13.8 943.3 1.86 15.1 6 943.3 1.88 17.4 943.3 1.87 19.4 8 943.3 1.85 21.2
please solve for the adjusted length and bearings please