Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN: 9781305970939
Author: Braja M. Das, Khaled Sobhan
Publisher: Cengage Learning
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Textbook Question
Chapter 2, Problem 2.2CTP
Refer to Problem 2.C.1. Results of the sieve analysis for Soils A, B, and C are given below. To obtain a more representative sample for further geotechnical testing, a ternary blend is created by uniformly mixing 8000 kg of each soil. Answer the following questions.
a. If a sieve analysis is conducted on the mixture using the same set of sieves as shown above, compute the mass retained (as a percentage) and cumulative percent passing in each sieve.
b. What would be the uniformity coefficient (Cu) and the coefficient of gradation (Cc) of the mixture?
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Q3
The results of a particle size analysis of a soil are given in the following
Table. No Atterberg limit tests were conducted.
Sieve No.
Percent finer
No.40 No.100 | No.200
49.8
No.4
No.10 No.20
70.2
62.5
9.53 mm
100
89.8
28.6
4.1
a) Would you have conducted Atterberg limit tests on this soil? Justify your
answer.
b) Classify the soil according to USCS and AASHTO.
c) Is this soil a good foundation material? Justify your answer
Mechanical analysis on four different samples designated as A, B, C and D were
carried out in a soil laboratory. The results of tests are given below. Hydrometer
analysis was carried out on samples C and D.
Sample C: liquid limit = 27%, plastic limit = 11%,
Sample D: liquid limit = 70%, plastic limit = 38%,
Classify the soils per the Unified Soil Classification System.
Samples
Sieve
75 mm
20.0 mm
4.75 mm
2.0 mm
600 μ
212 μ
75 μ
20 μ
6 μ
2μ
A
100
64
39
24
12
5
1
Sample
A
B
C
B
D10
0.47
0.23
0.004
с
Percentage finer than
100
98
90
9
2
D30
3.0
0.3
0.036
93
76
65
59
54
47
34
23
7
4
D60
20
0.41
2.4
D
100
95
69
46
31
Chapter 2 Solutions
Principles of Geotechnical Engineering (MindTap Course List)
Ch. 2 - For a gravel with D60 = 0.48 mm, D30 = 0.25 mm,...Ch. 2 - Prob. 2.2PCh. 2 - Prob. 2.3PCh. 2 - The following are the results of a sieve analysis....Ch. 2 - Repeat Problem 2.4 with the following data. 2.4...Ch. 2 - Repeat Problem 2.4 with the following data. 2.4...Ch. 2 - Repeat Problem 2.4 with the following data. 2.4...Ch. 2 - The following are the results of a sieve and...Ch. 2 - Repeat Problem 2.8 using the following data. 2.8...Ch. 2 - Repeat Problem 2.8 using the following data. 2.8...
Ch. 2 - The grain-size characteristics of a soil are given...Ch. 2 - Repeat Problem 2.11 with the following data. 2.11...Ch. 2 - Repeat Problem 2.11 with the following data. 2.11...Ch. 2 - A hydrometer test has the following result: Gs =...Ch. 2 - Repeat Problem 2.14 with the following values: Gs...Ch. 2 - Three groups of students from the Geotechnical...Ch. 2 - Refer to Problem 2.C.1. Results of the sieve...
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- Repeat Problem 2.4 with the following data. 2.4 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, and D60 for each soil. c. Calculate the uniformity coefficient Cu. d. Calculate the coefficient of gradation Cc.arrow_forwardRepeat Problem 2.8 using the following data. 2.8 The following are the results of a sieve and hydrometer analysis. a. Draw the grain-size distribution curve. b. Determine the percentages of gravel, sand, silt and clay according to the MIT system. c. Repeat Part b according to the USDA system. d. Repeat Part b according to the AASHTO system.arrow_forwardQ.2 Results of sieve analysis are shown in the table below. Check whether this soil can be classified as SP or not? 100 90 60 30 10 3. Finer (%) No. 200 No. 4 3 mm 0.42 mm 0.25 (0.075 *Sieve 12.5 mm (4.7 mm) mm)arrow_forward
- The results of the particle-size analysis of the soil are as follows: Percent passing No. 10 sieve = 100; Percent passing No. 40 sieve = 80; Percent passing No. 10 sieve = 58.The LL and PI of minus No. 40 fraction of the soil are 30 and 10 respectively. Classify the soil by USCS system a. CH b. SP-SC c. CL-ML d. CL-MLarrow_forwardSoil Engg.arrow_forwardFor the two systems shown below (not to scale), the following are given. • Ah is the same for both systems. • The length of k₂ soil is twice the length of k₁ soil. • The cross-sectional areas of the soil samples are the same. • k₂= 2k₁ (Permeabilities) • q₁ = 100 cm³/hour Find q2 9₁ Ah 921 k k₂arrow_forward
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