Consider the grain size gradation curve for the soil shown in the image below and classify the soil according to the USCS. Notice the sieve names/sizes given. This soil is inorganic. Any fines present are non-plastic (say PI=0). Estimate values on log scale if necessary. ### Grain Size Distribution Graph#### Description:This graph represents the grain size distribution of a soil sample. The x-axis is labeled as "Grain Diameter (mm)" and is marked with a logarithmic scale ranging from 0.001 to 100 millimeters. Along the x-axis, there are also labels indicating U.S. Standard Sieve Sizes, such as No. 200, No. 100, No. 40, No. 20, No. 10, No. 4, 1/2", 3/4", and 1".The y-axis represents "Percent Finer," indicating the cumulative percentage of material that passes through each sieve size. The scale ranges from 0% to 100% in increments of 10%.#### Key Observations:- The curve starts near the top left, indicating that most grains have a smaller diameter.- The percentage finer decreases steadily as the grain diameter increases.- At smaller sieve sizes, a high percentage of grains pass through, showing finer material.- As the grain diameter increases, around the 0.4 mm to 2 mm range, the percentage finer decreases significantly.- The curve flattens out as it approaches larger grain diameters, representing coarser soil particles that do not pass through larger sieve sizes.This type of graph is commonly used in soil mechanics and civil engineering to assess the particle size distribution of soil samples for construction and research purposes. Understanding the distribution helps in determining the suitability of soil for various engineering applications, such as foundation support, drainage, or filtration.

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

Consider the grain size gradation curve for the soil shown in the image below and classify the soil according to the USCS. Notice the sieve names/sizes given. This soil is inorganic. Any fines present are non-plastic (say PI=0). Estimate values on log scale if necessary.

Consider the grain size gradation curve for the soil shown in the image below and classify the soil according to the USCS. Notice the sieve names/sizes given. This soil is inorganic. Any fines present are non-plastic (say PI=0). Estimate values on log scale if necessary.

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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The results of the sieve analysis done on a soil are shown below. Sieve No. Diameter Soil A 4 4.760 96 6 2.380 91 10 2.000 89 20 0.840 85 40 0.420 81 60 0.250 78 100 0.149 67 200 0.074 64 LL 40 PL 30 Determine its classification using USCS. ML MH None of the above CL CH
The Name :- Answer the following questions :-( Choose 5 only) 1- .is the moisture content that defines where the soil changes from a plastic to a viscous fluid state. 2- The difference between the PL and the LL is defined as.... 3- Why is it not possible to use core cutter method for calculating the density of cohesion less soil? 4- The relative density of a soil is 5- In the sieve analysis test Place the sieve stack in the mechanical shaker and shake for . minutes . 6- Hydrometer testing is performed on soil passing from a sieve.
00 docs.google.com/forms/ A It is easier than the sieve analysis CH soil is formed by chemical 1/1 weathering * True False Neither true nor false The grain size distribution has 0/2 been conducted on a soil sample and the following have been obtained: Percentage of passing from sieve number 200 is 47%, LL = 30%, PL = 15% Using the above information, the soil classification according to the USCS is * :O A5
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2. The following data were recorded during particle size analysis tests in the laboratory. Plot the grading curve and determine the grading characteristics. Using the information, classify the soil according to BSCS. Soil 1 Fine-Mesh dry sieve analysis data: Sieve Size 425 (mm or um) Mass 1.18 600 300 212 150 63 retained 1.9 3.2 8.1 10.4 22.7 126.1 (g) Total sample = 212.0 g
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the results of the sieve analysis done on a soil are shown below. Determine its classification using USCS. a. None of the above b. ML c. MH d. CL e. CH
Mechanical analysis on four different samples designated as A, B. Cand Dwere carried out in a soil laboratory. The results of tests are given below. Hydrometer analysis was carried out on sample D. The soil is non-plastic. Sample D D. liquid limit w 42, plastic limit = 24. plasticity index = 18 Classify the soils per the Unified Soil Classification System. Samples ASTM Sieve Percentage finer than Designation 63.0 mm 100 93 20.0 mm 64 76 6.3 39 100 65 20 mm 600 212 63 A 20 24 98 59 12 90 54 9. 47 100 34 95 23 69 7. 46 2 31 B and 100 60 20 Sand Course lo medium Cotles l>762 mm Clay Silt Gravel Fine 0.001 001 0.075 0.1 10 100 Particle size (mm) Figure Ex. 3.21 Percentage finer
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
A sample soil is obtained and sieve analysis is performed to determined its particle distribution. Sieve Opening Mass Retained % Finer # 6 3.35 146 # 8 2.36 88 #16 1.18 125 # 40 0.425 56 #50 0.3 42 #140 0.106 127 # 200 0.075 126 Pan 29 Determine: 1. Effective Size (in mm and in 4 decimal places) 2. Uniformity Coefficient (in 2 decimal places) 3. Coefficient of Curvature (in 2 decimal places) 4. Sorting Coefficient (in 2 decimal places)
Question 5 The results of the particle-size analysis of a soil are as follows: Percent passing the No. 10 sieve = 100 %3D Percent passing the No. 40 sieve = 80 • Percent passing the No. 200 sieve = 38 The liquid limit and plasticity index of the minus No. 40 fraction of the soil are 40 and 15, respectively. Compute the group index. A) 1.75 1.0 2.5 2.0
Mechanical analysis on four different samples designated as A, B. Cand Dwere carried out in a soil laboratory. The results of tests are given below. Hydrometer analysis was carried out on sample D. The soil is non-plastic. Sample D D. liquid limit w 42, plastic limit = 24. plasticity index = 18 Classify the soils per the Unified Soil Classification System. Samples ASTM Sieve Percentage finer than Designation 63.0 mm 100 93 20.0 mm 64 76 6.3 39 100 65 20 mm 600 212 63 A 20 24 98 59 12 90 54 9. 47 100 34 95 23 69 7. 46 2 31 100 60 20 Sand Coles >762 mm Clay Silt Gravel Fine Course lo medium 0.001 001 0.075 0.1 10 100 Particle size (mm) Samples A,BandC Figure Ex. 3.21 Percentage finer