From your Atterberg limits tests on the -40 portion of these samples, you have obtained thefollowing dataPercentage finer by weight100%80%60%40%20%Soil SampleIdentification0%0.001HIJLFINES0.010HLiquidLimit604637PlasticLimit26nonplastic#2003418FineSANDMedium Course0.100Particle size (mm)1.000#4danastantsGRAVELFine10.000Coarse3"100.000

From your Atterberg limits tests on the -40 portion of these samples, you have obtained the following data Percentage finer by weight 100% 80% 60% 40% 20% 0% Soil Sample Identification 0.001 H I J L FINES 0.010 H Liquid Limit 60 46 37 Plastic Limit 26 nonplastic Fine #200 34 18 SAND Medium 0.100 Particle size (mm) 1.000 GRAVEL Coarse Pine #4 10.000 Coarse 3" 100.000 F

From your Atterberg limits tests on the -40 portion of these samples, you have obtained the following data Percentage finer by weight 100% 80% 60% 40% 20% 0% Soil Sample Identification 0.001 H I J L FINES 0.010 H Liquid Limit 60 46 37 Plastic Limit 26 nonplastic Fine #200 34 18 SAND Medium 0.100 Particle size (mm) 1.000 GRAVEL Coarse Pine #4 10.000 Coarse 3" 100.000 F

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 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.
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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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Geotechnical Engineering
SG is 2.66 and temperature is 18 degrees celcius
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4. Determine the USCS designation and soil description of a soil sample given the following sieve analysis test results: D10 = 0.2 mm = D30 0.73 mm D60 3.35 mm % passing #4 = 77.5 % passing #200 = 9.5 The fines are silty.
MY REGISTRATION NO. IS 120 REG NO.120 SIR,PLS GIVE ME PROPER AND CORRECT WITH FULL DETAILS. THANKS
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