Classify the soil sample using the U.S. Department of Agriculture textural classification chart. Sand =60 Silt = 35 Clay = 70. Classify the soil sample using the U.S. Department of Agriculture textural classification chart. Sand =60 Silt = 35 Clay = 70.1001090208073070 7Clay4060 7R50SiltyclaySandyclay6040Silty clayJoamClay loam Soi7030Sandy clayInam20LeamSilty Joain9010LoamySandy JoamSandsandSilt100100907060sa 4020Percentage of sandSelect one:a. Loamy sandb. Silty clayc. none of the choicesd. Sandy loame. Clay loamPercentage of sitPercentsge of clay

Draw the line from sand at 60% Draw the line from clay at 70% Draw the line from silt at 35% Find…

Classify the soil sample using the U.S. Department of Agriculture textural classification chart. Sand =60 Silt = 35 Clay = 70. 100 10 90 20 807 30 70 Clay 40 60 7 Silty clay Sandy clay 60 40 Silty clay Joam Clay loam Soi 70 30 Sandy clay Ioam 20 Leam Silty Joain 90 10 Loamy Sand Sandy Joam Silt 100 sand 100 90 70 60 s0 40 20 10 Percentage of sand Select one: a. Loamy sand b. Silty clay c. none of the choices d. Sandy loam e. Clay loam Perventage of sit Percents ge of clay

Classify the soil sample using the U.S. Department of Agriculture textural classification chart. Sand =60 Silt = 35 Clay = 70. 100 10 90 20 807 30 70 Clay 40 60 7 Silty clay Sandy clay 60 40 Silty clay Joam Clay loam Soi 70 30 Sandy clay Ioam 20 Leam Silty Joain 90 10 Loamy Sand Sandy Joam Silt 100 sand 100 90 70 60 s0 40 20 10 Percentage of sand Select one: a. Loamy sand b. Silty clay c. none of the choices d. Sandy loam e. Clay loam Perventage of sit Percents ge of clay

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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Calculate the coefficient of gradation, Cc. ? Classify the soil by AASHTO and USCS system. (LL = 30, PL = 10) ? Use this given M=Mass of soil retained in pan +∑Mass reatianed in each sieve⇒M=617 gM=Mass of soil retained in pan +∑Mass reatianed in each sieve⇒M=617 g Sieve No. Opening Size (mm) Mass retained (g) %Mass retained=MiM×100=MiM×100 %Cumulative mass retained % Finer 4 4.750 28 =28617×100%=4.54%=28617×100%=4.54% 4.54 95.46 10 2.0 42 =42617×100%=6.81%=42617×100%=6.81% 11.35 88.65 20 0.850 48 =48617×100%=7.78%=48617×100%=7.78% 19.13 80.87 40 0.425 128 =128617×100%=20.74%=128617×100%=20.74% 39.87 60.13 60 0.250 221 =221617×100%=35.82%=221617×100%=35.82% 75.69 24.31 100 0.150 86 =86617×100%=13.94%=86617×100%=13.94% 89.63 10.37 200 0.075 40 =40617×100%=6.48%=40617×100%=6.48% 96.11 3.89 Pan 24 =24617×100%=3.89%=24617×100%=3.89% 100 0 Now, to plot the GSD curve, the % finer is kept on the vertical axis and the opening size are kept on the horizontal axis -…
Classify the soil using AASHTO. The liquid limit and plastic limit are found to be 30 and 20, respectively. The following data are obtained: Percent passing No. 4 sieve: 100% Percent passing No. 10 sieve: 91% Percent passing No. 40 sieve: 60% Percent passing No. 200 sieve: 20%
soil mechanic. Classify the soil according to unified soil classification system USCS; if the following data is given : Passing sieve No. 200 =7% / Passing sieve No. 4 = 40% / D10= 0.18, D30= 0.69 and D60= 3/ LL-28%, Pl=19%
The sieve analysis data for a granular soil are given below. % passing or finer 19 mm sieve = 100%% passing or finer 4.75 mm (4 sieve) = 61%% passing or finer 0.075mm (200 sieve) = 16% What are the percentages of gravels, sands and fines within this soil, based on the Unified Soil Classification System?
For an inorganic soil, the following grain-size analysis is given:Sieve No. % Passing4 10010 9020 6440 3880 18200 13For this soil, LL = 23% and PL = 19%. Classify the soil according to:a. The AASHTO soil classificationb. The Unified soil classification system
40.) The given data shows a sieve analysis of a soil sample. Classify the soil using AASHTO Method. Tables are provided below. Sieve No. Diameter (mm) Percent Passing No. 4 4.760 96 No. 8 2.380 80 No. 10 2.000 78 No. 20 0.840 61 No. 40 0.420 56 No. 60 0.250 48 No. 100 0.150 39 No. 200 0.075 33 Characteristics of No. 40 Fraction LL 38 PL 29 A. A-2-4(0) B. A-1-b(0) C. A-1-a(0) D. A-3(0) E. A-2-5(0)
41.) The given data shows a sieve analysis of a soil sample. Classify the soil using AASHTO Method. Tables are provided below. Sieve No. Diameter (mm) Percent Passing No. 4 4.760 96 No. 8 2.380 80 No. 10 2.000 78 No. 20 0.840 61 No. 40 0.420 56 No. 60 0.250 48 No. 100 0.150 39 No. 200 0.075 33 Characteristics of No. 40 Fraction LL 37 PL 25 A. A-2-4(0) B. None of these. C. A-2-5(0) D. A-2-6(0) E. A-2-7(0)
The moist unit weights and degrees of saturation of a soil are given in the table. y (kN/m3) S(%) 16.62 50 17.71 75 Determine: a. e b. Gs Show handwritten solutions.
Sieve # Maas retained (gr) 4 0 10 11,5 16 11,7 20 12 30 12,9 40 17,8 50 19,1 70 35,1 100 38,1 200 118 Pan 335 Draw grain size distribution curve,find D60, D30, D10 Classify the soil
Classify the soil according to the soil classification system AASHTO: With the following data: Percentage passing: sieve No. 200 = 45% sieve No. 40 = 10% sieve No. 10 = 45% LP = 32 LL = 54
Calculate the Gs of soil sample - Mass of empty clean pycnometer= 41.76 gm Mass of empty clean pycnometer + dry soil = 67.52 gm Mass of pycnometer + dry soil + water = 157.62 gm Mass of pycnometer + water =? Gs =? Write your comment about types of soil according to the result of Gs .
1. Classify the following soils using USCS givegroup symbol and group name.a) A 55% of soil sample retained in sieve no. 4and 89% retained on sieve no. 200. The soilsample has a liquid limit = 28% and plasticLimit = 18% Cu = 4.2, Cc = 1.4 2. A saturated soil has the followingcharacteristics:Initial Volume = 25 cm3Final Volume = 16 cm3Mass of wet soil = 45 gMass of dry soil = 31 ga) Determine the Shrinkage Limitb) Determine the Shrinkage Ratioc) Determine the Specific Gravity of the soil