From the given data, shows a sieve analysis of soil samples A and B. Classify the soil according to AASHTO Method. Sieve No. Diameter (mm) Percent Passing (Percent Finer) Soil Sample A Soil Sample B 4 4.76 100 100 8 2.38 90 100 10 2.00 77 78 20 0.84 59 92 40 0.42 51 84 60 0.25 42 79 100 0.149 35 70 200 0.074 33 63 Characteristics of – 40 fraction LL 46 47 PL 29 24

From the given data, shows a sieve analysis of soil samples A and B. Classify the soil according to AASHTO Method. Sieve No. Diameter (mm) Percent Passing (Percent Finer) Soil Sample A Soil Sample B 4 4.76 100 100 8 2.38 90 100 10 2.00 77 78 20 0.84 59 92 40 0.42 51 84 60 0.25 42 79 100 0.149 35 70 200 0.074 33 63 Characteristics of – 40 fraction LL 46 47 PL 29 24

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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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 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 4.760 100 No. 8 2.380 100 No. 10 2.000 95 No. 20 0.840 81 No. 40 0.420 74 No. 60 0.250 69 No. 100 0.150 55 No. 200 0.075 51 Characteristics of No. 40 Fraction LL 48 PL 40 A. A-5(3) B. A-7-5(4) C. None of these. D. A-5(4) E. A-7-6(3)
The given data shows a sieve analysis of a soil sample. Classify the soil using AASHTO Method. Sieve No. Diameter Soil A 4 4.760 96 8 2.380 80 10 2.000 78 20 0.840 61 40 0.420 56 60 0.250 48 100 0.150 39 200 0.075 33 Characteristics of No. 40 Fraction LL 46 PL 31
The given data shows a sieve analysis of a soil sample. Classify the soil using AASHTO Method. Sieve No. Diameter Percent Passing 4 4.760 100 6 2.380 91 10 2.000 88 20 0.840 76 40 0.420 55 60 0.250 50 100 0.150 47 200 0.075 46 Characteristics of No. 40 Fraction LL 50 PL 34
From the given data, shows a sieve analysis of soil samples A, B and C. Classify the soil sample A using USCS method (Group Symbol only), D60=2.3, D10 = 0.17, D30 = 0.70 Classify the soil sample B using AASHTO method Classify the soil sample C using the AASHTO method
The given data shows a sieve analysis of a soil sample. Classify the soil using AASHTO Method. Sieve No. Diameter Soil A 4 4.760 80 8 2.380 71 10 2.000 69 20 0.840 63 40 0.420 57 60 0.250 43 100 0.150 40 200 0.075 39 LL 36 PL 21
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%
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 -…
Write each criterion that is satisfied for each classification. For the given data of soil A and B, DETERMINE THE FF AND SHOW COMPLETE SOLUTION. 1. Classify soil A using AASHTO method. 2. Classify soil B using USCS method. SIZE (MM) A B 4.75 100 100 2.36 97 100 2.00 92 98 0.850 87 92 0.425 59 88 0.180 42 79 0.150 26 70.1 0.075 20 61 No. 40 Fraction Characteristics LL 32 46 PL 20 24
Calculate the CEC of a 100g soil sample that was found to contain the following cations: Ca2+ = 0.12g Mg2+ = 0.024g K+ = 0.078g Al3+ = 0.027g NH4+ = 0.036g H+ = 0.002g
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 .