PROBLEM 8:The particle size distribution curve of aparticular soil sample has the followingvalues: Diameter corresponding to10% finer D10 = 0.96 mm, diametercorresponding to 30% finer D30 = 0.16mm and the diameter corresponding to60% finer Dso = 0.24 mm. Determinethe coefficient of uniformity and thecoefficient of gradation.SITUATION 2:From the following data of a givensieve analysis:SieveOpening Percent(mm)4.762.382.00.710.250.1490.074passing9064583022No.#4# 8#10# 25# 60#100# 200104

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Answered: PROBLEM 8: The particle size…

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 following results were obtained by a mechanical sieve analysis. classify the soil using the Unified Soil Classification System. Sieve analysis % passing by weight is as follows: No.4 = 100% No. 40 = 96% No.200 = 52% Liquid Limit= 29 Plastic Limit = 11
U B. ( A Soil Specimen has the following characteristics: % passing No.4 sieve = 85 % passing No.200 sieve = 11 D60 = 2 mm D30 = 0.35 mm D₁0 = 70 μm L.L = 36% P.L = 31% D8 = 2 µm 1. Classify the specimen according to the Unified Soil Classification System (USCS). Assign the group name and the group symbol. 2. Determine the soil activity. 3. Determine the soil Liquidity index and consistency index if we = 26%.
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
Question 44 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 No. 8 No. 10 No. 20 4.760 96 2.380 80 2.000 78 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 45 PL 36 General classification Granular materials (35% or less of total sample passing No. 200) A-I A-2 Group classification A-l-a A-1-5 A3 A-24 A-25 A-2-6 A-2-7 Sieve analysis (percent passing) No. 10 No. 40 No. 200 50 max. 30 max. 15 max. 50 max. 25 max. 51 min. 10 max. 35 max. 35 max. 35 max. 35 max. Characteristics of fraction passing No. 40 Liquid limit Plasticity index 40 max. 41 min. 40 max. 41 min. 6 max. NP 10 max. 10 max. 11 min. 11 min. Usual types of significant Stone fragments, gravel, and sand Fine Silty or clayey gravel and sand sand
The table shows the laboratory results of the sieve analysis of a sample. Plot the grain size curve of the soil. The soil has a liquid limit of 35% and plasticity index of 26%. SIEVE SIZE (mm) WEIGHT RETAINED 0.25 18.96 0.149 33.18 0.074 45.03 0.052 54.51 0.02 42.66 0.01 11.85 0.004 4.74 0.001 4.74 Pan 21.33 a.) Classify the soil using AASHTO method b.) Classify the soil using USCS method c.) Classify the soil using USDA method Please answer this asap. For upvote. Thank you in advance
Given below is the result of a sieve analysis done on a soil sample.Sieve No. Diameter Percent Passing %4 4.750 9510 2.000 8220 0.850 7340 0.425 5360 0.250 23100 0.150 14200 0.075 7Pan --- 0Determine the value of the coefficient of uniformity
The following results were obtained by a mechanical sieve analysis. classify the soil using the Unified Soil Classification System. Sieve analysis % passing by weight is as follows: No.4 = 100% No. 40 = 89% No.200 = 65% Liquid Limit= 44 Plastic Limit = 23
5. The following information is obtained from a sieve analysis to determine the range of particle sizes in a granular soil sample. Sieve Size Sieve Opening (4.76) Percent Finer#4 4.76 96#10 2.00 80#20 0.84 51#40 0.42 38#60 0.25 25#100 0.149 12#200 0.074 5 Present the information as grain-size curve on semi-logarithmic coordinates of percent finer against particle size diameter. From the plot, determine the uniformity coefficient Cu.
2.8 2.7 The following are the results of a sieve and hydrometer analysis. Sieve number/ grain size Analysis Sieve Hydrometer Analysis Sieve 40 80 170 200 0.04 mm 0.02 mm 0.01 mm 0.005 mm 0.0015 mm Hydrometer Sieve number/ grain size Percent finer than 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. Repeat Problem 2.7 with the following data: 20 30 40 60 80 200 0.05 mm 0.03 mm 0.015 mm 0.006 mm 0.004 mm 0.0015 mm 100 96 85 80 59 39 26 15 8 Cengage Learning 2014 Percent finer than 100 96 90 76 65 34 27 19 11 7 6 5 Ⓒ Cengage Learning 2014
Sr B. ( A Soil Specimen has the following characteristics: % passing No.4 sieve = 85 % passing No.200 sieve = 11 D60 = 2 mm D30=0.35 mm D₁0 = 70 μm D8=2 μm L.L = 36% P.L = 31% 1. Classify the specimen according to the Unified Soil Classification System (USCS). Assign the group name and the group symbol. 2. Determine the soil activity. 3. Determine the soil Liquidity index and consistency index if wc = 26%.
SITUATION 3. Following are the results of a sieve analysis. SIEVE NO. OPENING (mm) RETAINED 4 4.75 10 2.0 40 20 0.85 60 40 0.425 89 60 0.25 140 80 0.18 122 100 0.15 210 200 0.075 56 PAN 12 a. Find the effective particle size. b. Determine the uniformity coefficient. c. Find the coefficient of curvature.
Q # 1. Organic soils are typically characterized by high void ratio, low specific gravity, and high compressibility. Following are the results of a consolidation test on a sample of organic soil. Pressure, o (kN/m2) Change in dial reading, AH (mm) 0.284 12 0.150 24 0.315 48 0.564 100 0.823 200 2.25 400 5.34 Given that the initial height of the specimen = 20.6 mm, mass of dry specimen = 12 g, area of specimen = 31.67 cm2 and G, = 2.49. Plot the e, log o' curve. b. Determine the preconsolidation pressure. a.

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