Given the following Sieve Analysis and Atterberg limits results. Answer the following questions: 1- What is the effective diameter of this soil? 2- Calculate the uniformity coefficient and coefficient of gradation 3- Determine the percentage of Gravel, Sand, Fine soils according to USCS. 4- Classify the soil using AASHTO (with GI) & USCS (with group name).

Given the following Sieve Analysis and Atterberg limits results. Answer the following questions: 1- What is the effective diameter of this soil? 2- Calculate the uniformity coefficient and coefficient of gradation 3- Determine the percentage of Gravel, Sand, Fine soils according to USCS. 4- Classify the soil using AASHTO (with GI) & USCS (with group name).

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

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
Soil lab Particle size distribution (Sieve method) In addition to that, calculate the value of D25 and D75 and calculate it in curvature
Subject: Soil Mechanics A. Classify the Soil Using USCS (with explanation/ justification) Given data: D10 = ? D30= 0.0023mm D60= 0.032mm Liquid Limit= 49% Plastic Limit= 24% CHOICES: a. CL (lean clay with gravel) b. CL ( lean clay with sand) c. ML (silt with gravel) d. ML (Silt with sand)
The modified Proctor test (ASTM D 698, method C) was performed for a soil with G, = 2.70, and the results are as follows: Moist Weight in Mold, gf Water Content, % 6.5 3250 9.3 3826 12.6 4293 14.9 4362 17.2 4035 18.6 3685 (a) Plot the Ya versus w relation. (b) Determine Ya.max and wopt• (c) Calculate S and e at the maximum dry unit weight point. (d) What is Y at w. (e) What is the range of water content if the relative compaction (R.C.) is required to be 95% of the modified Proctor Ya.max? pt? opr?
Q1: Use the USCS to Classify the soil sample with the following data: % passing 4.75 mm Sieve = 64 % passing 75- Sieve =4 Uniformity Coefficient= 7.5 Coefficient of Curvature=2.7 Plasticity index = 2.5 Solution: Q2: Use the USCS to Classify the soil samples with the following data dalla Soil LL PL Group Symbol Soil Name 15 35 25 ABC 8. ८८ 25 45 65 2-Seepage velocity (m/s) = 70 3- Flow rate in (m3/s) = oss and dry ng www Parmbly & volume change decrease Campering wes Toughness and dry strength dease Presby & val change in 10 20 30 40 50- 60 Liepakt. 70 80 Atra -0.73 (W₂-201 90 Q3: For the soil shown below the hydraulic conductivity is (3 m/day). The sample diameter is 1.5 m. The porosity is (30%). Find: 1- Flow velocity (m/s)== 100 110 120 Soil 4 m ITE
SIEVE ANALYSIS 1. From the result of sieve analysis and the grain size curves for soils 1 and 2. a. Compute the value of uniformity coefficient Cu. b. Classify soil No. 1 using USCS Method. c. Classify soil No. 2 using USCS Method. Sieve Number Diameter Percent Passing mm Soils 1 Soils 2 4. 4.760 100 90 8. 2.380 99 64 10 2.000 98 58 20 0.850 92 35 40 0.425 70 22 60 0.250 46 15 100 0.150 25 10 200 0.074 4
The size characteristics of the soil particles are shown in the table below. Classify soil using: 1. USDA Method 2. AASHTO Method (Including group index) 3. USCS Method (Full Soil Name not Symbol only) Size (mm) Percent finer 4.7500 93 2.0000 77 0.0330 69 0.0180 55 0.0100 42 0.0062 33 0.0018 24 0.0010 12 From consistency limit test, the following results are determined: Liquid Limit 32% Plastic Limit 24%
47.) The results of the sieve analysis done on a soil are shown below. Sieve No. Diameter Soil A 4 4.760 89 6 2.380 70 10 2.000 63 20 0.840 55 40 0.420 31 60 0.250 20 100 0.149 18 200 0.074 4 LL 60 PL 40 Determine its classification using USCS. GW SP None of the above SW GP
The results of the sieve analysis of a non plastic soil is tabulated below. Classify the soil according to AASHTO classification system and unified soil classification systems (USCS). Diameter (mm) 4.75 % Passing 100 2.0 0.85 0.425 98 92 12 0.150 0.075 4 2 D10 D30 D60 0.39 0.51 0.62
Answer ALL Questions Time: 2 Hours Q3: Classify the following soil by using the Unified soil classification system. Give the group symbols and the group names. Table 1. Sample 1 Grain Size Results (ASTM D422). Table 2. Atterberg Limits Results (ASTM D4318). Sleve No. Dia. (mm) % Passing Sample PL LL % in 12.7 100 G0 40 14 4.75 97 #10 2 94 #20 0.85 84 # 40 0.425 57 #60 0.25 32 140 0.106 15 #200 0.075 100 90 Atterberg Limit Results: LL = 60%, PL = 40% E 80 * 70- 60 - 50 40 30- a 20 10 1 30 a40 100 0.1200 10 80 0.01 Particle Diameter (mm) Answer ALL Questions Time: 2 Hours 70 CH or OH 60 A-line PI = 0.73(LL - 20) Or 50 40 30 CL MH OL 20 or CL-ML OH ML 10 - or OL 10 16 20 30 40 50 60 70 80 90 100 Liquid limit Figure 5.3 Plasticity chart Percent (%) Finer by Weight Plasticity index U-line PI = 0.(LL - 8) O Cengage Learning 2014
Classify the given soils using the Unified Soil Classification System (USCS). Give the group symbols and group names. (Show the path of the classification.) Sieve Analysis (percent finer) Soil No. No. 4 No. 200 LL PL Comments A 94 3 NP Cy = 4.48 and C = 1.22 - В 100 77 63 25 B.
How field identification of clayey soil can be done from silty soil using simpler field tests. Find the group index value according AASTHO soil classification system by using following data. % passing 75 micron sieve = 58%, Liquid limit = 50% Plastic Limit = 35 %