Following results are obtained from particle size analysis: Uniformity coefficient = 8 Coefficient of curvature = 2.8 Percentage of soil passing through 4.75 mm IS sieve = 70% If liquid limit and plastic limit for the soil are 38% and 34.2% respectively, the soil can be classified as per IS soil classification?

Following results are obtained from particle size analysis: Uniformity coefficient = 8 Coefficient of curvature = 2.8 Percentage of soil passing through 4.75 mm IS sieve = 70% If liquid limit and plastic limit for the soil are 38% and 34.2% respectively, the soil can be classified as per IS soil classification?

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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1. From the given soil samples below, (a) compute the effective size, uniformity coefficient, and coefficient of gradation (b) Determine the percent of gravel, sand, silt and clay of soil A using AASHTO Sieve size. 0.425 0.033 0.018 0.01 0.0062 0.0035 0.0018 0.001 Soil A Passing Finer (%) 100 90 80 70 60 50 40 35
A soil sample retains 40% and 30% of total soil sample on 4.75 mm and 0.075 mm sieve respectively. It passes 10% when sieved through 0.002 mm sieve. Find the ratio of coefficient of curvature to coefficient of uniformity.
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
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
In a soil specimen, 70% of particles are passing through 4.75 mm sieve and 40% of particles are passing 0.075 mm sieve. Its uniformity coefficient is 8 and coefficient of curvature is 2. As per USCS, this soil is classified as: a.SP b.GP c.GW d.SW
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
Soil lab Particle size distribution (Sieve method) In addition to that, calculate the value of D25 and D75 and calculate it in curvature
Q3- A sieve analysis on soil sample showed the following results; |Size, mm | 12 4.75 2.00 0.84 0.42 0.25 0.15 0.075 0.001 % finer 100 63 37 49 | 31 25 14 10 | 4 Plot the gradation curve and determine the weight of the gravel, sand, silt and clay if the total weight of the investigated soil mass is (1.3xN) kN. Show the plot and all calculation clearly. (N = 16 )
(b) Classification tests for the determination of the particle size distribution of a soil produced the following results: Particle Size % finer (mm) 99 2.36 80 1.3 63 0.6 33 0.2 10 0.075 3 0.02 Then using cone penetrometer for Liquid Limit test, the results were recorded as follows. Cone penetration 15.9 17.7 19.1 20.3 22.5 (mm) Water 32.6 42.9 51.6 59.8 66.2 content (%) While the plastic limit test conducted for the same soil recorded as below:- Test No. 1 2 3 Wet soil 10.01 11.23 9.72 Dry soil 8.02 9.01 7.72 SKAA 1713 4 i) Plot the grain size distribution curve on the log graph that has been provided (Figure 2b). Calculate the uniformity coefficient, Cu and the coefficient of gradation, C, for the soil sample. Determine the plastic limit, the liquid limit and the plasticity index of the soil. Plot the graph to help you obtain the results (if necessary). ii) iii)
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%
Given that coefficient of curvature = 1.4, D30 = 3 mm, D₁0 = 0.6 mm. 10 Based on this information of particle size distribution for use as subgrade, this soil will be taken to be (a) uniformly-graded sand (b) well-graded sand (c) very fine sand (d) poorly-graded sand
A synthetic soil is made by mixing: 400 plastic spheres with Gs = 0.8 and diameter = 5mm, and 250 glass spheres with Gs = 2.65 and diameter = 4mm. %3D %3D According to the USCS, would the synthetic soil classify as a sand or as a gravel? Show work to support your answer.