Q1: Use the USCS to Classify the soil sample with the following data:% passing 4.75 mm Sieve = 64% passing 75- Sieve =4Uniformity Coefficient= 7.5Coefficient of Curvature=2.7Plasticity index = 2.5Solution:Q2: Use the USCS to Classify the soil samples with the following data dallaSoilLLPLGroup SymbolSoil Name153525ABC8. ८८2545652-Seepage velocity (m/s) =703- Flow rate in (m3/s) =oss and dry ng wwwParmbly & volume changedecreaseCampering wesToughness and dry strength deasePresby & val change in102030 4050- 60Liepakt.7080Atra-0.73 (W₂-20190Q3: For the soil shown below the hydraulic conductivity is (3 m/day). The sample diameter is 1.5 m. Theporosity is (30%). Find:1- Flow velocity (m/s)==100 110 120Soil4 mITE

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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:

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:

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%.
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