Q5: Results from liquid and plastic limit tests conducted on soil are given below. No. of blows, N Moisture content (%) 14 16 20 28 38.4 36.5 33.1 27.0 If the plastic limit is PL = 13.4% a. Draw the flow curve and obtain the liquid limit. b. What is the plasticity index of the soil? c. Determine the liquidity index of the soil if Winsi = 32%

Q5: Results from liquid and plastic limit tests conducted on soil are given below. No. of blows, N Moisture content (%) 14 16 20 28 38.4 36.5 33.1 27.0 If the plastic limit is PL = 13.4% a. Draw the flow curve and obtain the liquid limit. b. What is the plasticity index of the soil? c. Determine the liquidity index of the soil if Winsi = 32%

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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Results from liquid and plastic limit tests conducted on a soil are given below. Liquid limit tests: Number of blows, N Moisture content (%) 14 38.4 16 36.5 20 33.1 28 27.0 Plastic limit tests: PL = 13.4% a. Draw the flow curve and obtain the liquid limit. b. What is the plasticity index of the soil? © Cengage Learning 2014
P2, 4.3 List the given, required, graph and box the final answer.
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PART IV: PERMEABILIT Y PROBLEM A For a constant head laboratory test on a fine sand, the following are the given values : • Length of the specimen • Di ameter of the specimen 14.0 inches 3.25 inches He ad difference 25.0 inches Water collected in 2.50 mins Void ratio A. Compute for the hydraulic conductivity of the soil in in/sec. B. Determine the discharge velocity through the soil in in/sec. c. Determine the seepage velocity in in/sec. 0.065 cubic inches 0.45
The following results were recorded during a Falling head permeability test: Sample diameter = 120 mm Sample length = 140 mm Initial head = 510 mm Final head = 310 mm Standpipe diameter = 6 mm Duration of flow = 3.4 mins Determine the coefficient of permeability for the soil in m/s. Based on the coefficient of permeability estimate the soil type
Using Unified Soil Classification System, determine the following: (a) Coefficient of Gradation (b) Group Symbol. Passing No. 4 = 47 % Passing No. 200 = 10 % D10 = 3.45 mm %3! D30 = 6.29 mm De0 = 13.72 mm Plasticity for the minus No. 40 fraction: Liquid Limit = 28 % Plastic Limit = 17 % O GC-GM GP-GC O GC O Coef. Of Gradation = 0.84 O GP O Coef. Of Gradation = 3.97
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Soil mechanics
Q5) A: Laboratory test results on the soil (L.L=35% and P.L-21%) is given in the following Table below. Classify the soil by using the unified classification system (10 mark) (showing all steps). Sieve Size(mm) Weight retained (gm) 4.75 2 0.425 0.25 0.075 Pan 135 70 60 30 142.5 462.5 B: A fine-grained soil has a liquid limit of 200% and a plastic limit of 45%, The natural water content of the soil in the field is 60% and the clay content is 63%, Calculate the plasticity index, the liquidity index and the activity. (10 mark)
4. The results of liquid limit testing on 3 samples of a soil are shown below. sample 1. 2 3 mass (g) 20 mass mass (g) emply can (M.) wet soil + can (M.) dry soil + can (M) moisture content: w (%) Number of cranks 20 20 35 35 28.82 36 28.15 30.38 4 10 30 Calculate the moisture content of cach of the 3 samples. Show your calculations below and then put your answers in the chart above. M,- M, м, - м, x100%
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