please answer all 4.1Results from liquid and plastic limit tests conducted on asoil are given here.Liquid limit tests:Number of blows, NMoisture content (%)1438.41636.52033.12827.0Plastic limit tests: PL = 13.4%a. Draw the flow curve and obtain the liquid limit.b. What is the plasticity index of the soil?4.2Determine the liquidity index of the soil in Problem 4.1 ifWin situ32%.

It is required to find 1. a. Liquid limit b. Plasticity index 2. Liquidity index

Answered: Results from liquid and plastic limit…

Principles of Geotechnical Engineering (MindTap Course List)

9th Edition

ISBN:9781305970939

Author:Braja M. Das, Khaled Sobhan

Publisher:Braja M. Das, Khaled Sobhan

Chapter7: Permeability

Section: Chapter Questions

Problem 7.17P

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The total volume of a soil specimen is 80,000 mm^3 and has a mass of 152 g. the dry mass of the specimen is 122g, and the density of the soil is 2.65 Mg/m^3. Compute the following index properties: (a) e; (b) n; (c) S; (d) ρt; and (e) ρd. Include phase diagrams.
A sample of clay (weight = 13.2 N) is brought back from the field and trimmed to the following dimensions: height = 150 mm and diameter = 75 mm. The water content has been determined to be 25%. The Gs value for the solids is 2.75. Show and complete your phase diagram and find the following parameters: a. Draw a phase diagram b. Complete the phase diagram by finding the parameters of VT, Va, Vw, Vs, MT, Ma, Mw, and Ms. Volume should be expressed in the unit of m' and mass in grams. Show all calculations to receive full credits.
Permeability tests were performed on a soil sample, under different void ratio and different temperatures and the following results were obtained. Test No. 1 2 Void ratio (e) 0.65 1.02 Temperature °℃ 25° 40⁰ Estimate the coefficient of permeability at a temperature of 20°C for a voids ratio of 0.80. Given the following physical properties of water: At 20°C, n = 10.09 × 10+ and p„ = 0.998 g/cm³ At 25°C, n = 8.95 × 104 g sec/cm² and p k(cm/s) 0.4x10 1.9x10-¹ At 40° C, n = 6.54 × 10-¹ g sec/cm² and p $0.997 g/cm³ = = 0.992 g/cm³
QI: The following are the results of a laboratory consolidation test on a soil specimen obtained from the field: dry mass of specimen = 128 gm, height of specimen at the beginning of the test = 2.54 cm, Gs = 2.75, and area of the specimen = 30.68 em?. Effective pressure, ớ(kN/m²) Final height of the specimen at the end of the consolidation (cm) 2.540 2.488 50 100 200 2.465 2.431 400 800 1600 3200 2.389 2.324 2.225 2.115 Required: Make necessary calculations and draw an e versus log ở curve.
1. During Atterberg limit tests in the soil mechanics laboratory, the students obtained the following results from a clayey soil. Liquid limit tests: Number of Moisture blows, N content (%) 14 38.4 16 36.5 20 33.1 28 270 Plastic limit tests: Students conducted two trials and found that PL = 172% for the first trial and PL = 17.8% for the second trial. a. Draw the flow curve and obtain the liquid limit. b. What is the plasticity index of the soil? Use an average value of PL from the two plastic limit trails.
SITUATION 3: A 100 cc proctor mold was used to collect a wet soil sample. Its weight was found to be 200 gms. Its oven dried weight was recorded as 189 grams. G = 2.7 7. The moisture content is most likely: a. 0.058201 c. 0.069841 8. Its void ratio is: a. 0.771429 c. 0.514286 b. 0.064021 d. 0.046561 b. 0.471429 d. 0.428571
The results of a laboratory consolidation test on a clay specimen are the following. Pressure, ir (lb/ft²) 500 1.000 2.000 4.000 8,000 16.000 Total height of specimen at end of consolidation (in.) 0.6947 0.6850 0.6705 0.6520 0.6358 0.6252 Given the initial height of specimen = 0.748 in.. G specimen = 95.2 g. and area of specimen = 4.91 in.²: a. Plot the e-log or curve = 2.68. mass of dry b. Determine the preconsolidation pressure c. Calculate the compression index, C.
QI: The following are the results of a laboratory consolidation test on a soil specimen obtained from the field: dry mass of specimen = 128 gm, height of specimen at the beginning of the test = 2.54 cm, Gs = 2.75, and area of the specimen = 30.68 cm². Final height of the specimen at the end of the consolidation (cm) 2.540 2.488 Effective pressure, ở(kN/m?) 50 100 2.465 200 2.431 2.389 2.324 400 800 1600 2.225 2.115 3200 Required: Make necessary calculations and draw an e versus log ở curve.
Pressure (kN/m²)27 54 107 214 429 214 107 54 Void ratio1.243 1.217 1.144 1.068 0.994 1.001 1.012 1.024 The table above shows results obtained from an oedometer test on a specimen of saturated clay. A layer of this clay 2.5 m thick lies below a 10 m depth of sand, the water table being at the surface. The saturated weight for both soils is 19 kN/m³. A 10-m depth of fill of unit weight of 21 kN/m³ is placed on the sand over an extensive area. If the fill was to be removed some time after the completion of consolidation, what heave (mm) would eventually take place due to swelling of the clay? (0 d.p). Use unit weight of water of 9.8 kN/m3. Use log base 10.
Q1/ specimen of the fine-grained soil, 75 mm in diameter and 20 mm thick was tested in an oedometer in a laboratory. At applying pressure (240 kN/m²), the time with dial gauge readings are: Time (min) 0 Settlement (mm) 0 0.25 1 0.22 0.42 4 9 0.6 0.71 16 0.79 36 64 100 0.86 0.91 1.04 d 1- Find Coefficient of consolidation in cm²/s using root time and TV90% = 0.848. Scanned by TapScanner
Permeability tests were performed on a soil sample, under different void ratio and different temperatures and the following results were obtained. Test No. 1 2 Void ratio Temperature °C (e) 0.65 1.02 25° 40⁰ Estimate the coefficient of permeability at a temperature of 20°C for a voids ratio of 0.80. Given the following physical properties of water: At 20°C, n = 10.09 × 10+ and p„ = 0.998 g/cm³ At 25°C, n = 8.95 × 104 g sec/cm² and p k(cm/s) 0.4x10 1.9x10¹ At 40° C, n = 6.54 × 10 g sec/cm² and p 8 0.997 g/cm³ = 0.992 g/cm³
Question 8 (1) - Oedometer test What are the boundary conditions applied to a soil sample tested in an Oedometer? Z H a) Pore water pressure at z = H (top of the sample) is PH = Yw. H b) Ez = Ex c) &z=&v (₂ is the volumetric strain of the sample) d) At time t=0, 4p = 0

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