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MindTap Engineering for Das/Sobhan's Principles of Geotechnical Engineering, 9th Edition, [Instant Access], 2 terms (12 months)
9th Edition
ISBN: 9781305971257
Author: Braja M. Das; Khaled Sobhan
Publisher: Cengage Learning US
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Chapter 4, Problem 4.8P
To determine
Find the liquid limit value by the one-point method.
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A consolidated-undrained tri-axial test was conducted on a normally conslidated clay sample and the results are follows:
Chamber confining pressure= 119kpa
Deviator Stress at failure=90 kpa
Pore water pressure= 58kpa
These results were used to determine the drained friction angle of the soil. Compute the deviator stress (kpa) at failure when the drained test was conducted with the chamber confining pressure changed to 156 kpa.
Use stored value. Answer to 5 decimal places.
A constant head permeability test is performed on soil that is 2 cm x 2 cm square, and 2.5cm long. The head difference applied during the test is 20 cm and 7 cm3 is collected over a time of 100 sec. Provide step by step solution.
a) Compute the permeability based on these test conditions and results.
b) A falling head test is conducted on the same soil specimen at the same time (t1 – t2 = 100 sec), and the standpipe diameter is 0.8 cm. If the average head during the test should be 18 cm, what are h1 and h2 values, provide step by step solution solving h1 and h2 values.
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
Chapter 4 Solutions
MindTap Engineering for Das/Sobhan's Principles of Geotechnical Engineering, 9th Edition, [Instant Access], 2 terms (12 months)
Ch. 4 - During Atterberg limit tests in the soil mechanics...Ch. 4 - Prob. 4.2PCh. 4 - Prob. 4.3PCh. 4 - Results from a liquid limit test conducted on a...Ch. 4 - The following data were obtained by conducting...Ch. 4 - Refer to the soil in Problem 4.5. Using the...Ch. 4 - Following results are obtained for a liquid limit...Ch. 4 - Prob. 4.8PCh. 4 - Prob. 4.9PCh. 4 - Prob. 4.10P
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- solve only 4.2 please. 4.2 Refer to the soil in Problem 4.1. A second group of students conducted only one test and found that the groove on the soil sample closed 12.5 mm when N = 21 and w = 30.4%. Estimate the liquid limit by the one-point method. 4.1 problem is provided below: During Atterberg limit tests in the soil mechanics laboratory, the students obtained the following results from a clayey soil. Plastic limit tests: Students conducted two trials and found that PL = 17.2% 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.arrow_forwardQ1/ 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 Vertical pressure (kPa) 15 Settlement (mm) 0.1 Void ratio 1.88 4 0.6 30 0.11 1.87 9 0.71 16 0.79 And the void ratio and settlement for each load increment is shown in the table below. 60 0.21 1.85 120 1.13 1.72 100 1.04 36 64 0.91 0.86 240 2.17 1.57 480 3.15 1.43 1- Find Coefficient of consolidation in cm²/s using root time and TV90% = 0.848. 2- Find Preconsolidation pressure (oc) and compression index (Cc). Scanned by TapScannerarrow_forwardA consolidated undrained tri-axial test was conducted on a normally consolidated clay sample and the results are as follows: chamber confining pressure = 110kpa deviator stress at failure = 96kpa pore water pressure = 66kpa these test results were used to determine the drained friction angle of the soil. compute the deviator stress in (kpa) at failure when the drained test was conducted with the chamber confining pressure changed to 163kpa.arrow_forward
- Q3): (A): The result of constant head permeability test for sand sample having a diameter of 100 mm and a length of 200 mm are as follows: Constant head difference= 300 mm, time of collection of water= 3 min, volume of water collected 250 cc. Find the coefficient of permeability for the soil.arrow_forward(e). A field density test was conducted by core cutter method and the following data was obtained: Weight of the empty core-cutter = 25 N; Weight of soil and core cutter = 50 N Inside diameter of core-cutter = 90 mm; Height of the core-cutter = 180 mm Weight of wet sample for moisture determination = 0.53 N; Weight of over dried sample=0.48 N Specific gravity of solids = 2.72 Determine 1. Dry unit weight, Ya 2. Void ratio, e 3. Degree of saturation, S,arrow_forwardPermeability 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/cm³arrow_forward
- A consolidated undrained triaxial test was conducted on a normally consolidated clay sample and the results are as follows: chamber confining pressure = 118 kpa, Deviator stress at failure 93 kpa, pore water pressure = 52 kpa. These results were used to determine the undrained friction angle of the soil. Compute the deviator stress (kPa) at failure when the drain test was conducted in the chamber confining pressure change to 168 kpaarrow_forwardIn a falling permeameter, the sample used is 20 cm long having a cross-sectional area of 24 cm2. The sample of soil is made of three layers. The thickness of the first layer from the top is 8 cm and has a value of k1 = 2 x 10-4 cm/s, the second layer of thickness 6 cm has k2 = 5 x 10-4 cm/s and the bottom layer of thickness 4 cm has k3 = 7 x 10-4 cm/s. Assume the flow is taking perpendicular to the layers. Estimate the average hydraulic conductivity of soil in the vertical direction. a.3.12 x 10-4 cm/s b.5.69 x 10-4 cm/s c.2.35 x 10-4 cm/s d.4.28 x 10-4 cm/s e.6.69 x 10-4 cm/sarrow_forwardProblem 3) Shown below is a cross-section through a soil layer. A Shelby tube sample was taken from a depth of 25 ft. The sample was extruded in a laboratory, thus subjected to zero total stress. Assume Ko = 0.5 and Ā = -0.1, determine the effective stress on the sample after extrusion. Use the basic Skempton (1954) and Ladd and Lambe (1963) approach to determine the change in pore pressure.arrow_forward
- 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/cm³arrow_forward49. The following data are given for the laboratory sample. o=175 kPa; e = 1.1; +Ao = 300 kPa; e = 0.9 If thickness of the clay specimen is 25 mm, the value of coefficient of volume compressibility is x 10-4 m²/kN A. 3x10-³ m²/kN B. 5x10-6 m²/kN C. 12x10-4 m²/kN D. 7.61×10-4 m²/kNarrow_forwardIn a falling permeameter, the sample used is 20 cm long having a cross sectional area of 24 cm2. The sample of soil is made of three layers. The thickness of the first layer from the top is 8 cm and has a value of k1 = 2 x 10-4 cm/s, the second layer of thickness 6 cm has k2 = 5 x 10-4 cm/s and the bottom layer of thickness 4 cm has k3 = 7 x 10-4 cm/s. Assume the flow is taking perpendicular to the layers. Calculate the time required for the drop of head from 25 to 12 cm if the cross sectional area of the stand pipe is 2 cm2. a.71.5 min b.62.9 min c.87.8 min d.66.7 min e.58.81 minarrow_forward
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