A Liquid Limit test was carried out on a sample of Clay with the following results: Number of taps 6. 12 26 28 31 Moisture Content 53.4 52.2 48.3 40.0 38.8 37.1 (36) The results from the Plastic Limit test gave the Plastic Limit of 28%. Determine: a) the Liquid Limit b) the Plasticity Index of the soil c) and Classify the soil
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- 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/cm³Permeability 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/cm³3- A series of standard Proctor compaction tests were performed on soil and the obtained results are presented in Figure below. The volume of the standard mould was 1 x 10³ m³ and Gs = 2.65. 1.82 1.80 1.78 1.76 What is the degree of saturation of the sample compacted to the maximum dry density? Estimate the volume of air (in m3) in the sample compacted to 15% water content. The void ratio of this soil in its loosest state is 1.74 1.72 1.70 1.68 1.66 0.623. Estimate the relative density of the soil sample compacted to 11% water content? 10 11 12 13 14 15 16 17 18 19 20 21 22 Water content (%) Dry density (g cm-3)
- In falling head permeability test the initial head was 350 mm and the final head was 60 mm. The cross-sectional area of the burette was 200 mm2. The sample was 50 mm in diameter and 105 mm long. If the test took 8 hours and 30 minutes, what was the coefficient of permeability of the soil in m/s?A liquid limit test on a clay was performed with the following results. The natural water content of the clay is 38% and plastic limit is 21%. Number of Blows 6 12 20 28 32 Water content (%) 52.5 47.1 42.3 38.6 37.5 What is the liquidity index of this clay? (Use Interpolation) Group of answer choices 0.895 0.96 1.08 0.90 Please answer this asap. For upvote. ThanksSand cone equipment is used to perform Field density test on a compacted arth fil. Soil sample dug from the test hole = 19.88 N Dry weight of soil sample = 17.92 N Ottawa sand used to fill the hole weighs 16.05 N and is known to have a density of 15.74 HN/m. Which of the following gives the water content of the tested soil? O a. 15.45% O b. 12.08% O c. 14.96% O d. 10.94% e. 11.12%
- 3. A dry sand is placed in a container 9 cm3. The dry weight of the sample is 15 g. Water is carefully added to the container so as not to disturbed the condition of the Sample. When the container is filled the combined weight of soil and water is 18 g. Determine the following: a) Void ratio of sample in the container b) Porosity c) Water constant d) Specific gravity of the soil particle e) Saturated unit weight f) Dry unit weightA 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.3- Following are the results of a laboratory consolidation test on a sample of undisturbed clay obtained from the field. Final height of specimen (in) Pressure, o' (lb/ft') 1 0.9917 1,000 2,000 4,000 0.9843 0.956 0.9142 8,000 0.8685 16,000 0.8228 32.000 The height of the specimen at the beginning of the test was 1.0 inch, and the diameter was 2.5 inches. The moist specimen weighed 0.31 lb and the water content was determined to be 19%. Estimate the compression index and the preconsolidation pressure from the e-log o' curve. (Plot in Excel) Given: Gs = 2.7.
- 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%.Afalling-head permeability test is performed on a fine-grained soil. The soil sample has a length of 120 mm and a cross-sectional area of 600 mm2. The water in the standpipe flowing into the soil is 0.60 m above the top of the sample at the start of the test. It falls 50 mm in 30 minutes. The standpipe has a cross-sectional area of 200 mm2. (a) Make a sketch of the described conditions. (b) What is the coefficient of permeability in millimeters per second? (c) What is the coefficient of permeability in feet per minute? (d) On the basis of the calculated value for khyd, what is the probable soil type?1. A soil with a liquidity index of -0.20 has a liquid limit of 56 percent and a plasticity index of 20 percent. What is its natural water content? 2. Volume of a soil sample obtained was 150 cm3 and its total mass was found to be 250 g. In the laboratory the dry mass of the sand alone was found to be 24 0 g. Tests on the dry sand indicated e-max = 0.80 and e-min = 0.48. Estimate ps, w, e, S, pd and Dr of the sand in the field. Given Gs = 2.67. CUO tendord