Calibration of Pycnometer (1) Mass of dry clean pycnometer, Mp = 317.11 g (2) Mass of pycnometer + water at calibration temperature, Mpw,c = 903.03 g (3) Observed temperature of water at calibration, T = 26.4°C Secondary data for the determination of Specific Gravity is as follows: 1 966.17 22.3 Determination No: Mass of pycnometer + soil + water at test temperature, Mpws,t (9) Test Temperature, T₁ (°C) Mass of pycnometer + water at T₁, Mpw,t (9) Evaporating dish no. Mass of evaporating dish, Md (9) Mass of evaporating dish + oven-dried soil, Mds (9) Mass of solids, M, (g) Conversion factor, K Specific gravity of soil, G₁@20°C 1 15.52 115.51 Determine the Mass of pycnometer + water at Tt, Mpw,t- (Units of g, 2 decimal places) Note: Find Pw,t using table 6-1.

Calibration of Pycnometer (1) Mass of dry clean pycnometer, Mp = 317.11 g (2) Mass of pycnometer + water at calibration temperature, Mpw,c = 903.03 g (3) Observed temperature of water at calibration, T = 26.4°C Secondary data for the determination of Specific Gravity is as follows: 1 966.17 22.3 Determination No: Mass of pycnometer + soil + water at test temperature, Mpws,t (9) Test Temperature, T₁ (°C) Mass of pycnometer + water at T₁, Mpw,t (9) Evaporating dish no. Mass of evaporating dish, Md (9) Mass of evaporating dish + oven-dried soil, Mds (9) Mass of solids, M, (g) Conversion factor, K Specific gravity of soil, G₁@20°C 1 15.52 115.51 Determine the Mass of pycnometer + water at Tt, Mpw,t- (Units of g, 2 decimal places) Note: Find Pw,t using table 6-1.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

See similar textbooks

Similar questions

Initial data for determining the specific gravity of a soil sample is given below: Calibration of Pycnometer (1) Mass of dry clean pycnometer, M₁ = 317.11 g (2) Mass of pycnometer + water at calibration temperature, Mpw,c = 903.03 g (3) Observed temperature of water at calibration, Tc = 26.4°C Secondary data for the determination of Specific Gravity is as follows: Determination No: Mass of pycnometer + soil + water at test temperature, Mpws,t (9) Test Temperature, T₁ (°C) Mass of pycnometer + water at T₁, Mpw,t (9) Evaporating dish no. Mass of evaporating dish, Md (9) Mass of evaporating dish + oven-dried soil, Mds (g) Mass of solids, Ms (9) 1 966.17 22.3 1 15.52 115.51 Conversion factor, K Specific gravity of soil, Gs@20°C Determine the Mass of Solids, Ms. (Units of g, 2 decimal places)
Initial data for determining the specific gravity of a soil sample is given below: Calibration of Pycnometer (1) Mass of dry clean pycnometer, Mp= 317.11 g (2) Mass of pycnometer + water at calibration temperature, Mpw,c = 903.03 g (3) Observed temperature of water at calibration, T = 26.4°C Secondary data for the determination of Specific Gravity is as follows: Determination No: Mass of pycnometer + soil + water at test temperature, Mpws,t (9) Test Temperature, T₁ (°C) Mass of pycnometer + water at T₁, Mpw,t (9) Evaporating dish no. Mass of evaporating dish, M. (g) Mass of evaporating dish + oven-dried soil, Mas (9) Mass of solids, M, (g) Conversion factor, K Specific gravity of soil, Gs@20°C Determine the conversion factor, K (5 decimal places) 1 966.17 22.3 1 15.52 115.51
Find the missed data and give the result of Fineness Modulus of sample soil data given below. Weight Percentage Cumulative Percentage finer Sieve retained Weight retained percentage N = (100 - C) S.NO size retained (C) Select one: (gm) (gm) O a. 7.824 1 75mm O b. 9.438 63mm O c. 8.142 3 50mm O d. 7.649 4 37.5mm 28mm 123 6.15 6.15 6 20mm 107 5.35 11.50 7 14mm 208 10.4 21.9 8 10mm 56 2.8 24.7 6.3mm 3.9 10 5.0mm 92 4.6 33.2 11 3.35mm 308 15.4 48.6 12 2.0mm 223 11.15 59.75 13 1.18mm 188 9.4 69.15 14 600µ 282 14.1 83.25 15 300µ 96 4.8 88.05 16 150µ 88 4.4 92.45 17 75µ 103 5.15 97.6 18 pan 48 2.4 100
Let us assume that a soil sample taken in the field is taken and it must be identified by the SUCS Method and AASHTO Method. Moisture content = 5.75% Limit liquid = 53.5 Plastic limit = 31.6 Granulometry Sieve % Pass No. 4 100 # 40 68 # 200 54.3 Do you select the best answer? a) Inorganic silt and sand, with low plasticity b) Inorganic clay with low plasticity c) Clay silt, traces of gravel, dark Brown
The same soil sample has a natural moisture content of 30% and was subject to the Atterberg Limits test. The plastic limit was measured as 27 while the liquid limit test shows the results in the table below. Number of Blows 42 26 10 Moisture Content 20 32 45 a. Determine the liquid limit of the soil sample. Show and briefly explain how you got the answer. b. Solve and describe the Plasticity Index of the soil sample. c. Determine value and soil state of the liquidity index of the soil sample. d. Determined and describe the clay activity of the soil sample.
4. The table below shows the laboratory results of the sample. Determine the soil classification using Unified Soil Classification System (USC). Sieve No. Opening (mm) Percent Passing 4 4.760 90 8 2.380 64 10 2.000 58 20 0.840 35 40 0.420 22 60 0.250 15 100 0.149 0.074 10 200 4 D10 = 0.15mm, D30 = 0.65mm, D60 = 2.20mm %3D
2a. The following data was obtained from Liquid Limit Test conducted at a Soil Lab. Using the data given in Table 01, determine the following; Table 01: Liquid Limit Test data for the particular soil sample Liquid Limit Determination Specimen No. No. of Blows Wt. of Container Wt. of Container + Wet Soil Wt. of Container + Dry Soil i. ii. Unit 1 14 7.11 (g) (g) (g) 21.31 27.91 2 20 7.31 27.31 21.21 3 28 7.71 28.41 22.31 4 32 10.91 30.11 24.61 5 39 7.21 26.51 21.11 Determine the Liquid Limit and Flow Index If Plastic Limit of this soil is 22. Classify the soil based on BS Soil Classification system. I
The following are the data of a soil sample: Sample Size – 2-inch diameter, 4 inches high. Wet weight – 0.401 lbs Oven-dry weight – 0.334 lbs Specific Gravity – 2.68 Calculate the following: a. Void Ratio b. Water content c. Degree of Saturation d. Dry Unit Weight in pcf e. Unit weight in pcf f. Saturated Unit Weight in pcf g. Dry Density in English h. Density in English i. Saturated Density in English
5. The results of field unit weight determination of a soil sample using the sand cone method is as follows: Volume of soil: 0.00134 m³ Mass of soil: 2250 grams Dry mass of soil: 1734 grams In a laboratory, this soil yields a dry density of 1.425 g/cc at optimum moisture content of 13.8% a. What is the field unit weight of soil? b. What is the in-situ water content of soil? c. Determine the relative compaction of the soil.
5. The following information is obtained from a sieve analysis to determine the range of particle sizes in a granular soil sample. Sieve Size Sieve Opening (4.76) Percent Finer #4 4.76 96 #10 2.00 80 #20 0.84 51 #40 0.42 38 #60 0.25 25 #100 0.149 12 #200 0.074
The laboratory test results of a soil sample are given below Percentage finer than 4.75 mm 60 ; Percentage finer than 0.075 mm = 30 Liquid Limit = 35%; Plastic Limit = 27% The soil classification is
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