Given the tabulated data for an Atterberg Limit Test.A.LIQUID LIMITTest NumberNo. of blowsWeight of Wet soil + container 22.24g 21.19gWeight of Dry soil + container 20.10g| 19.28gWeight of Container1.2393012.74g 13.24g. Test NumberNo, of blowsWeight of Wet soil + container 21.27g 26.12gWeight of Dry soil + container 19.25g 22.60gWeight of Container3423122/213.06g 13.27gB. PLASTIC LIMITTest Number2Weight of Wet soil + container 22.12g 21.84g1Weight of Dry soil + container 20.42g 20.19gWeight of Container13.07g 13.18gC. NATURAL WATER CONTENTTest Number12Weight of Wet soil + container 17.53g 16.97gWeight of Dry soil + container 14.84g| 14.36gWeight of container7.84g 7.50ga. Compute the liquid limit.b. Compute the plastic limit.c. Compute the natural water content.d. Compute the plasticity index.e. Compute the liquidity index

As per the guidelines of Bartleby we are supposed to answer only the first three subparts of the…

Given the tabulated data for an Atterberg Limit Test. A. LIQUID LIMIT Test Number 1. No. of blows Weight of Wet soil + container 22.24g 21.19g Weight of Dry soil + container| 20.10g| 19.28g Weight of Container 39 30 12.74g 13.24g Test Number 3 4 No, of blows Weight of Wet soil + container 21.27g|26.12g Weight of Dry soil + container 19.25g 22.60g Weight of Container 23 12 13.06g .27g B. PLASTIC LIMIT 2 Weight of Wet soil + container 22.12g 21.84g Weight of Dry soil + container 20.42g 20.19g 13.07g 13.18g Test Number Weight of Container C. NATURAL WATER CONTENT Test Number 2 Weight of Wet soil + container 17.53g 16.97g Weight of Dry soil + container 14.84g| 14.36g 7.84g 7.50g Weight of container a. Compute the liquid limit. b. Compute the plastic limit. c. Compute the natural water content.

Given the tabulated data for an Atterberg Limit Test. A. LIQUID LIMIT Test Number 1. No. of blows Weight of Wet soil + container 22.24g 21.19g Weight of Dry soil + container| 20.10g| 19.28g Weight of Container 39 30 12.74g 13.24g Test Number 3 4 No, of blows Weight of Wet soil + container 21.27g|26.12g Weight of Dry soil + container 19.25g 22.60g Weight of Container 23 12 13.06g .27g B. PLASTIC LIMIT 2 Weight of Wet soil + container 22.12g 21.84g Weight of Dry soil + container 20.42g 20.19g 13.07g 13.18g Test Number Weight of Container C. NATURAL WATER CONTENT Test Number 2 Weight of Wet soil + container 17.53g 16.97g Weight of Dry soil + container 14.84g| 14.36g 7.84g 7.50g Weight of container a. Compute the liquid limit. b. Compute the plastic limit. c. Compute the natural water content.

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

SITUATION 1: The following tabulated data are results of plastic limit and liquid limit test. Plastic Limit test: Test Number Wt. of wet soil + 20.58g 21.76g container Wt. of dry soil + 19.70g 19.85g container Wt of container 11.12g 11.78g Liquid Limit test: Test Number Wt. of wet soil + container Wt. of dry soil + 4 22.31 21.12 22 89 21.95 19.15 19.21 21.18 20.30 container Wt of container 12.17 14.19 14.02 14.87 # of blows 33 28 22 17 Natural water content test: Test Number Wt. of wet soil + container Wt. of dry soll + 15.75g 15.98g 14.05g 14.25g container Weight of container 8.55g 8.12g Prepe
Subj- civil Q. An undisturbed soil sample that is cylindrical in shape and measures 3 in (76.2 mm) in diameter by 6 in (152.4 mm) in length after oven drying weighs 2.57 lb (1.166 kg bulk) as opposed to 3.04 lb (1.38 kg mass). What are the densities and the weights in (a) wet and dry units? Additionally, find out the moisture content.
For a soil specimen, given the following: % passing sieve #4-92 %, sieve #10 -81%, sieve # 40-78%, LL-48%, PL-16%, GI-18.2 classify the soil by AASHTO classification system. GI = (F200-35)[0.2 + 0.005 (LL - 40)] +0.01 (F20015)(PI – 10)
|Laboratory testing was performed on two soil samples (A and B). (a) Determine the USCS classification symbol for Sample A. (b) Determine the AASHTO classification for Sample B. Sieve No. Sieve Opening (mm) A- Percent Passing B - Percent Passing 3 inch 76.2 100 1.5 inch 38.1 98 0.75 inch 19.1 96 4 4.75 77 100 10 2.00 96 20 0.85 55 94 40 0.425 73 100 0.15 30 200 0.075 18 55 Liquid Limit 32 52 Plastic Limit 25 32 A Silty Clay (CL) sample was extruded from a 6-inch long tube with a diameter of 2.83 inches and weighed 1.71 lbs. (a) Calculate the wet density of the CL sample. (b) A small piece of the CL sample had a wet weight of 140.9 grams and a weight of 85.2 grams after drying. Compute the water content. (c) Compute the dry density of the CL sample A Standard Proctor test was performed on a soil with a specific gravity of 2.71. For the water content and wet soil unit weight in the following table: a) Plot the moisture-dry unit weight curve b) Find the maximum dry density and optimum…
PROBLEM 7: The laboratory results of a soil test are as follows: Sieve No. # 4 #10 #40 #200 LL PL Percent passing 100 93.20 81 55 45 17 Find the group index according to AASHTO classification system. Hint: GI = (F-35)[0.20 + 0.005(LL – 40)] + 0.01(F – 15)(PI – 10)
Problem C: For Questions 12 to 15. Following are the results from the liquid and plasticlimit tests for a soil. Liquid Limit Test Number of Moisture Blows,N Content (%) 16 42.150 20 40.775 28 39.115 12) From the flow curve what is the liquid Limit ? a) 39.683% b) 40.555 % c) 38.992 % d) 39.875 % 13) If the Plasticlimit is 18.750 % what is its plasticity index? a) 22.431 % c) 21.344 % b) 20.918 % 14) If in-situ moisture content is 31% determine its liquidity index ? c) 0.585 d) 20.819 % a) 0.665 b) 1.59 15) If the percentage of clay-size fraction is 4.831. Based on its Activity,this mineral is more likely to be a) Illites b) kaolinite d) 20.588 c) Attapulgite d) Smectites
Type 11/111 sedimentation analysis: A laboratory settling test was performed on a suspension exhibiting flocculent settling characteristics with the results given in the following table. The initial TSS concentration of the suspension was 400 mg/L. Time, min 20 30 TSS concentration at indicated depth 0.5 m 1.0 m 1.5 m 2.0 m 156 286 340 390 116 148 180 310 112 148 156 40 76 108 132 23 80 104 56 80 19 32 56 2.5 m 405 375 172 148 40 76 40 50 60 70 80 17 29 128 100 76 Estimate the removal efficiency for a sedimentation basin 3.5 m deep with an overflow rate, V. = 4.2 m/h (State all assumptions and show your work).
1. The following data shows the results of the liquid limit and plastic limit test using Casagrande cup method. Calculate the following: (Hint: Solve the moisture content of each test) a. Liquid Limit b. Plasticity Index c. Liquidity Index Liquid Limit Test: TEST NO. No. of Blows Wt. of Wet Soil + Container Wt. of Dry Soil + Container Wt. of Container Plastic Limit Test: TEST NO. Wt. of Wet Soil + Container Wt. of Dry Soil + Container Wt. of Container Natural Water Content: TEST NO. Wt. of Wet Soil + Container Wt. of Dry Soil + Container Wt. of Container 1 35 22.46 g 19.44 g 12.76 g 1 22.10 g 20.42 g 13.07 g 1 17.94 g 14.84 g 7.84 g 2 21 21.33 g 18.75 g 13.06 g 2 21.77 g 20.19 g 13.28 g 2 17.39 g 14.36 g 7.50 g 3 16 21.29 g 18.78 g 13.26 g
pls ans soon as posssible During a constant head permeability test on a sand sample, 260ml of water was collected in 2 min. If the sample had a length of 100 mm, a diameter of 40 mm and a maintained head of 200 mm, what is the coefficient of permeability?
Problem F: For Problems number 23 to 26. Given the results of a standard Proctor Test: Weight of Moisture wet soil in Content mold, (Ibs) (%) 3.77 5.0 4.12 7.5 4.32 10.0 4.425 12.5 4.45 15.0 4.35 17.5 4.20 20.0 23) What is the maximum dry unit weight of compaction? a)115.25 Ibs/ft b)117.90 Ibs/ft c)118.05 Ibs/ft d) 118.00lbs/ft 24) The corresponding Optimum moisture content is ? b) 11.95 % c) 12.50 % 25) Given G,=2.72 Determine the void ratio at its optimum moisture content ? b) 0.438 26) In relation to # 25 what is its equivalent degree of saturation? a) 12.59% d) 12.55 % a) 0.440 c) 0.483 d) 0.441 a) 77.626 % b) 77.635 % c) 78.905 % d) 77.265 %
Problem F: For Problems number 21 to 24. Given the results of a standard Proctor Test: Weight of Moisture wet soil in Content mold, (Ibs) (%) 3.92 5.0 4.12 7.5 4.29 10.0 4.37 12.5 4.45 15.0 4.35 17.5 4.20 20.0 21) What is the maximum dry unit weight of compaction? Name:
A specific gravity test was performed on three soil samples, and the following data were collected: Material weight of empty pycnometer weigt of dry soil weight of weight of pycnometer pycnometer Temperature + water +water+soil g Sand 1 106.6 Sand 2 106.6 Sand 3 106.6 2-a) Briefly describe the test procedure and discuss its applications in practice 2-b) The corrected specific gravity of SAND 1 G5,20 2-c) The corrected specific gravity of SAND 2 G5,20 2-d) The corrected specific gravity of SAND 3 G5,20 55.4 50 60 === === g 385.5 371 392 351 340 355 с 18 24 20