Problem (6)The laboratory test data for a standard Proctor test are given below.Volume ofProctor moldMass of wet soil inthe mold (kg)Moisture conient(%)(ст3)943.3943.31.65101.7512943.31.831.8114943.316943.31.7618943.31.7020a) Find the maximum dry unit weight and optimum moisture content.b) Calculate the zero-air-void unit weights for a soil with G, =2.68 at moisture content of5, 10, 15, 20, and 25%. Plot a graph of Yrav versus moisture content.

Answered: Image /qna-images/answer/ea3dba86-b24d-4d26-919f-be3aa4d389eb.jpg

Problem (6) The laboratory test data for a standard Proctor test are given below. Volume of Mass of wet soil in the mold (kg) Proctor mold Moisture conient (%) (ст3) 943.3 943.3 1.65 10 1.75 12 943.3 1.83 1.81 14 943.3 16 943.3 1.76 18 20 943.3 1.70 a) Find the maximum dry unit weight and optimum moisture content. b) Calculate the zero-air-void unit weights for a soil with G, =2.68 at moisture content of 5, 10, 15, 20, and 25%. Plot a graph of Yrav versus moisture content.

Problem (6) The laboratory test data for a standard Proctor test are given below. Volume of Mass of wet soil in the mold (kg) Proctor mold Moisture conient (%) (ст3) 943.3 943.3 1.65 10 1.75 12 943.3 1.83 1.81 14 943.3 16 943.3 1.76 18 20 943.3 1.70 a) Find the maximum dry unit weight and optimum moisture content. b) Calculate the zero-air-void unit weights for a soil with G, =2.68 at moisture content of 5, 10, 15, 20, and 25%. Plot a graph of Yrav versus moisture 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

The result of the liquid limit tests where N is the no of blows and w is the water content in % are given as follows: Trial 1 N=11 w=53.9; Trial 2 N=15 w=50.6; Trial 3 N=23 w=48.1; Trial 4 N= 30 w= 46; Trial 5 N= 46 w=43.3; Trial 6 N= 53 w=41. The PL= 25 and the in situ moisture content is 45%. Determine the liquid limit. a.41 b.43 c.50 d.47
A fully saturated soil sample has a volume of 28 cc. The sample was dried in oven and the weight of the dry soil pat was found to be 48.76 g. Given G= 2.68. Determine the weight of water in grams. Choose the correct answeer below: a.1.745 b.9.77 c.2.09 d.34.90
Plastic Limit Test: Weight of wet soil + container = 22.12 g Weight of dry soil + container = 21.42 g Weight of container = 13.07 g Natural Water Content Test: Weight of wet soil + container = 17.53 g Weight of dry soil + container = 14.84 g Weight of container = 7.84 g a.) Compute the liquid limit. (Tip: Use normal scale) b.) Compute the plastic limit. c.) Compute the natural water content. d.) Compute the plasticity index. e.) Compute the liquidity index. d.) Compute the consistency index.
- Find the moisture content, using dry weight per trial based on the table. - Solve the mean, sd, and rsd of the measurements. Description Trial 1 Trial 2 Trial 3 Trial 4 Weight of empty watch glass (g) 38.1151 38.1474 38.9473 38.7400 Weight of watch glass and soil sample (g) 90.1440 85.7554 86.6980 85.1235 Weight of watch glass and dry soil 67.0170 64.1915 65.1110 63.8859
Determine the moisture content, the moisture content by dry weight per trial based on the table below. Description Trial 1 Trial 2 Trail 3 Trial 4 Weight of empty aluminum dish (g) 38.1151 38.1474 38.9473 38.7400 Weight of aluminum dish and soil sample (g) 90.1440 85.7554 86.6980 85.1235 Weight of aluminum dish and dry soil 67.0170 64.1915 65.1110 63.8859 Then calculate the mean, sd, and rsd of the measurements.
The result of the liquid limit tests where N is the no of blows and w is the water content in % are given as follows: Trial 1 N=11 w=53.9; Trial 2 N=15 w=50.6; Trial 3 N=23 w=48.1; Trial 4 N= 30 w= 46; Trial 5 N= 46 w=43.3; Trial 6 N= 53 w=41. The PL=25 and the in situ moisture content is 45%. Determine the liquid limit. O a. 41 O b. 47 O c. 43 O d. 50
In a liquid limit test on a fine-grained soil, using the Casagrande apparatus, the following results were recorded: Test # 1 2 3 4 Number of blows 50 36 24 15 Container number J11 F12 C20 C2 Wet soil + container 90 92 91 91 Dry soil + container 79 82 83 87 Container mass 82.00 81.00 83.00 82.00 dry soil moisture loss Moisture content, w% If the plastic limit on the same soil was found to be 25%, determine the liquid limit and plasticity index of the soil.
Sample 2: Mass with empty mold M1 Mass of soil and mold M2 Hydraulic Head h Time interval t Volume V 145 g 250 g 950 mm 0.34 min 0.0046 L 145 g 260 g 900 mm 0.33 min 0.0049 L 145 g 228 g 922 mm 0.30 min 0.0051 L 145 g 281 g 915 mm 0.29 min 0.0044 L 145 g 219 g 908 mm 0.39 min 0.0042 L The diameter of the mold = 2.5 cm, Length of the sample = 0.3 m can u pls Find the permeability coefficient Mass of the soil Bulk Density Dry Density thank you Void Ratio
The result No of blows(N) and water content in % (w) of liquid limit tests are given as follows: Trial 1 N= 48 w=32.1 Trial 2 N=38 w=35.1; Trial 3 N= 29 w=40.7; Trial 4 N=20 w=46.1; Trial 5 N=15 w=52.8. Determine the plasticity index if PL=23. CHOICES: a.20 b.52 c.34 d.43
Determine youngs modulus
43.) The results of the sieve analysis done on a soil are shown below. Sieve No. Diameter Soil A 4 4.760 100 6 2.380 98 10 2.000 91 20 0.840 85 40 0.420 84 60 0.250 79 100 0.149 61 200 0.074 55 LL 60 PL 40 ML None of the above. CL CH MH
In a liquid limit test, using cone penetrometer, the following readings are recorded. Plastic Limit test Weight of wet soil+container= 22.12g Weight of dry soil+container=21.42g Weight of container = 13.07g Natural water content test Weight of wet soil+container= 17.53g Weight of dry soil+container=14.84g Weight of container = 7.84g A.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 F.Compute the consistency index 7. Define the consistency of the soil based on the liquidity index