The nominal compactive effort (NCE) used for standard proctor test is  600 kJ/m3   300 kJ/m3   6000 kJ/m3   1000 kJ/m3

Q: Bulk Unit Weight (kN/m3) 13.0 14.0 17.0 16.0 16.0 Water Content (%) 16.0 18.5 21.0 23.5 26.0 1. Plot…

A: Use the basic soil mechanics relation between the dry unit weight of the soil with the bulk unit…

Q: constant head permeability test is performed on soil that is 2 cm x 2 cm square, and 2.5cm long. The…

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Q: Draw the compaction curve using the following given data from a standard Proctor test. 2. From…

A: Objective: To draw the compaction curve for the given Standard Proctor Test results, and determine…

Q: The Table 4 shows the results of a Standard Proctor Compaction test. Based on the information…

A: "Hi! Thank you for the question, As per the honor code, we are allowed to answer three sub-parts at…

Q: Calculate the percentage of air for a Given porosity n and the saturation S ?

A: Given specific gravity G=2.65  we know γd=Gγw1+e⇒void ratio e=Gγwγd-1porosity  n=e1+eand   Se=wG   ⇒…

Q: Q1-A construction site requires 100m3 of soil compacted at the maximum dry density estimated from…

A: The volume of compacted soil is, The optimum moisture content is,  = 0.12The degree of saturation at…

Q: The results of a standard and a modified Proctor tests ate given in the table below. a) Generate a…

A: * zero air voids curve is mention in step 3 and 4 in soil sample, moisture content=weight of water…

Q: Problem 4. A sample of sand is tested in a constant head permeameter. The sample diameter is 50 mm…

A: Given data : Diameter of sample = 50 mm Length of sample = 120 mm Applied head =  60 cm  Discharge…

Q: 1. Using phase relationships and the assumption that the volume of solids is 1.0, show WGs = Se.

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Q: 2. (a) 2 kg of dry sand is poured into a Eureka can, where it displaces 755 cm³ of water. Calculate…

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Q: (a) The following data were recorded in a constant Head Permeability Test: Internal diameter of…

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Q: Determine the efficiency of removing particles of size 50 μm and density 1,100 kg/m², settling in a…

A: Particles size, d = 50 μm or 50 x 10-6 m Density of particle,  = 1,100 kg/m²Length = 10 m .Flowrate,…

Q: The soil profile shown below is typical of Miami-Dade County. Estimate the equivalent permeabilities…

A: Each layer permeability are given and so we can find the equivalent permeability in horizontal and…

Q: Table 1 below is the result of compaction results. Please evaluate the correct combination of…

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Q: Question 1: Earth whose in situ weight is 112 pcf, whose loose weight is 95 pcf and whose compacted…

A: Given:- Unit weight of the earth = 112 pcf The loose weight = 95 pcf The compacted weight = 120 pcf…

Q: about 0.8 mm. Estimate the variation of the relative density with depth for sand. Use Eq. (3.23).

A: water table located 12 m below surfacedry unit weight of sand (soil) depth upto 12 m = 17.6…

Q: A sample of sand is subjected to a constant head permeability test. The sample diameter is 60 mm and…

A: Given data  Diameter of sample (D) = 60 mm  Length of sample (L) = 130 mm  Applied head (h) = 60 cm…

Q: Afalling-head permeability test is performed on a fine-grained soil. The soil sample has a length of…

A: Note: Since you have posted a question with multiple subparts, we will solve the first three…

Q: A 50 cc of water sample is passed through a filter paper whose initial weight is found to be 1.243…

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Soil Compaction Test

Soil compaction is the major area of geotechnical engineering. Soil is composed of multiple pores present on its surface, while the soil is mainly composed of gravel, sand, rocks, etc. The arrangement of these rocks and gravels is such that they build up voids (spaces) between them, which allows spaces for the roots of the plants and crops to spread and absorb the required amount of nutrients. Soil compaction is a major problem primarily caused by human activities, micro-organisms, and other multiple factors.