A clayey sand (SC) was compacted into a 1/30th cubic foot cylindrical Proctor steel mold with the following results: Weight of dry soil = 4.18 lb ,Weight of wet soil = 6.37 lb ,Specific Gravity, Gs = 2.73. Determine the percent saturation of the compacted soil.

Q: The results from a soil compaction test are shown in the table below. The solid grains of the soil…

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Q: A loose compacted sand fill is 1.81 m in depth and has a relative density of 41%. Laboratory tests…

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Q: Following are the results of a field unit weight determination test on the same soil with the sand…

A: Soil compaction is a process in which the soil retains mechanical pressure and is compacted. The…

Q: A sample of soil compacted according to the standard Proct or test has a density of 2.06 g/cm³ at…

A: Dry Unit weight is given by ; γdry = γ1 + w where ; γ = ρg = 2.06 × 9.81 = 20.209 KN/m2 w = Water…

Q: Problem 1: The table below present the results of a standard proctor test of a borrow pit soil. FEB…

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Q: A sand cone test has been performed in a recently compacted fill. The test results obtained are as…

A: Relative compaction is 95.23%Explanation:

Q: 1. Sand cone equipment is used to determine an in-place unit weight (field density test) on a…

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Q: The results of a laboratory test of a sand are emax= 0.91, emin= 0.48. a. what is their relative…

A: Given data emax= 0.91, emin= 0.48 Find relative compactness ?

Q: 737 grams of dry sand completely fills a 500 cm³ permeameter tube. The specific gravity of the sand…

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Q: The volume of a soil sample at the liquid limit (L.L = 50%) was 329.7 cm³ and the mass 570 gm. The…

A: Given data of soil sample :-  Liquid limit = 50% = 0.5  Volume of soil at liquid limit = 329.7 Cm3…

Q: Modified Proctor compaction test results for excavated soil at a construction site indicated that…

A: Solution:

Weight-Volume relationships

Weight is defined as the quantity of matter contained in a body. Weight can also be defined as the force acting on a body due to gravity. Mathematically, weight can be defined as-