4.5 A saturated soil used to determine the shrinkage limit has initial volume V₁ = 20.2 cm³, final volume V, 14.3 cm³, mass of wet soil M₁ = 34 g, and mass of dry soil M₂ = 24 g. Determine the shrinkage limit and the shrinkage ratio. 4.6 Repeat Problem 4.5 with the following data: V₁ = 16.2; V, 10.8 cm³; M₁ = 44.6 g, and mass of dry soil, M₂ = 32.8 g.

4.5 A saturated soil used to determine the shrinkage limit has initial volume V₁ = 20.2 cm³, final volume V, 14.3 cm³, mass of wet soil M₁ = 34 g, and mass of dry soil M₂ = 24 g. Determine the shrinkage limit and the shrinkage ratio. 4.6 Repeat Problem 4.5 with the following data: V₁ = 16.2; V, 10.8 cm³; M₁ = 44.6 g, and mass of dry soil, M₂ = 32.8 g.

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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Concept explainers

Soil classification and characteristics

Soil classification is a method of categorizing soil based on its properties, types, and uses. The motive of soil classification is to arrange various types of soils and divide them into groups for various engineering and agricultural purposes. Soil classification is carried out by various agencies as per their purposes. From an engineering perspective, soil classification is done to find out the suitability of soil for various construction purposes like the construction of dams, foundations, highways, infrastructures, and so on. For engineering purposes, the soil is classified by the following systems:

Question

P2.4.6 List the given, required, graph and box the final answer Note: 4.5 is just a reference since as stated at the problem of 4.6

4.5 A saturated soil used to determine the shrinkage limit has initial volume
V₁ = 20.2 cm³, final volume V, 14.3 cm³, mass of wet soil M, = 34 g, and mass
of dry soil M₂ = 24 g. Determine the shrinkage limit and the shrinkage ratio.
4.6 Repeat Problem 4.5 with the following data: V; = 16.2; V, 10.8 cm³; M₁ = 44.6 g,
and mass of dry soil, M₂ = 32.8 g.

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