A soil profile is shown bellow. Given: H1 =1.83 m, H2= 0.91 m, H3=1.83 m.Determination of Unit Weight of each layers.Calculate σ, u, and σ’.Plot the variation of σ, u, and σ’ with depth. G, - 2.65e-0.5HZone of capillary riseG, 2.65: e0.5Degree of saturationS= 50%Groundwater tablew - 42% (moisture content)G, = 2.71SandSaturated clayRock

Given data Specific gravity Void ratio Height Degree of saturation Cappilary rise

Answered: G, - 2.65 e-0.5 Zone of capillary rise…

G, - 2.65 e-0.5 Zone of capillary rise G, 2.65: e0.5 Degree of saturationS= 50% Groundwater table w- 42% (moisture content) G, = 2.71

Fundamentals of Geotechnical Engineering (MindTap Course List)

5th Edition

ISBN:9781305635180

Author:Braja M. Das, Nagaratnam Sivakugan

Publisher:Braja M. Das, Nagaratnam Sivakugan

Chapter11: Ground Improvement

Section: Chapter Questions

Problem 11.9P

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Concept explainers

Sediment transport

Sediment transport or Soil transport is the movement of solid particles, usually due to the combination of gravitational forces acting on the sediment, and the movement of the liquid at which the sediment enters. Soil erosion occurs in natural systems where the particles are sandstones (sand, rocks), mud, or clay; the liquid is air, water, or ice; and the gravitational force that works to move particles to the landing surface. Sediment transportation or soil migration due to the movement of liquids occurs in rivers, lakes, lakes, oceans, and other bodies of water as a result of currents and waves. Transportation is also caused by glaciers as they flow, and to places on earth under the influence of wind. Soil erosion due to gravity alone is possible in most sloping areas, including hills, scarves, cliffs, and the continental shelf — the continental boundary of the slope.

Question

100%

A soil profile is shown bellow. Given: H₁ =1.83 m, H₂= 0.91 m, H₃=1.83 m.

1. Determination of Unit Weight of each layers.
2. Calculate σ, u, and σ’.
3. Plot the variation of σ, u, and σ’ with depth.

G, - 2.65
e-0.5
H
Zone of capillary rise
G, 2.65: e0.5
Degree of saturationS= 50%
Groundwater table
w - 42% (moisture content)
G, = 2.71
Sand
Saturated clay
Rock

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