   Chapter 9, Problem 9.14P Principles of Geotechnical Enginee...

9th Edition
Braja M. Das + 1 other
ISBN: 9781305970939

Solutions

Chapter
Section Principles of Geotechnical Enginee...

9th Edition
Braja M. Das + 1 other
ISBN: 9781305970939
Textbook Problem

9.13 Through 9.14 Figure 9.29 shows the zone of capillary rise within a clay layer above the groundwater table. For the following variables, calculate and plot σ, u, and σ′ with depth.  Figure 9.29

To determine

Find and plot the total stress, pore water pressure, and effective stress for various depth.

Explanation

Given information:

The depth H1 of dry sand is 4.0 m.

The specific gravity of the dry sand Gs is 2.69.

The void ratio (e) of the dry sand is 0.47.

The depth H2 of clay zone is 2.5 m.

The specific gravity of the clay zone Gs is 2.73.

The void ratio (e) of the clay zone is 0.68.

The depth H3 of clay is 4.5 m.

The specific gravity of the clay Gs is 2.70.

The void ratio (e) of the clay is 0.89.

The degree of saturation (S) is 60 %.

Calculation:

Determine the dry unit weight γd(sand) of the sand using the formula.

γd(sand)=Gsγw1+e

Here, γw is the unit weight of the water.

Take the unit weight of the water as 9.81kN/m3.

Substitute 2.69 for Gs, 9.81kN/m3 for γw, and 0.47 for e.

γd(sand)=2.69(9.81)1+0.47=17.95kN/m3

Determine the saturated unit weight γsat(clay.capillaryzone) of the clay in capillary zone using the formula.

γsat(clay.capillaryzone)=γw(Gs+Se)1+e

Substitute 9.81kN/m3 for γw, 2.73 for Gs, 60 % for S, and 0.68 for e.

γsat(clay.capillaryzone)=9.81(2.73+(60100)0.68)1+0.68=30.7841.68=18.32kN/m3

Determine the saturated unit weight γsat(clay) of the clay using the formula.

γsat(clay)=γw(Gs+e)1+e

Substitute 9.81kN/m3 for γw, 2.70 for Gs, and 0.89 for e.

γsat(clay)=9.81(2.7+0.89)1+0.89=35.221.89=18.63kN/m3

Determine the total stress at 0 m depth using the relation.

σ=0

Determine the pore water pressure at 0 m depth using the relation.

u=0

Determine the effective stress at 0 m using the relation.

σ=σu=0

Since, there is no soil and water table in the ground level, the value of total stress, pore water pressure and the effective stress values becomes zero automatically.

Thus, the total stress σ at 0 m depth is zero_.

Thus, the pore water pressure (u) at 0 m depth is zero_.

Thus, the effective stress σ at 0 m depth is zero_.

Determine the total stress at 4.0 m depth using the relation.

σ=γd(sand)×H1

Substitute 17.95kN/m3 for γd(sand) and 4.0 m for H1.

σ=17.95×4.0=71.8kN/m2

Determine the pore water pressure at 4.0 m depth using the relation.

u=γw×h(S×γw×H2)

Here, h is the depth of water table.

Substitute 9.81kN/m3 for γw, 0 ft for h, 0.6 for S, and 2.5 m for H2.

u=(62.4×0)(0.6×9.81×2.5)=014.71=14.71lb/ft2

Determine the effective stress at 4.0 m depth using the relation.

σ=σu

Substitute 71.8kN/m2 for σ and 14

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