Principles of Geotechnical Engineering (MindTap Course List)
Principles of Geotechnical Engineering (MindTap Course List)
9th Edition
ISBN: 9781305970939
Author: Braja M. Das, Khaled Sobhan
Publisher: Cengage Learning
Question
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Chapter 12, Problem 12.4P

(a)

To determine

Find the drained angle of friction for the silty clay soil.

(a)

Expert Solution
Check Mark

Answer to Problem 12.4P

The drained angle of friction for the silty clay soil is 19.2°_.

Explanation of Solution

Given information:

The diameter (d) of clay sample is 63.5 mm.

The height (h) of the clay sample is 32 mm.

Calculation:

Calculate the area of specimen (A) as follows:

A=πd24

Substitute 63.5 mm for d.

A=π(63.5)24=12,661.264=3,165.32mm2(1m103mm)2=0.00316m2

Find the normal stress (σ) for Test No. 1 using the formula as follows:

σ=NA

Here, N is the normal force and A is the area of specimen.

Substitute 84N for N  and 0.00316m2 for A.

σ=840.00316=26,582.27N/m2(1kN1,000N)=26.58kN/m2

Similarly calculate the normal stress (σ) for remaining Tests as shown in Table 1.

Test noNormal force, N(N)Normal stress (σ)(kN/m2)
18426.58
216853.16
325480.38
4360113.92

Table 1

Find the shear strength (τf) of the failure for Test No. 1 as follows:

τf=SA

Substitute 28.9N for S and 0.00316m2 for A.

τf=28.90.00316=9,145.5N/m2(1kN1,000N)=9.14kN/m2

Find the angle of friction (ϕ) for Test No. 1 as follows:

ϕ=tan1(τfσ)

Substitute 26.58kN/m2 for σ and 9.14kN/m2 for τf.

ϕ=tan1(9.1426.58)=18.976°

Similarly calculate the angle of friction (ϕ) and shear strength (τf) for remaining tests as shown in Table 2.

Test noShear force, S (N)Shear strength, τf(kN/m2)Angle of friction (ϕ°)
128.99.1418.97
259.618.8619.53
389.128.1919.33
4125.339.6519.19

Table 2

Calculate the drained angle of friction (ϕ) for the silty clay soil using Table 2.

The average value of angle of friction of all tests is the drained angle of friction.

Refer to the Table 2,

Find the drained angle of friction as follows:

ϕ=18.97°+19.53°+19.33°+19.19°4=77.02°4=19.2°

Thus, the drained angle of friction for the silty clay soil is 19.2°_.

(b)

To determine

Find the shear strength of the clay in the field at location A.

(b)

Expert Solution
Check Mark

Answer to Problem 12.4P

The shear strength of the clay in the field at location A is 30.92kN/m2_.

Explanation of Solution

Given information:

The specific gravity (Gs) of silty sand is 2.69.

The specific gravity (Gs) of silty clay is 2.72.

The unit weight (γw) of water is 9.81kN/m3.

The void ratio (e) is 0.72.

The water content (w) of silty clay is 22%.

The depth (h1) of silty sand above water table is 2m.

The depth (h2) of silty sand below water table is 2.2m.

The depth (h3) of silty clay is 3.5m.

Calculation:

Calculate the dry unit weight (γd) for silty sand using the relation as follows:

γd=Gsγw1+e

Substitute 2.69 for Gs, 9.81kN/m3 for γw, and 0.72 for e.

γd=2.69×9.811+0.72=26.38891.72=15.342kN/m3

Calculate the saturated unit weight [(γsat)s] for silty sand using the relation as follows:

(γsat)s=(Gs+e)γw(1+e)

Substitute 0.72 for e, 9.81kN/m3 for γw, and 2.69 for Gs.

(γsat)s=(2.69+0.72)9.81(1+0.72)=33.4521.72=19.448kN/m3

Calculate the saturated unit weight [(γsat)c] for silty clay using the relation as follows:

(γsat)c=(1+w)Gsγw(1+wGs)

Substitute 22% for w, 2.72 for Gs, and 9.81kN/m3 for γw.

(γsat)c=(1+0.22)2.72×9.81(1+0.22×2.72)=32.551.5984=20.36kN/m3

Determine the normal stress (σA) at point A as shown below:

σA=h1×γd+h2×[(γsat)sγw]+h3×[(γsat)cγw]

Substitute 2m for h1, 15.342kN/m3 for γd, 2.2m for h2, 19.448kN/m3 for (γsat)s 9.81kN/m3 for γw, 3.5m for h3, 20.36kN/m3 for (γsat)clay.

σA=2×15.34+2.2×(19.459.81)+3.5×(20.369.81)=30.68+21.208+36.925=88.81kN/m2

Find the shear strength (τf) of the clay in the field at location A as follows:

τf=σAtanϕ

Here, ϕ is the angle of friction.

Substitute 88.81kN/m2 for σA and 19.2° for ϕ.

τf=88.81×tan19.2°=88.81×0.348=30.92kN/m2

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