Fundamentals of Geotechnical Engineering (MindTap Course List)
Fundamentals of Geotechnical Engineering (MindTap Course List)
5th Edition
ISBN: 9781305635180
Author: Braja M. Das, Nagaratnam Sivakugan
Publisher: Cengage Learning
Question
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Chapter 13, Problem 13.27CTP

(a)

To determine

Find the maximum shear stress developed within the soil.

(a)

Expert Solution
Check Mark

Answer to Problem 13.27CTP

The maximum shear stress developed within the soil is 33.3kN/m2_.

Explanation of Solution

Given information:

The depth (H) of slope is 5 m.

The angle (β) of bed slope is 23°.

The unit weight of the soil (γ) is 18.5kN/m3.

The cohesion (c) is 15kN/m2.

The angle (ϕ) of friction is 20°.

Calculation:

The maximum shear stress developed at soil rock interface (5 m depth).

Find the maximum shear stress 20° developed within the soil using the equation:

τd=γHsinβcosβ

Substitute 18.5kN/m3 for γ, 5 m for H, and 23° for β

τd=18.5×5×sin23°×cos23°=33.3kN/m2

Thus, the maximum shear stress developed within the soil is 33.3kN/m2_.

(b)

To determine

Find the maximum shear strength available within the soil.

(b)

Expert Solution
Check Mark

Answer to Problem 13.27CTP

The maximum shear strength available within the soil is 43.5kN/m2_.

Explanation of Solution

Given information:

The depth (H) of slope is 5 m.

The angle (β) of bed slope is 23°.

The unit weight of the soil (γ) is 18.5kN/m3.

The cohesion (c) is 15kN/m2.

The angle (ϕ) of friction is 20°.

Calculation:

Find the maximum shear strength (τf) available within the soil using the equation:

τf=c+γHcos2βtanϕ.

Substitute 15kN/m2 for c, 18.5kN/m3 for γ, 5 m for H, 23° for β, and 20° for ϕ.

τf=15+(18.5)(5)cos2(23°)tan(20°)=15+28.5=43.5kN/m2

Thus, the maximum shear strength available within the soil is 43.5kN/m2_.

(c)

To determine

Find the factor of safety of the slope.

(c)

Expert Solution
Check Mark

Answer to Problem 13.27CTP

The factor of safety of the slope is 1.31_.

Explanation of Solution

Given information:

The depth (H) of slope is 5 m.

The angle (β) of bed slope is 23°.

The unit weight of the soil (γ) is 18.5kN/m3.

The cohesion (c) is 15kN/m2.

The angle (ϕ) of friction is 20°.

Calculation:

Find the factor of safety (FSs) of the slope using the equation:

FSs=cγHcos2βtanβ+tanϕtanβ

Substitute 15kN/m2 for c, 18.5kN/m3 for γ, 5 m for H, 23° for β, and 20° for ϕ.

FSs=15(18.5)(5)cos2(23°)tan(23°)+tan(20°)tan(23°)=0.45+0.86=1.31

Therefore, the factor of safety of the slope is 1.31_.

(d)

To determine

Find the maximum possible depth for the soil before it becomes unstable.

(d)

Expert Solution
Check Mark

Answer to Problem 13.27CTP

The maximum possible depth for the soil before it becomes unstable is 15.8m_.

Explanation of Solution

Given information:

The depth (H) of slope is 5 m.

The angle (β) of bed slope is 23°.

The unit weight of the soil (γ) is 18.5kN/m3.

The cohesion (c) is 15kN/m2.

The angle (ϕ) of friction is 20°.

Calculation:

The slope becomes unstable then the factor of safety (FSs) against sliding is 1.0.

Find the maximum possible depth (Hcr) for the soil before it becomes unstable using the equation:

Hcr=cγ1cos2β(tanβtanϕ)

Substitute 15kN/m2 for c, 18.5kN/m3 for γ, 23° for β, and 20° for ϕ.

Hcr=(1518.5)1cos223°(tan23°tan20°)=0.811(19.51)=15.8m

Thus, the maximum possible depth for the soil before it becomes unstable is 15.8m_.

(e)

To determine

Find the factor of safety with respect to cohesion when the friction is fully mobilized.

(e)

Expert Solution
Check Mark

Answer to Problem 13.27CTP

The factor of safety with respect to cohesion when the friction is fully mobilized is 3.1_.

Explanation of Solution

Given information:

The depth (H) of slope is 5 m.

The angle (β) of bed slope is 23°.

The unit weight of the soil (γ) is 18.5kN/m3.

The cohesion (c) is 15kN/m2.

The angle (ϕ) of friction is 20°.

Calculation:

The developed angle of friction is equal to the angle of friction when the friction is fully mobilized. Therefore, ϕd=ϕ.

Find the developed cohesion in the soil using the equation:

τd=cd+γHcos2βtanϕdcd=τdγHcos2βtanϕd

Substitute ϕ for ϕd.

cd=τdγHcos2βtanϕ

Substitute 33.3kN/m2 for τd, 18.5kN/m3 for γ, 5 m for H, 23° for β, and 20° for ϕ.

cd=33.3(18.5)(5)cos2(23°)tan(20°)=33.328.5=4.8kN/m2

Find the factor of safety (FSc) with respect to cohesion when the friction is fully mobilized using the equation:

FSc=ccd

Substitute 15kN/m2 for c and 4.8kN/m2 for cd.

FSc=154.8=3.1

Thus, the factor of safety with respect to cohesion when the friction is fully mobilized is 3.1_.

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Fundamentals of Geotechnical Engineering (MindTap...
Civil Engineering
ISBN:9781305635180
Author:Braja M. Das, Nagaratnam Sivakugan
Publisher:Cengage Learning