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 13.1P

(a)

To determine

Whether the statement “A lower safety factor implies that a greater fraction of the shear strength is mobilized along the failure surface” is true or false.

(a)

Expert Solution
Check Mark

Answer to Problem 13.1P

The given statement is True_.

Explanation of Solution

Write the equation of factor of safety (FSs) with respect to strength.

FSs=τfτdτd=τfFSs (1)

Here, τf is the average shear strength of the soil along the potential failure surface and τd is the average shear stress mobilized along the potential failure surface.

Equation (1) clearly shows that the average shear stress mobilized (τd) along the potential failure surface is inversely proportional to the factor of safety (FSs) with respect to strength.

Hence, the lower value of safety factor implies that the greater fraction of the shear strength.

Therefore, the given statement is True_.

(b)

To determine

Whether the statement “An infinite slope of cohesive soil cannot remain stable when β>ϕ” is true or false.

(b)

Expert Solution
Check Mark

Answer to Problem 13.1P

The given statement is False_.

Explanation of Solution

Condition for stability:

The factor of safety is greater than 1.

Write the equation of factor of safety (FSs) with respect to strength.

FSs=cγHcos2βtanβ+tanϕtanβ

Consider cohesionless soil, the effective cohesion (c) is zero.

FSs=0γHcos2βtanβ+tanϕtanβ=tanϕtanβ (2)

From Equation (2), when β>ϕ, the term tanϕ/tanβ is less than 1. Therefore, cohesionless is not stable when β>ϕ.

But in case of cohesive soil, effective cohesion (c) is greater than zero. Therefore, the term [c/(γHcos2βtanβ)] compensated the term tanϕ/tanβ and it makes that the factor of safety is more than 1. Hence, the infinite slope of cohesive soil remains stable when β>ϕ.

Therefore, the given statement is False_.

(c)

To determine

Whether the statement “In an undrained slope of 1.0 (horizontal): 1.5 (vertical), the critical circle is always a toe circle.” is true or false.

(c)

Expert Solution
Check Mark

Answer to Problem 13.1P

The given statement is True_.

Explanation of Solution

For slope angle (β) greater than 53.8°, the critical circle is always a toe circle.

Find the slope angle (β) using the equation:

tanβ=VerticalriseHorizontalrise

Substitute 1.5 for vertical rise and 1.0 for horizontal rise.

tanβ=1.51.0β=tan1(1.5)β=56.3°

Here, the slope angle (56.3°) is greater than 53°. Hence, the critical failure circle is a toe circle.

Therefore, the given statement is True_.

(d)

To determine

Whether the statement “The mid-point circle passes through the toe” is true or false.

(d)

Expert Solution
Check Mark

Answer to Problem 13.1P

The given statement is False_.

Explanation of Solution

  • In case of base failure, the failure circle is called a mid-point circle.
  • The surface of sliding that passes at some distance below the toe of the slope at the time of failure is called base failure. Hence, the mid-point circle does not pass through the toe.

Therefore, the given statement is False_.

(e)

To determine

Whether the statement “Taylor’s stability charts can be used only for homogeneous soils” is true or false.

(e)

Expert Solution
Check Mark

Answer to Problem 13.1P

The given statement is True_.

Explanation of Solution

  • Taylors stability chart is used to take the value of stability number (m) with corresponding friction angle of soil (ϕ) and the slope angle (β).
  • Taylor’s stability chart is developed for slopes in homogeneous soil with friction angle (ϕ).greater than zero. Hence, the Taylor’s stability charts can be used only for homogeneous soils.

Therefore, the given statement is True_.

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Students have asked these similar questions
Q4: For the soil element shown, compute the stresses acting on the plane inclined by 40° with the horizontal plane then draw Mohr circle and place the stresses with respect to O.P. 20kPa 35kPa 100kPa 300 40°
Q2: For the soil element shown in the figure. Draw Mohr circle, locate OP (origin point), then find the stresses inclined at: 1- 75° from the horizontal plane. 2- With the horizontal plane.
The location of trial failure surface on a slope is shown in Figure and the stress components for each slice are listed in Table. Slice Shearing component (kN.m') -0.280 -0.227 0.383 3.214 6.543 8.368 9.792 Normal Length of trial failure surface=11m Soil friction angle 6° Cohesion, c=28kPa No. component (kN.m') 1.911 7.745 13.139 1 2 4 16.344 17.625 16.718 12.125 7 8. 4.228 0.486 Trial Failure Surface (a) Compute the driving stress. (b) Compute the stabilizing stress. (c) Analyze the safety of the slope.
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