EBK FUNDAMENTALS OF GEOTECHNICAL ENGINE
EBK FUNDAMENTALS OF GEOTECHNICAL ENGINE
5th Edition
ISBN: 8220101425829
Author: SIVAKUGAN
Publisher: CENGAGE L
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
2. Find the equivalent concentrated load(s) for the bags of cement stacked on the dock as shown here. Each bag weighs 100 lbs and is 12 inches long. Draw the loading conditions for each showing the equivalent concentrated load(s). 1 bag = 100lbs L= 12 ft L= 6 ft L= 8ft
I have a question for this problem in the first one wouldn't it be finding the total weight of the bags which =4800lbs and the multiply that by 12ft to find the concentrated load?? but if this is the case the load would end up as lbs/ft so I'm not too sure that is right.
There are 2 parts A) L=12ft B) L1= 6ft, L2= 8ft
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