Bundle: Principles Of Geotechnical Engineering, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card
Bundle: Principles Of Geotechnical Engineering, Loose-leaf Version, 9th + Mindtap Engineering, 2 Terms (12 Months) Printed Access Card
9th Edition
ISBN: 9781337583817
Author: Braja M. Das, Khaled Sobhan
Publisher: Cengage Learning
Question
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Chapter 15, Problem 15.12P
To determine

Find the factor of safety Fs with respect to sliding.

Expert Solution & Answer
Check Mark

Answer to Problem 15.12P

The factor of safety Fs with respect to sliding is 2.25_.

Explanation of Solution

Given information:

The slope with an inclination β is 55°.

The unit weight γ of the soil is 121lb/ft3.

The angle of friction ϕ is 17°.

The cohesion c is 1,200lb/ft2.

The height (H) of the retaining wall is 45 ft.

Calculation:

Trial 1:

Consider the factor of safety as 2.

Determine the cohesion cd develop along the potential failure surface using the relation.

cd=cFs

Substitute 1,200lb/ft2 for c and 2.0 for Fs.

cd=1,2002.0=600lb/ft2

Determine the angle ϕd of friction that develops along the potential failure surface using the relation.

ϕd=tan1(tanϕFs)

Substitute 17° for ϕ and 2.0 for Fs.

ϕd=tan1(tan17°2.0)=tan1(0.153)=8.69°

Determine height of the slope that will have a factor of safety of 2.0 against sliding using the formula.

Hcr=4cdγ[sinβcosϕd1cos(βϕd)]

Substitute 600lb/ft2 for cd, 121lb/ft3 for γ, 55° for β, and 8.69° for ϕd.

Hcr=4(600)121[sin55°cos8.69°1cos(55°8.69°)]=19.835(0.810.31)=51.83ft

The height of the retaining wall is not equal to the calculated height of the wall.

51.83ft45ft

Hence, the assumption is incorrect.

Trial 2:

Consider the factor of safety as 2.5.

Determine the cohesion cd develop along the potential failure surface using the relation.

cd=cFs

Substitute 1,200lb/ft2 for c and 2.5 for Fs.

cd=1,2002.5=480lb/ft2

Determine the angle ϕd of friction that develops along the potential failure surface using the relation.

ϕd=tan1(tanϕFs)

Substitute 17° for ϕ and 2.5 for Fs.

ϕd=tan1(tan17°2.5)=tan1(0.1222)=6.97°

Determine height of the slope that will have a factor of safety of 2.0 against sliding using the formula.

Hcr=4cdγ[sinβcosϕd1cos(βϕd)]

Substitute 480lb/ft2 for cd, 121lb/ft3 for γ, 55° for β, and 6.97° for ϕd.

Hcr=4(480)121[sin55°cos6.97°1cos(55°6.97°)]=15.867(0.810.33)=38.94ft

The height of the retaining wall is not equal to the calculated height of the wall.

38.94ft45ft

Hence, the assumption is incorrect.

Trial 3:

Consider the factor of safety as 2.25.

Determine the cohesion cd develop along the potential failure surface using the relation.

cd=cFs

Substitute 1,200lb/ft2 for c and 2.25 for Fs.

cd=1,2002.25=533.33lb/ft2

Determine the angle ϕd of friction that develops along the potential failure surface using the relation.

ϕd=tan1(tanϕFs)

Substitute 17° for ϕ and 2.25 for Fs.

ϕd=tan1(tan17°2.25)=tan1(0.136)=7.74°

Determine height of the slope that will have a factor of safety of 2.25 against sliding using the formula.

Hcr=4cdγ[sinβcosϕd1cos(βϕd)]

Substitute 533.33lb/ft2 for cd, 121lb/ft3 for γ, 55° for β, and 7.74° for ϕd.

Hcr=4(533.33)121[sin55°cos7.74°1cos(55°7.74°)]=17.631(0.8120.32)=44.74ft45ft

The height of the retaining wall is equal to the calculated height of the wall.

45ft=45ft

Thus, the factor of safety with respect to sliding is 2.25_.

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