   Chapter 15, Problem 15.13P Principles of Geotechnical Enginee...

9th Edition
Braja M. Das + 1 other
ISBN: 9781305970939

Solutions

Chapter
Section Principles of Geotechnical Enginee...

9th Edition
Braja M. Das + 1 other
ISBN: 9781305970939
Textbook Problem

The inclination of a finite slope is 1 vertical to 1 horizontal. Determine the slope height, H, that will have a factor of safety of 2.8 against sliding. Given: ρ = 1950 kg/m3, ϕ ′ = 23 ° , and c′ = 40 kN/m2. Assume that the critical sliding surface is a plane.

To determine

Find the height (H) of the slope that will have a factor of safety of 2.8 against sliding.

Explanation

Given information:

The density ρ is 1,950kg/m3.

The angle of friction ϕ is 23°.

The cohesion c is 40kN/m2.

The factor of safety Fs is 2.8.

The inclination β of a finite slope is 1 vertical to 1 horizontal.

Calculation:

Determine the inclination angle β using the relation.

β=tan1(11)=45°

Determine the unit weight γ using the relation.

γ=ρg1,000

Here, g is the acceleration due to gravity.

Take the acceleration due to gravity as 9.81m/s2.

Substitute 1,950kg/m3 for ρ and 9.81m/s2 for g.

γ=(1,950)(9.81)1,000=19.12kN/m3

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

cd=cFs

Substitute 40kN/m2 for c and 2.8 for Fs.

cd=402.8=14.28kN/m2

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

ϕd=tan1(tanϕFs)

Substitute 23° for ϕ and 2

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