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Chapter 15, Problem 15.9P
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### Principles of Geotechnical Enginee...

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

#### Solutions

Chapter
Section
BuyFindarrow_forward

### Principles of Geotechnical Enginee...

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

# Refer to the slope in Problem 15.7. Assume that the shear strength of the soil is improved by soil stabilization methods, and the new properties are as follows: γ = 22 kN/m3, ϕ ′ = 32 ° , and c′ = 75 kN/m2. What would be the improved factor of safety against sliding along the trial failure surface AC?15.7 Figure 15.49 shows a slope with an inclination of β = 58°. If AC represents a trial failure plane inclined at an angle θ = 32° with the horizontal, determine the factor of safety against sliding for the wedge ABC. Given: H = 6 m; γ = 19 kN/m3, ϕ ′ = 21 ° , and c′ = 38 kN/m2.Figure 15.49

To determine

Find the factor of safety Fs against sliding along the trial failure surface AC.

Explanation

Given information:

The height (H) of the soil and rock core is 6.0 m.

The slope with an inclination β is 58°.

The failure plane inclined at an angle θ of 32°.

The angle of friction ϕ is 32°.

The cohesion c is 75kN/m2.

The unit weight γ of the soil is 22kN/m3.

Calculation:

Draw the free body diagram of the wedge ABC as in Figure 1.

Determine the weight W of the wedge ABC using the formula.

W=12γH2[sin(βθ)sinβsinθ]

Substitute 22kN/m3 for γ, 6.0 m for H, 58° for β, and 32° for θ.

W=12(22)(6)2[sin(58°32°)sin58°sin32°]=386.28kN

Determine the normal component Ta using the relation.

Ta=Wsinθ

Substitute 386.28 kN for W and 32° for θ.

Ta=386.28sin32°=204.69kN

Determine the tangential component Na using the relation.

Na=Wcosθ

Substitute 386.28 kN for W and 32° for θ

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