   Chapter 15, Problem 15.9P 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

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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