   Chapter 15, Problem 15.7P 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

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 for the wedge ABC.

Explanation

Given information:

The vertical height (H) of the soil 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 21°.

The cohesion c is 38kN/m2.

The unit weight γ of the soil is 19kN/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 19kN/m3 for γ, 6.0 m for H, 58° for β, and 32° for θ.

W=12(19)(6.0)2[sin(58°32°)sin58°sin32°]=333.6kN

Determine the normal component Ta using the relation.

Ta=Wsinθ

Substitute 333.6 kN for W and 32° for θ.

Ta=333.6sin32°=176.78kN

Determine the tangential component Na using the relation.

Na=Wcosθ

Substitute 333.6 kN for W and 32° for θ

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