Chapter 15, Problem 15.8P

### Principles of Geotechnical Enginee...

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

Chapter
Section

### Principles of Geotechnical Enginee...

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

# Refer to Problem 15.7. With all other conditions remaining the same, what would be the factor of safety against sliding for the trial wedge ABC if the height of the slope were 9 m?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 for the trial wedge ABC if the height of the slope is

9 m.

Explanation

Given information:

The vertical height (H) of the soil is 9.0 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, 9.0 m for H, 58° for β, and 32° for θ.

W=12(19)(9)2[sin(58°32°)sin58°sin32°]=750.62kN

Determine the normal component Ta using the relation.

Ta=Wsinθ

Substitute 750.62 kN for W and 32° for θ.

Ta=750.62sin32°=397.76kN

Determine the tangential component Na using the relation.

Na=Wcosθ

Substitute 750.62 kN for W and 32° for θ.

Na=750

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