   Chapter 13, Problem 13.27P 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 Figure 13.27. Given: H = 7.5 m, θ = 10°, α = 5°, γ = 17.9 kN/m3, ϕ' = 28°, δ'= 1 2 ϕ', kh = 0.4, kv = 0. Determine the active force, Pae, per unit length of the retaining wall. Also find the location of the resultant line of action of Pae. To determine

Find the active earth force Pae on the wall and the location z¯ and direction of the resultant.

Explanation

Given information:

The height (H) of the retaining wall is 7.5 m.

The soil friction angle ϕ is 28°.

The angle of wall friction δ is 12ϕ.

The unit weight γ of the material is 17.9kN/m3.

The angle θ of the wall is 10°.

The angle α is 5°.

The horizontal inertial kh force is 0.4.

The vertical inertial kv force us 0.

Calculation:

Determine the angle of wall friction using the relation.

δ=12ϕ

Substitute 28° for ϕ.

δ=12×28°=14°

Determine the ratio of angle of wall friction to the soil friction angle.

δϕ=14°28°=0.5

Determine the active force Pae per unit length of the wall using the formula.

Pae=12γH2(1kv)Ka (1)

Here, Ka is the coefficient of active force.

Refer Table (13.10) “values of Ka with θ=10° and kv=0” in the text book.

For the horizontal inertial force of 0.4 and 12ϕ of angle of wall friction, and the angle α of 5°, the coefficient of active force can be taken as 1.216.

Substitute 17.9kN/m3 for γ, 7.5 m for H, 0 for kv, and 1.216 for Ka in Equation (1).

Pae=12(17.9)(7.5)2(10)(1.216)=612.18kN/m

Determine the active earth pressure Pa on the wall using the formula.

Pa=12KaγH2

Here, Ka is the active earth pressure coefficient

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