There is a 2D “on-plane” state of displacements which allow only an outward movement of the wall, i.e

W.Brząkała: Appendix to the Lecture #5/6 CEB (& Project #2)

THE COULOMB-PONCELET active ultimate earth pressure

1. Assumptions for a basic case

A very long wall is rigid, its surface AB=L is flat (linear), usually not vertical (∠β to the vertical line), usually not smooth (∠δ2 to the normal line). There is a 2D “on-plane” state of displacements which allow only an outward movement of the wall, i.e. to the left in Fig.1.

Note the anti-clockwise sign convention for the angles; in particular, both the angles ε,β presented in Fig. are positive (generally, they can be negative as well), there is always ϕ ≥ δ2 ≥ 0, χ > 0, but -ϕ = δ1 < 0.

Soil parameters c= 0 [kPa], ϕ >0 [deg or rad], γ>0 [kN/m³] – it is a backfill, generally.

A continuous vertical live load q ≥ 0 [kPa] distributed on the ground surface can happen as well.

Fig.1. Decomposition of the weight G of triangle ABC plus the linear (live) load q·AC

You can take in the design practice:

• δ² = 0 for a perfectly smooth surface (not realistic),

• δ2 ∼ +ϕ/3 for a „rather smooth” case (pre-cast concrete elements first of all),

• δ2 ∼ +2ϕ/3 for a „rather rough” case (in-field made concrete works),

• δ2 = ϕ for a perfectly rough surface (possible but not recommended).

3. Solution

The ultimate (extreme) condition dEa/dχ = 0 gives ∙ ∙ ∙ ∙ ∙ in [kN/m]

where ^{ϕ β}

β ∙

ϕ _{∙ !"}

#$ % ∙#$ "!%&

and ^'()

*$

+,- . / .

Clearly, 0 1₄³ 2 , therefore 1 2 ⁄ and finally 12 ∙ ∙ ∙ in [kPa].

Conclusion: the active earth stress ea(L) increases linearly along the wall, it is of the trapezoidal shape – - with a constant part due to q and a triangular part due to γ; in both cases with the same angle δ2 to the normal.

+

χ A

B

C

β

ε Ea

R GABC

δ1

δ2

∠ = χ+δ1

∠ = 90^o-β-δ²

L

q

q

2. Study on the wall roughness

A

B1

B2

q

h1

4. Non-trivial example

1. Consider a multi-linear profile of the wall AB1B2 and so on.

2. Solve on AB1 as above on AB.

3. Draw a new virtual ground surface from the point B1 as previously you did from the point A; so, the angle ε is the same and the lines are parallel.

4. Solve a new task on B1B2 as above on AB1 but for a new loading q1 =q+h1·γ·cos(ε).

5. Note that ∠β is evidently different (∠δ2 sometimes tooC);

also L=0 starts anew at B1, not at A.

q1

Ea

GABC