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STABILITY AGAINST SLIDING ON THE FOUNDING LEVEL

The retaining wall (or other structure) is checked being treated as a rigid body. There are no significant differences in calculation between a massive wall and a light cantilever-wall with a rigid triangular wedge (Poncelet)

or a rigid trapezoid wedge (Rankine), attached to it. Design condition is as follows:

so in the format of DA-2*, two forces are analyzed on the (inclined) founding level:

destabilizing tangent forces T

d

[kN/m] which act to shift the wall to the left,

stabilizing tangent forces T

fd

= T

f

R

[kN/m] caused by friction and adhesion on this level.

Consider, for example, GEO stability against sliding for a cantilever retaining wall in the Rankine model:

soil pressure E

a

acts only on the virtual surface AB, which is inclined at an angle β to the vertical; here β = 0, δ

2

= ε as usually for the Rankine model, Fig.1. The resultant Q [kN/m] comes from the vertical live load q [kPa], it is taken from the section to the left of the point A; base slope in the Fig.1 equals α = -11

o

(1: 5).

Design values of destabilizing forces

Partial safety factors for loads γ

G

= 1,35 (1,00) and γ

Q

= 1,50 (0,00).

I Method – overestimation of unfavorable actions + underestimation of favorable actions (envelope):

=

,

∙ 1,35 +

,

∙ 1,50 ∙ cos − +

,

∙ 1,00 +

,

∙ 1,00 + ! ∙ 0,00 ∙ cos 90

#

− more clear when omitting negative sign of α:

=

,

∙ 1,35 +

,

∙ 1,50 ∙ cos + | | −

,

∙ 1,00 +

,

∙ 1,00 + ! ∙ 0,00 ∙ sin | | . II Method – on the effects of actions, separately for γ and q:

= 1,35 ∙ '

,

∙ cos − +

,

+

,

∙ cos 90

#

− ( + 1,50 ∙ '

,

∙ cos − + ! ∙ cos 90

#

− ( more clear when omitting negative sign of α:

= 1,35 ∙ '

,

∙ cos + | | −

,

+

,

∙ sin | | ( + 1,50 ∙ '

,

∙ cos + | | − ! ∙ sin | | (.

The advantage of the I Method is that it overestimates (but in a rather unrealistic way) the impact, making it safe.

The disadvantage of I Method is that it overestimates (but in a rather unrealistic way) the impact, making it uneconomical.

The advantage of II Method is that it simplifies the matter by separating the effects of dead and live loads - for example, there are no coefficients 1.00 or 0.00.

The disadvantage of II Method is that it simplifies the matter by separating the effects of dead and live loads - making this strange, for example, the increasing negative component -1,35 ⋅G

k

⋅ sin( | |) - 1,50⋅Q

k

⋅ sin( | |).

It should be emphasized, however, that this is not a big problem, because overall actions in brackets [ ] are evidently positive.

Eurocode EC7-1 does not decide which method to use.

N T

E

G

z

G

b

α <0

ε

δ

2

+ β - α W

B A

Fig.1. Acting loads.

Dead loads: G, E

aγ

Live loads: q, Q, E

aq

Eccentric W

k

: e

B

(not shown)

Angle of E

a

to the base: δ

2

+ β - α = ε - α Angle of G, Q to the base: 90

o

-α.

Q

q

E

aq

δ

2

+β-α

(2)

Design values of stabilizing forces

)

=

*+,

-

,

)

= . ∙ /0 1

2

∙ 3 + 4 ∙ 5′

Partial safety factor for the resistance against sliding equals γ

R

= 1,1.

The symbol a

k

denotes adhesion [kPa] on a contact between concrete and soil;

it depends on the subsoil cohesion c

k

; assume a

k

= η

c

⋅c

k

.

Both reduction factors η ≤ 1 are not an additional safety reserve, but they take into account construction

technologies of the slab and possible material weakening of the soil. Concreting directly on natural soil provides a contact strength such as the strength of the soil itself ( η = 1.0), but this does not always have to be at the interface of structural concrete and lean concrete, especially for pre-cast elements with smooth surfaces.

When the wall is moved, more or less horizontally, there is usually an destabilizing effect of soil shearing deformation:

• soil loosening may occur if it is very dense and this reduces the angle ϕ ; most often η

ϕ

= 0.8 ÷ 1.0,

• as a rule, soil cohesion decreases, because its permanent value is less than the peak one; Polish National Code PN-83/B-03010 recommends η

c

= 0.2 ÷ 0.5, while EC7-1 recommends even η

c

= 0.0 in some situations;

there may also be a weakening of the thin top-layer of cohesive soil due to environmental impacts

1

; if there are no special difficulties, the value of η

c

= 0.5 seems appropriate.

There exist, however, some situations when η

c

= 1.0 is justified, i.e. when just the cohesion – not the adhesion – is more adequate. This is the case of a foundation slab with an anchoring element, Fig.2.

It is believed that the eccentricity e

B

of the resultant load W

k

on the founding level reduces the resistance against sliding and hence B’= B - 2e

B

. Certainly, it is true, but it's not like that.

The Meyerhof condition B’= B - 2e

B

was taken from the GEO for bearing capacity. The basic Terzaghi formula concerns foundations loaded centrally, without an eccentricity; if the load is on the eccentricity e

B

, but the width of the foundation B is cut exactly by 2e

B

, then the same load works in the middle of the reduced foundation, therefore the Terzaghi solution can be used.

It's not clear how this relates to the sliding.

A small eccentricity e

B

does not reduce the adhesion resistance, for a larger eccentricity e

B

, a gap occurs under foundation, which is also acceptable by EC7-1 (to some extent). The creation of such gap under the foundation, of course, does not affect the loading N

k

, while the adhesion resistance vanishes on the detached part of foundation.

Therefore,

1

Frost heaves or flooding with water for a longer time may require replacement of weakened cohesive soil.

ε

α <0

A B

Fig.2. Foundation slab with a sur anchor CD.

C E D

The slab itself is horizontal, but the spur anchor CD creates a rigid triangular wedge G

o

and de facto this system, as a whole, works like an inclined-base

foundation DE. Therefore, the sliding line DE is included

in the subsoil interior, this is not a contact between

subsoil and concrete, so the cohesion c

k

is adequate.

(3)

the value of B’ is determined by the gap width s, if it occurs, so:

5

7

= 5 … 9:; 0 < =

>

< 5/6

5

7

= 3 ∙ 5/2 − =

>

… 9:; 5/6 < =

>

< 5/3

Fig.3. There is: 3 ⋅ (B/2 - e

B

) = B – s = B’

B e

B

s

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