Ejemplo 1: IP sobre IP con GRE

Dry slope in sand

Referring to Fig. 4.1, assume that the stresses on the vertical sides of the slice are equal and opposite. Resolving forces parallel to the slope give the disturbing force W sin. Resolving forces normal to the slope give the resultant normal force W cos.

The factor of safety against sliding is then Fˆ Resisting force Disturbing forceˆ W cos tan 0 W sin ˆ tan0 tan

whence, for limiting equilibrium, maxˆ0

Fully submerged slope in sand

Referring to Fig. 4.2, consider a slice of width a and depth d. The buoyant weight of the slice is:

W0ˆ ad 0  0_{ˆ ÿ}

w†

The effective stress normal to the base of the slice is: 0

nˆ ad 0cos cos

a

The shear stress along the sloping base of the slice is:  ˆ ad 0

sincos a

The factor of safety against sliding may be derived as Fˆd

0_cos2 _tan0

d sin  cos  ; whence Fˆtan

0

tan

Note that the expression for factor of safety for the fully submerged slope is identical to that for a dry slope.

W

β

Fig. 4.1 De®nition sketch for a dry slope in sand

W′ a

d β

γ

Semi-in®nite slope: = 0 analysis

Referring to Fig. 4.3, consider a slice of width b and depth z. The disturb- ing force is zb sin . The restoring force is

su a cos

The factor of safety against sliding is therefore

Fˆ su

z cos  sin 

a

z _β

γ = bulk unit weight

Fig. 4.3 De®nition sketch for a semi-in®nite slope, ˆ 0 analysis

WORKED EXAMPLE

Consider a natural slope in over-consolidated London clay near limiting equilibrium.

Take zˆ 7 m, suˆ 100 kPa, ˆ 20 kN/m3, ˆ 108 which are typical values. Then Fˆ su z cos  sin  ˆ 100 20 7  cos 108 sin 108 ˆ 4:2

This is a nonsense for London clay slopes which can fail at this inclina- tion! This calculation serves to illustrate how inappropriate a  ˆ 0 analysis is when applied to natural slopes.

Now consider a natural slope in soft normally consolidated Norwegian clay.

Take zˆ 3 m, suˆ 10 kPa, ˆ 18 kN/m3, ˆ 158, as typical values for a slope about to fail. Now Fˆ 0:74.

This calculation again con®rms that a =0 analysis should not be applied to assess the stability of a natural slope whether in an over- consolidated ®ssured clay or a normally-consolidated intact clay.

 = 0 analysis for vertical cut: no tension crack

In Fig. 4.4, the disturbing force is 1=2 H2sin= tan . The restoring force is suH=sin .

Taking the critical angle of sliding as critˆ 458, then Hcˆ 4su= . Alternatively, assuming a slip circle, the corresponding result would be Hcˆ 3:85su= . These predictions would be unsafe in practice because real clay soils are weak in tension.

 = 0 analysis for vertical cut: with tension crack

Refer to Fig. 4.5. Assume the depth of tension crack Dˆ Hs=2, following Terzaghi (1943).

The disturbing force is 3

4Hc Hc

2  sin 458 The restoring force is

su Hc

2 

1 sin 458

At failure when Fˆ 1, the critical height is Hcˆ 2:67su= . This gives a more realistic prediction for the height of a temporary vertical cut in clay.

H

H

tan α

α

Fig. 4.4 De®nition sketch for a vertical bank, ˆ 0 analysis, no tension crack

Hc

45˚

D = Hc/2

γ = bulk unit weight s_u = undrained shear strength

Fig. 4.5 De®nition sketch for a vertical bank, ˆ 0 analysis, with tension crack

WORKED EXAMPLE

Consider the different approaches to analysing the short term stability of a vertical bank illustrated in Fig. 4.6.

Case (a) may be summarized as follows.

608: Disturbing ˆ1 2 10  5:774  20  sin 608ˆ 500 kN Restoring ˆ 60  11:55 ˆ 693 kN F ˆ 693=500 ˆ 1.30 458: Disturbing ˆ1 2 10  10  20  sin 458 ˆ 707 kN Restoring ˆ 14:14  60 ˆ 848 kN F ˆ 848/707 ˆ 1.2 308: Disturbing ˆ1 2 10  17:3  20  sin 308 ˆ 865 kN Restoring ˆ 60  20 = 1200 kN F ˆ 1200=865 ˆ 1.39

For case (b) we have: For a dry slope:

Disturbing ˆ 5  7:5  20  sin 458 ˆ 530 kN Restoring ˆ 7:07  60 ˆ 4.24 kN Fdry ˆ 424/530 ˆ 0.80 45_˚ 45˚ 30˚ 60˚ 10 m 10 m 10 m 10 m 5 m 5 m Qw 4·24 m (a) (b) (c) γ = 20 kN/m3 su = 60 kN/m2

Fig. 4.6 De®nition sketches for worked example (a) assuming different angles of slip planes, (b) taking a 458 slip plane and a water-®lled tension crack, (c) assuming slip plane is the quadrant of a circle

Semi-in®nite slope: e€ective stress analysis

Refer to Fig. 4.7. For ¯ow parallel to slope, an effective stress analysis gives the following.

Soil strength is sˆ c0‡0tan0. Shear stressˆ ˆ z sin …1=cos †

ˆ z sin  cos 

Normal stressˆ ˆ z cos …1=cos † ˆ z cos2

Pore water pressureˆ u ˆ mz wcos cos  ˆ mz wcos2

The normal effective stress on the base of the slice is: 0_{ˆ … ÿ m}

w†z cos2

Whence the factor of safety against sliding is Fˆc

0_{‡ … ÿ m}

w†z cos

2 _tan0

z sin  cos 

For a wet slope (tension crack ®lled with water)

Water force ˆ1

2 5  5  10 ˆ 125 kN

Extra disturbing ˆ 125  sin 458 ˆ 88.4 kN

Fwet ˆ 424=530 ‡ 88:4 ˆ 0.69

It can be seen that the ®lling of the tension crack with water has a signi®cant effect on stability.

For case (c) we have (noting the centroid of a quadrant of a circle radius R is located 4R=3 from the centre of the circle):

Area ˆ  102=4 ˆ 78.5 m2

Perimeter ˆ  10=2 ˆ 15.71 m

Taking moments gives:

Disturbing ˆ 78:5  20  4:24 ˆ 6656.8 kN/m

Restoring ˆ 15:71  60  10 ˆ 9424 kN/m

F ˆ 9424/6656.8 ˆ 1.42

In summary we have:

case (a) Fˆ 1.39 (608), 1.20 (458), 1.39 (308). case (b) Fˆ 0:80 (dry), 0.69 (wet)

If c0ˆ 0, the critical slope is given by tancˆ

… ÿ m w† tan0

If in addition mˆ 1, i.e. the water table is at the ground surface, then tancˆ

0

tan0

If ¯ow is horizontal, however, then uˆ wz and Fˆ 1ÿ w cos2   tan0 tan compared with Fˆ 1ÿ w   tan0 tan for parallel ¯ow and c0ˆ 0.

1 _β Seepage parallel to slope Z mZ U γw γz γz cos β cos β γz sin β 1

Fig. 4.7 De®nition sketch for a semi-in®nite slope with seepage parallel to the slope, effective stress analysis

Anisotropy of permeability

The effect of anisotropy with respect to permeability on the stability of a long slope in a frictional soil (i.e. c0ˆ 0), for a condition of seepage parallel to the slope surface and the water table at the ground surface may be assessed using the equation of Telling (1988):

Fˆ ‰1 ÿ … w= † sec2iŠ tan0= tan i WORKED EXAMPLE

Referring to Fig. 4.8, take ˆ 19 kN/m3, c0ˆ 2 kPa. Assume the slope is on the point of sliding, i.e. Fˆ 1. What is the value of the angle of shearing resistance with respect to effective stress,0?

Now the factor of safety against sliding is:

Fˆc

0_{‡ … ÿ m}

w†z cos2 tan 0

z sin  cos  ; whence

tan0ˆ 27:045 ÿ 2 69:838 ˆ 0:3586 0_{ˆ 19:78} If c0ˆ 0,0ˆ 21:28 and if z is in®nite, 0ˆ 24:28. 1 m 6 m 12_˚ u = γw × 6 × cos2 12˚

Fig. 4.8 De®nition sketch for a semi-in®nite slope with seepage parallel to the slope, worked example

wherein  ˆ 1=…1 ‡ k2tan2i†; k ˆ kH=kV p

, where kH is the horizontal permeability, and kVis the vertical permeability; i is the slope angle and w, and 0are as usually de®ned.

By way of illustration, the variation of F with k is as follows for iˆ 148 and0ˆ 278:

kˆ 1 Fˆ 1:02

kˆ 10 Fˆ 1:56

kˆ 20 Fˆ 1:73