D_Soils Level 2 Parts IV-V 2008

8/8/2019 D_Soils Level 2 Parts IV-V 2008

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DAC UniSA 2005 Rock & Soils

TOPIC SUMMARY ASSESSMENT

I. Soil stresses due toexternal loads

Simple analyses for uniformly loaded areas resting on deeplayer of homogeneous and isotropic material material.Stress bulb concept, Girouds chart, Newmarks chart.Exercises: Try 4.1 to 4.4

II. Consolidation 1D consolidation testing, consolidation parameters, pre-consolidation pressure, excess pwp, rate of averageconsolidation, isochrones, examples.Exercises: Try 9.4 and 10.1 to 10.4

Practical 1

III. Strength andstiffness

Direct shear test. Triaxial test with pwp measurement,drained and undrained tests, vane shear, typical strengths ofsands and claysExercises: Try 3.1 to 3.7

Practical 2

IV. Non-embeddedretaining walls

Rankine earth pressure states; horizontal backfill, influenceof water. Simple coulomb wedge analysis. Elements ofgravity wall design. Coulomb approach. Limit stateapproach.Exercises: Try 7.1 to 7.2.

V. Analysis of slopeswith pore water

pressures

Stability of an infinite slope. Taylors charts. Pore pressuresin embankment dams (A & B pore pressure parameters form only). Peak, critical state and ultimate strengths.Bishops simplified method. Managing pore water

pressures.Exercises: Try 5.1 to 5.7.

Practical 3

VII. Design of shallowfootings for bearingcapacity

Allowable and safe bearing pressure, Brinch-Hansengeneral bearing capacity equation for vertical or inclinedloading, eccentric loading, and with depth and shapecorrection; local shear failure, effect of water table,influence of soil layering.Exercises:

Quiz

VIII. Design of deepfootings Types of piles. Static analysis of vertically loaded piles inclay and in sand. Dynamic analysis. Group action.Exercises:

IX. Immediatesettlements

Ueshita & Meyerhof method. Layered soils. Field tests(SPT, CPT) and empirical approaches.Exercises:

X. Total settlement Skempton and Bjerrums consolidation settlementcorrection. Stress path testing for total settlement. Drains.Exercises:

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IV HORIZONTAL EARTH PRESSURE AND RETAINING WALLS

SMITH CHAPTER 7

Content:

Rankine earth pressure states; horizontal backfill, influence of water; simple coulomb wedgeanalysis; elements of gravity wall design; Coulombs limit state approachExercises: Try 7.1 to 7.2.

IV.1 INTRODUCTION

The horizontal stresses in a soil mass are not the same as the vertical stresses, as in a fluid.Structures sitting against soil must be designed to resist these horizontal stresses or pressures, so we

need to be able to estimate them. The Rankine earth pressure approach allows estimation of thesestresses under a range of conditions, but does have some limitations. The Coulomb approachprovides more general solutions, but requires a graphical approach to resolve the design force andhence pressure. Both methods are presented and then the design considerations for retaining wallsare discussed.Three earth pressure states are possible and can be highlighted by looking at the behaviour of soilfill behind a retaining wall that moves (refer Figure IV.2). If the wall stays completely still, the atrest pressure state is appropriate. However, most walls allow some movement as they are pushed bythe supported soil. Typically a small rotation or translation occurs relaxing the pressures behind thewall. It doesnt take much movement for the Active earth pressure state to be reached. If for somereason the wall was pushed into the soil fill, the soil would react increasing the resistance to

movement, just like a compressed spring. With considerable movement, the Passive earth pressurestate is reached.

K

Kp

Ka

Lateral Wall movement

Active

State Passive

StateKo

Figure IV.2 Earth pressure states as a function of soil movement

IV.2 RANKINE EARTH PRESSURES

IV.1.1 Earth Pressure at Rest

Consider the stresses on an element of soil at depth, z, as shown:

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Effective horizontal stress

o

'=H

'=Ko z'

where oK is the coefficient of earth

pressure at rest.

NOTE : u oo =

Figure IV.1 Stresses in a soil mass

For normally consolidated (NC) sands, Ko may be calculated from Jakys equation;

= sin1Ko

So, the denser a sand, the greater the angle of internal friction, and so the lesser is the value of Ko.For example, dense sand has a Ko of 0.35, while loose sand has a Ko of 0.6.Some other typical values for clay soils are given in Table IV.1.

Table IV.1 Typical values of Ko for clays

Soil K o

Norwegian NC clays 0.5 0.6London clay OCR = 3.5 1.0London clay OCR = 20 2.8

IV.2.2 Active Earth Pressure

Consider the case of a frictionless wall of height H, supporting a dry soil with a level soil surfacebehind it. If the wall is allowed to move away from the soil horizontally, the vertical stress wouldremain constant, but the horizontal stress would reduce until a minimum is reached. This lateral

effective stress, H , is called the Rankine Active pressure, which is a soil failure state that can be

defined by the equation:

AAzH K2cK =

where KA is the coefficient ofActive earth pressure.

NOTE :+

=

=

sin1

sin1

245tanK

2

A

For a friction angle of 30, KA is a third (0.333). However for a dense soil with a friction angle of

45, KA drops to 0.17.

3

z

h

Z


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In a uniform or homogeneous soil (or backfill), the effective horizontal stress distribution behind agravity retaining wall is as shown in Figure IV.3.

H

PA

(45o- /2)

zc

Tensionisneglected

A

Figure IV.3 Rankine active earth pressures in uniform soil

QUESTIONS:

a) The LHS of the diagram indicates a failure plane in the soil mass inclined at an angle of (45 -

( /2)). Why is this so? Justify your answer with consideration of Mohrs circle for the activepressure state.

b) What is the value of the ordinate of the pressure diagram at the base of the wall?

c) What is the value of depth, zc, from the surface to the depth at which the horizontal stress istheoretically zero?

NOTE: Since soil is weak in tension, cracks will usually develop to the theoretical depth of z c. So zcis commonly referred to as the depth of cracking orthe critical depth.

The resulting force or thrust at the back of the wall, PA, per metre length of wall can be found fromthe horizontal effective pressure distribution. This thrust is needed for design of the retaining wall.

PA = Height x (average pressure over this height)

( ) ( )( )AAZoHoA Kc2KzH21zH

21P ==

If the soil is below water, the pore water pressures must be considered separately.

IV.2.3 Passive Earth Pressure

If the wall is pushed into the soil, the horizontal stress would increase, until a maximum is reached.Considerable displacement is required to generate the maximum resistance in the soil. Thismaximum pressure is called the Rankine Passive pressure, which is the soil failure state defined

by the equation:

PPzH K2cK +=

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where PK is the coefficient of Passive earth pressure.

NOTE:

A

2

PK1

24 5t a nK =

+=

The passive pressure coefficient is simply the inverse of the active pressure coefficient.

The resulting force per metre of wall, PP, can be found from the distribution of horizontal stressshown below.

QUESTIONS:

a) Why is the failure plane in the soil mass now inclined at an angle of (45+ ( /2))?

b) What is the value of PP in the diagram?

Passive pressure does not involve a tension zone. Soil cohesion adds to the

horizontal presure. Passive pressure is unlikely to have much impact on thedesign of gravity retaining walls, but is an essential element of the design offootings and of embedded retaining walls such as driven sheetpile walls.

IV.3 STRESS STATES IN THE SOIL

Picture a simple retaining wall with a horizontal (non-sloping) backfill surface. The soil has

constant vertical stress, z, despite what happens to the wall. In Figure IV.5, a Mohrs circle is

given which assumes an initial horizontal stress, o (at rest pressure). The failure envelope for the

backfill soil is given. Failure can occur if o is decreased to A (Rankine Active pressure). Failurecan also occur if o is increased to P (Rankine Passive pressure). In this case, horizontal stress,

P, exceeds vertical stress, z, and so becomes the major principal stress.

5

H

PP

45o

+/2

H P Z PK c K= +2

H c= 2 Kp

Figure IV.4 Rankine passive earth pressures in uniform soil


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Consideration of these circles explains the difference in the orientations of the failure planesgenerated in the soils. The position of the pole changes between the two states. For the Active case,

the pole is at A (horizontal and minor principal stress), but for the Passive state, the pole is at P(horizontal and major principal stress). Although the angle between the horizontal and the failure

plane is (45+ /2) for the Active state, it becomes (45 - /2) for the Passive state.

+= tanc eakppeak

A 0 Z P

45 - /245 + /2

Figure IV.5 Rankine earth pressures as Mohrs' circles

IV.4 EFFECT OF PORE WATER PRESSURES ON WALL DESIGN

The thrust due to water behind a wall is calculated and added to the effective stresses calculatedby the previous analyses to give total horizontal stresses. So the pore water pressures due to highwater tables have a detrimentaleffect on retaining walls. Initially they may seem to be beneficial as

the effective lateral stress is smaller than the total lateral stress by Kau as 'H = Ka 'z. However,the water pressure acts equally in all directions (K = 1!) and so the total lateral pressure is:

H = Ka 'z + u

So effective vertical stress is used to get effective horizontal stress, but the designer must design forthe total lateral stress.

Thrusts can be almost doubled where high water tables are allowed to develop. Suitably designeddrains can reduce or eliminate the water pressures (Figure IV.6).

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Weep

holes

Granular zone or

geofabric drain

Granular zone or

geofabric drain

Figure IV.6 Drainage provisions for retaining walls

IV.5 EFFECT OF SURCHARGE LOADS ON WALL DESIGN

If a surcharge load is applied to the surface of a soil, it increases the thrust on the wall as follows;

IV.5.1 Uniform surcharge

This increases the vertical and hence the lateral stresses in the soil. It is good design practice toassume a small surcharge on the backfill surface (say 10 kPa) for all walls.

IV.5.2 Concentrated load (information only)

R

hQ

Q

222

5

2

hQ

zhxR

R

zxQ48.0

++=

Figure IV.7 The influence of a vertical point load, Q, on the soil surface

IV.5.3 Line Load (information only)

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z

x

R

hq

QL

22

5

2Lhq

zxR

RzxQ63.0

+=

Figure IV.8 The influence of a vertical line load, QL, on the soil surface

IV.6 EFFECT OF NON-UNIFORM SOIL CONDITIONS

The earth pressure coefficients are dependent on the effective angle of friction for the soil.Therefore the horizontal effective stress may not be the same at a soil boundaryit may differabove and below the boundary, despite the vertical effective stress being constant. Consider thefollowing hypothetical situation of a two-layered soil shown below.

Total stress Effective stress

Vertical stress distribution

pwp

c1 = 0, 1, 1

c2= 0, 2, 2

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pwp

PA1

PwPA2

PA3

z

c

Horizontal stress distribution

Stronger and

densersoil

PA1

PwPA2

PA3

Figure IV.9 Rankine earth pressure diagrams for a two-layered soil

IV.7 EFFECT OF WALL FRICTION

Friction (tan ), and for that matter adhesion, ca, can be developed between the material of thewalls and the backfill soil. The shear strength at the wall-soil interface can be represented by theCoulomb equation;

tanc na +=

This shear resistance can be of considerable benefit to gravity retaining walls.

For walls with a reliable angle of friction, , between the soil and the wall, the resulting force willact at an angle, , to the wall. The zone of failure will be as shown in Figure IV.10. is usually

less than the friction angle of the soil (e.g. tan 0.67(tan )). The coefficient of wall friction

will depend on the material of construction of the wall and the soil type and its condition.

The Rankine approach cannot deal with wall friction. It has been found that for small walls withrelatively smooth faces, the Rankine states, which ignore wall friction, have small errors. For high

walls ( 10 m) and rough faces, the errors can be large.

However in earthquake-susceptible regions, wall friction is often chosen to be ignored in design.

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Figure IV.10 Active and passive thrust directions with frictional walls

IV.8 COULOMB METHOD FOR ACTIVE AND PASSIVE THRUSTS

IV.8.1 Coulomb Method for Active Thrust

Rankine suits the case of no-friction between wall and soil, uniform surcharge and non-slopingbackfill. The Coulomb approach, which considers Limit Equilibrium, has generally widerapplicability. The likely failure mechanism is assumed and the force required to achieve stability, orequilibrium, is predicted. This force is the resultant of the lateral earth pressures and the shear stressalong the wall developed through friction, and indeed adhesion along the interface between the wallmaterial and the soil.

For the general case in cohesive soils, a trial wedge method is used as shown in Figure IV-11. A

series of trial wedge angles, , are used, the objective being to find the worst case (the angle which

yields the highest value of thrust). Thrust on the wall per unit length of wall, P, is found byequilibrating forces acting on and within the soil wedge. So for this purpose, P is taken as a reactionprovided by the wall in supporting the wedge, and is of the same magnitude as the thrust, but it actsin the opposite direction.

Forces per unit length of the wall include the weight of the wedge, W, which is simply the area ofthe wedge by the total unit weight of the soil, and the reaction, R, on the sliding surface, whichsupports the wedge weight. Force R is unknown but its direction is known. In a frictionless soil itacts normal to the sliding plane; however if the soil has an apparent angle of friction, the resultant

is offset from the normal by . The offset of the resultant reaction is due to the sliding,frictional shear resistance, Ntan , which acts upward along the failure plane, where N is the

normal component of R. Likewise the thrust against the wall, P, may be offset by the angle offriction between the wall surface and the backfill surface, . Since the wall in the Active statetends to move downwards, the shear support is upward and so P is inclined upwardly.

A force vector diagram is drawn for the set of forces for each trial wedge ( n) as shown to the rightof Figure IV.11. In this chosen case, the wedge is not simply triangular, as a potential tension crackhas been introduced to a depth of zo from the surface. The only two forces not already discussedthat are in the vector diagram arise from shear resistance along the sliding plane due to soilcohesion, cL, where L is the length of the sliding plane, and adhesive shear resistance along the

back of the wall, cah, where ca is the adhesion between the wall and the backfill, and h is the length

of wall over which the soil is in contact with the wall. Not all these forces may be needed.

10

PA

PP

Failureline

Failure

line


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The known forces and their directions are plotted to scale end to end. Although only the directionsof R and P are known, this is sufficient information to complete the vector diagram and to scale offthe value of the thrust for the trial wedge.

The trial wedge angle, , which gives the greatest thrust P, represents the inclination of the sliding

failure surface. Pmax may be found by plotting P against trial wedge angle, .

Then Pmax for this wedge angle is equivalent to the active thrust, Pa.

h P

zo

W

cL

R

L

W

R

cL

P

Force Polygon

cwh

cwh

Figure IV.11. Active pressure considerations for Coulomb method

Horizontal water forces in the cracks (U = 0.5 wzo2) could be added to the force polygon by thedesigner.

IV.8.2 Coulomb Method for Passive Resistance

The use of Coulomb's wedge method is not recommended for passive resistance determinationbecause it overestimates the resistance. This is due to the actual slip surface being curved (whichCoulomb recognized). Therefore a curved slip surface (log-spiral usually) has to be assumed toestimate the resistance. Some solutions are available in the literature, e.g. Terzaghi and Peck"SoilMechanics in Engineering Practice".

IV.9 RETAINING WALLS

A retaining wall is a permanent, relatively rigid structure that supports a mass of soil. It substitutesthe steep face of the wall for the natural slope of the earth surface, e.g. for use in highway orrailroad cuts.

IV.9.1 Types of Walls

The diagrams in Figure IV.12 show some of the types of walls used. Most of the examples rely onthe self-weight of the wall to resist active thrust. For instance gabion walls are made out ofrectangular wire cages filled with large rock fragments. The blocks are simply stacked up on top of

one another. Water readily drains from the backfill through the large void spaces in the gabions.Geofabric often needs to be placed between the wall and the soil to stop migration of fines from thesoil.

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Both the crib wall and the concrete cantilever wall use the dead weight of the soil to advantage.Crib walls are comprised of short inter-locking beams of concrete or timber, stacked to formrectangular hollow walls, which are subsequently filled with soil. The back structure of thecantilever wall captures some of the earth fill behind it, which will move with the wall.

Masonry Gravity walls

Gabion

gravity

wall

Crib gravity

wall

(concrete,

timber)

Reinforced

concrete

cantilever wall

(can be pre-cast,

and different

sections)

Figure IV.12 Types of shallow-seated retaining walls

IV.10 EXCAVATION BRACING (information only)

Excavation bracing is used as temporary supports for open cuts and trenches. Some of the types areshown below.

Possible

failure

shapeTrench

Strut

Steel

sheeting

WaleSupport systems may use

soldier beams (vertical)

and shuttering between

them instead of steel

sheeting

Figure IV.14 Braced excavations

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IV.10.1 Design Earth Pressure for Support Systems

Since pressures are very difficult to analyse and change during construction, empirical stressdistributions are used. For example, a simplified pressure distribution envelop is shown in FigureIV.15. Strut force is found as pressure by contributing area.

Stabilty No. N

0.25H

Sand Clay N =H / CuN = 2 No overstress

N = 4 May just stand

H N > 4 Bulges

0.25H

.65KaH .2H For N


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IV.12 EXAMPLES1. A wall 8 m high retains sand weighing 15 kN/m3 dry and 19 kN/m3 saturated. The water table

is permanently 3 m below the top of the wall. Assuming = = ' 36 :

a) Sketch the effective and total active earth pressure diagrams (assuming no capillary riseto effect pore pressures and soil densities above the water table).

b) Find the location and magnitude of the resultant of both pressure distributions.

c) How much reduction in overturning moment about the base of the wall would occur ifthe ground water level could be lowered to the base of the wall.

2. A retaining wall 8 m high supports a dry sand fill whose = 34 and = 19 kN/m

3

. Theback of the wall slopes at an angle of 75 to the horizontal, while the front is vertical, with thebase width 6 m.

The top of the fill rises at a gradient of 1:3 from the top of the wall.

a) Compute the resultant of the active earth pressure and its direction and line of action.

b) Find the location and magnitude of the reaction on the base of the wall given the wall is

concrete with = 24 kN/m3.

3. For the retaining wall shown in Figure 1, compute:

1) The horizontal pressure distribution along the line TB, assuming a Rankine activepressure state.

2) The active thrust on TB assuming a metre length of wall.

3) The height h such that the passive thrust along the line RA will be equal to the activethrust.

Given :

+

==sin1

sin1

K

1K

p

A

PPzP

AA'zA

Kc2K

Kc2K

=

=

14

3m

3m

Loose sand

c=0,=30o,=16 kN/m3

Dense sand

c=0,=36o,=18 kN/m3h

T

A

R

B


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V SLOPE STABILITY

SMITH CHAPTER 5

V.1 INTRODUCTION

Ground at an inclination is not necessarily stable. It may appear to the casual viewer that it is, but itmay be steadily creeping downslope. Take for example a spoil pile of dry sand loosely dumpedfrom a truck. What happens to the grains of sand when it is dumped?

The soil ravels until it comes to a stable gradient or slope angle. Can we predict this angle?If we look at the mechanics of this situation, i.e. the forces driving the soil and those holding thesoil together, we can begin to understand the concept of slope stability.

This topic requires an understanding of soil shear strength and the changes in shear strength due to

soil deformation or strain. It also requires understanding of pore water pressure development insoils and the role of effective stresses.

Natural slope instability in glacial clays Hummocky ground above a small farm dam

Failure scarp in glacial till Stabilized slope by the Torrens River

Figure V.1 Slope instability examples

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V.2 TYPES OF SLOPES

Slopes can be natural or they can be man-made (artificial). Natural slopes (refer Figure V.1) maybe worn or cut, as in a hillside and valley, or along a coastline or river. Natural slopes may be builtor deposited (screes, piedmont slopes, slide and flow slopes).

Artificial slopes arise from engineering works, cutting of roads into hillsides, elevating road and railwith embankments, forming dams, tips or spoil piles.

V.3 LONG SLIDING SURFACE OR PLANE TRANSLATIONAL SLIP

Examination of an infinite slope is a good introduction to slope stability in geomaterials. This caseis illustrated in Figure V.2. This type of analysis is most suited to clean granular soils or soils (androck masses) with laminations dipping parallel to the slope. The mechanics of the dashed linesection is depicted in the lower half of the Figure

WSlid

ingsurf

ace

resist

ance

h

b

Vertical

slice

WSlid

ingsurf

ace

resist

ance

h

b

Vertical

slice

W

WP

WN

C+WN

tan

Equilibrium of slice

WN = Wsin

Figure V.2 Infinite slope stability consideration

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The forces acting on the slice consist of:

W = weight of the slice of soil per unit length of slope

This weight can be resolved parallel to the slope, WP = Wsin

When resolved normal to the slope we get, WN = Wcos

WP drives the block along the potential sliding surface in the soil at the base of the block. The shearstrength along the sliding surface resists the sliding and the frictional component of this strength isaided by WN.

So, provided the underlying soil is frictional, it can be said that the weight of the block generatesthe instability, but also assists in preventing sliding.

The concept of stability can now be introduced:

Stability is guaranteed IF WP (F + C)

where F = frictional resistance = WNtan 'and C = resistance due to apparent soil cohesion

The concept of a factor of safety can be introduced:

FoS =actiondisturbing

actionrestoring

The action is usually a force or moment; in the case of a translational slide, it is a force.

V.3.1 Case 1:Define the factor of safety of a purely frictional soil (clean sand or gravel)

C = 0 = c (no cohesion)F = WN tan

But WN = Wcos

Therefore F = Wcos .tan

Also WP = Wsin

Therefore FoS =PW

F

Or FoS =

sinW

tancosW

FoS =

tan

tan

Therefore, the point of instability occurs when FoS = 1 or when = tantan

So we now have another concept termed the natural angle of repose.

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Dry granular material will come to rest at a slope angle equal to the angle of friction of the soil.

For loosely dumped dry sand, 30

V.3.2 Case 2:

Cohesionless Soil with water flow (seepage) down the slopec = 0 = c (no cohesion)If the top flow line (or phreatic line) is above the sliding zone, then a pore force, U, due to the pore

pressure, u, at the base of the sliding soil must be introduced.

If the vertical height of the sliding soil is h and the phreatic line is coincident with the ground

surface, then u = wh

Taking the breadth of the block to be b, u acts over the inclined length, l = b/(cos )

Thereforecos

bhU w=

which suggests that only half the level of stability exists for this situation when compared to a dryslope.

Effective normal force becomes WN - U = Wcos -cos

bh w

Or (WN U) = ( hb.cos - cos

bh w

)

So (WN U) = ( - w)hb (cos - cos 1

)

Or (WN U) = ( hb) (cos - cos 1

)

But F = (WN- U)tan

Or F = ( hb) (cos -cos

1)tan

And WP = Wsin = ( hb)sin

Now FoS =PW

F

So FoS =( )

( )

cossin

tancos cos2

V.3.3 Stable Slope Angles in Cohesionless Soils

a) For dry soils and a factor of safety on stability of 1.3:

soil strength slope angle = 30 to 40 = 24 to 30.5

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However, if seepage of water occurs down slope and phreatic line (the top flow line) is coincident

with the surface, then reduces to just 12.5 to 18, which is about one half the slope angle for adry soil!

V.4 CIRCULAR SLIDING FAILURES

These types of failures (refer Figure V.3) are particularly relevant to short term deformations of

slopes in saturated cohesive soils with little or no friction ( u = 0or undrained analysis).

In a homogeneous, isotropic soil of semi-infinite depth, the analysis is quite simple. The factor ofsafety is based on moments rather than forces, and the moments are taken about the centre of thecircle.

The big question is where is the sliding surface likely to develop? So we have to guess where thesurface may be and trial hundreds, if not thousands, of plausible circular slip surfaces. The one withthe lowest factor of safety is regarded as the critical slip surface, and hopefully there isnt one wehave missed with a lower factor of safety.

Instead of estimating the total weight, W, of the moving soil each time, the soil mass isconveniently broken up into vertical slices of width b and the factor of safety is then deduced as:

=

sinW

secbcFoS u

where W is now the weight of a slice

b is the breadth of the sliceand is the mean inclination of the sliding surface within the slice to the horizontal

slope

toe of slope

crest of

slope

sliding

surface

sliding

surface

centre of

circle

centre of

circle

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centre of

circle

A potential

sliding

surface

A potential

sliding

surface 13

5

7

2

4

6

13

5

7

2

4

6

Figure V.3 Circular slope instability and vertical slices for slope stability analysis

General warning: very few soils are truly homogeneous. Often soil profiles contain thin, weakseams of material, along which slides are likely to develop.

IV.1.2 Design considerations

a) A tension crack can develop at the top of the slope and the depth of the crack can be estimatedas:

= uc

c2z .

b) This same crack can fill up with water and so a hydrostatic force (acting horizontally) can beintroduced which adds to the potential instability;

2

z

P

2cw

W

=

c) Any external forces or pressures must be considered.

V.5 UNDRAINED ANALYSIS STABILITY CHARTS - Taylor (1937)

( u = 0 analysis)The charted solution by Taylor (refer handout) provides the stability number, Ns, for variouscombinations of slope angle and relative depth, D, of top of slope to an underlying rigid base. The

soils is assumed to be isotropic, homogeneous and non-frictional, and the slip failure is assumed tobe circular.

Use of the chart requires the following definitions:

Relative depth, D:H

DH

slopeofheight

boundarytodepth=

The stability number is defined as

=H

1

F

cN us

where F is the factor of safety.

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V.6 EFFECTIVE STRESS ANALYSES OF SLOPE STABILITY

Where pore water pressures need to be considered or where the soil shear strength has a frictionalcomponent, the simple methods previously discussed become inadequate. Take the example of a

circular slip in a c- soil. In the previous cu analysis, we were able to adopt a single value of shear

resistance along the entire slip surface. With a frictional component, the shear resistance variesalong the slip surface length as the normal force varies markedly. At the entry and exit points on theslope, the frictional resistance is almost zero as there is little weight from the soil above.

Likewise, seepage analysis of water movement will invariably indicate that the pore water pressurealong the sliding surface varies.

So in each of these cases it is best to divide the slope up into vertical slices, as illustrated in FigureV.3. There have been a number of people who have approached the method of slices in differentways; in order to provide a solution to the mechanics of the case, some approximations must bemade. Bishops simplified method of slices (1955) is regarded as fairly reliable and robust(Fellenius method isnt!).

IV.1.3 Bishops Simplified Method of Slices

Each vertical slice has a normal and shear force (E and X) acting on its side boundaries. TheFellenius method assumes that the resultant of all 4 forces (two per side) is zero. In contrast, Bishopassumed that the resultant of the side forces on the slice simply does not have a vertical component.

Therefore to obtain a solution to the Bishop problem, all vertical components of the forces acting

on the sliding surface are summed. Resisting forces due to the shear strength of the soil arereduced by an initial assumed value of the Factor of Safety (F1) for the slope. Since the sum of the

vertical forces must be zero, the effective normal force1 (N = N U = N - ul) acting on the slidingsurface may be found in terms of the Factor of Safety, F1.

Moment equilibrium demands that restoring moments = disturbing moments and for design it mustbe greater by a desired factor of safety:

( )

+=

s i nW

t a nNlcF o S

All terms are as previously defined, except l, which is the length of the sliding surface at the base ofa slice.

By substituting for N , a final estimate of the FoS may be made (F 2). If a value of F1 can be foundthat matches the value of F2, then the analysis for the potential sliding surface is acceptable. Aniterative process is usually required to make this happen. Then, other sliding surfaces may besimilarly analysed to find the lowest FoS and therefore the most likely or critical failuresurface.

1 N' is just the soil reaction normal to the sliding surface, N, less the reducing force due to the pore water pressure (u)against the surface.

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IV.1.4 XSLOPE (University of Sydney) Bishops simplified method of slices

Modern day computers allow thousands of slope stability analyses to be performed on a particularslope problem to define the critical sliding surface in seconds (the one with the lowest factor ofsafety).

Still we must understand what the program assumes, what inputs it needs and in particular the mostappropriate choice of shear strength parameters for the soil. The programs also permit the user tofind how sensitive the analysis is to the range of material properties possible for the soil profile.

You will be given an exercise to gain some familiarity with XSLOPE and possibly GALENA in theprac periods after the term break. Figure V.4 presents some of the screens you will see when inXSLOPE:

a) The geometry

b) Soil properties

Figure V.4 XSlope stability program examples of input/output

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c) Analysis options

d) Results for the most critical surfaces

Figure V.4 XSlope stability program examples continued

V.7 RECOMMENDATIONS FOR SLOPES IN CLAY

1. Intact stiff clays: Failures have occurred in slopes of OC clay, well after the dissipation ofexcess pwp should have occurred (drained analysis, c and ). Backanalyses of these failures hasindicated that the strength parameters are much less than the peak strength parameters. The reasonsfor the softening of stiff clays include:

a. Normal stress at the ends of the slide is low, shear strength is low

overstressing (Bishop)

strain softening of OC clays

dilation opens up voids and allows more water in softening the soil further

b. Opening-up of closed fissures, with stress reliefwhen a slope is cut from a hillside

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Slopes in these soils are best designed with critical state (fully softened, or constant volume, cv, ccv) parameters.

2. Fissured clays: The influence of fissures on the clay mass strength will depend on:

nature of fissuring

orientation continuity

spacing

In extreme cases, the shear strength along the fissure may need to be determined and applied (it

should be close to the residual strength ( r, cr) for intact soil as fissures are just sheared surfaces

within the soil). However, usually the soil mass strength tends to the fully softened value ( cv, ccv) as the orientation of the fissures is unlikely to line up precisely with the sliding surface. This isthe normal design strategy for fissured clays.

3. Pre-existing slides in clayey soils or rock with clay-filled joints, should be evaluated withresidual strength parameters. Where residual behaviour is appropriate, either the strength may beback-figured from the field slip (needs geometry of sliding surface), or from large displacement

shear box tests (e.g. ring shear, or cyclic shear box).

All shear strength parameters must be evaluated for the correct

stress levels.

IV.1.5 Estimates of Shear Strength Parameters from

Correlations with Soil Indices

There are a number of broad first estimates available from theliterature. Some of these are indicated below.

1. from relationship with plastic index(after Kenney 1959) refer the handout

Clays with PIs of 50%, > 20!

2. Residual friction angle

ult is difficult to assess in the laboratory. An estimate of ult ( referred to as r)may be basedon percent clay in the soil, after Skempton (1964).

Fell & Jeffrey proposed a relationship between ult and dominant clay mineral:Kaolin 15 Illite 10 Montmorillonite 5

Summary: use

, cCompacted soils earthembankments

r, cr

Natural

Documents

D_Soils Level 2 Parts IV-V 2008