CE5108 Part 1 - Introduction and Earth Pressure Theories

CE5108 Earth Retaining Structures

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Part 1 – Introduction and Earth Pressure Theories

Learning Objectives 1. Introduction to Different Types of Earth Retaining Structures 2. At-rest Earth Pressures 3. Active and Passive Earth Pressures 4. Rankine Earth Pressure Theory 5. Coulomb Earth Pressure Theory 6. Culmann’s Method of Solution 7. Factors Affecting Earth Pressures


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1. INTRODUCTION

In situations where soils and geologic materials as well as materials with

similar engineering characteristics e.g. granular, powder or particulate materials,

need to be retained at a larger height or steeper slope than they would naturally

assume, some form of retaining structures will be needed. The main function of the

retaining structure is to withstand the earth pressure which the retained material

would exert onto it.

A variety of retaining structures have been used in practice. Commonly used

retaining structures can be classified into several broad categories as follows:


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1. Gravity Retaining Walls – These are retaining walls that rely solely on the self

weight of the structure and wall geometry for stability. These retaining walls are

analysed using classical retaining wall mechanics. That is, the stabilizing forces on

the structure is greater than the disturbing forces trying to destabilise the retaining

wall. The gravity retaining walls are generally constructed of discrete segmental

units. For example concrete or timber crib walls use interlocking crib units to bind

the infill material into a coherent structure to increase mass and stability. Solid

blockwork uses solid block units. Segmental blockwork uses segmental blockwork

units as a facing system. Gabion retaining walls use rock filled baskets forming

discrete units making up the wall structure. Rubber tyre retaining walls use infilled

rubber tyres in the same way as gabions. Figure 1.1 shows some of the gravity

retaining walls which have been used.

2. Flexural Retaining Walls – Flexural retaining walls are walls that rely on the

flexural action of the structure, coupled with cantilever anchorage of the

foundation soil alone, or together with some props, to support the retained

material. They may be broadly classified into two categories depending upon their

flexural rigidity. Walls with lower flexural rigidity consist commonly of sheet piles

and soldier piles. Sheet pile walls (Figure 1.2) consist of continuously interlocked

corrugated sheet pile segments adequately embedded into the soil to resist

horizontal pressures, primarily from earth and water loads. The wall is typically

constructed by driving individual sheet piles into the soil. The wall obtains its

stability from the resistance it develops against the soil it is being driven into.

Figure 1.1 Examples of Gravity Retaining Walls


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Another type of commonly used flexible wall system is soldier piles, which often

consists of steel I-beams driven into the ground at fairly close intervals of between

1 to 2 m. The intervening exposed face of the earth is then normally covered up by

timber boards or planks known as timber lagging (shown on Figure 1.3). In some

cases, concrete in-fill may also be used.

Figure 1.2 Sheet Pile Walls


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More rigid flexural wall systems include diaphragm walls, contiguous bored pile and

secant pile walls (Figure 1.4).

Figure 1.3 Soldier Pile Walls

Figure 4 Diaphragm Wall


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3. Reinforced Soil or Geosynthetics Retaining Walls - Retaining walls that consist

of a front facing system anchored to a composite block of soil and geosynthetic

reinforcement. The reinforced block increases the mass and therefore stability of

the structure. The front facing system can consist of segmental blockwork,

segmental rubber tyre or gabions for example. The reinforcement is generally

connected to the rear of the front facing system using a mechanical connection,

pin or key, or relies on friction between courses. The basic principles of a

reinforced soil wall is similar to those of gravity retaining walls, that is, they make

use of the self-weight of the reinforced soil portion to generate stabilizing forces on

the wall. The main difference between a reinforced concrete and reinforced soil

wall is that the internal integrity of the former is a matter of structural design

whereas the internal integrity of the latter is a matter of geotechnical design. An

example of a reinforced earth wall is shown on Figure 1.5.

Figure 4 (continued) Diaphragm, CBP and Secant Pile Walls


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Figure 1.5 Reinforced Earth Wall


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2 EARTH PRESSURE THEORIES

The significance of earth pressure theory to the design of retaining structures

comes about because the main function of a retaining structure is to retain soil

which, if unsupported, would collapse. In order to do so, the retaining structure has

to withstand the earth pressure (mainly lateral) imposed by the soil mass. For this

reason, the lateral earth pressure from the soil often constitutes a major part of the

loading on a retaining structure. In the study of earth pressure, the main focus is

often the lateral earth pressure since this is often the more difficult component to

evaluate; the vertical earth pressure is often (correctly or incorrectly) expressed as a

product of the unit weight of the overlying soil and the depth of the point in question.

In earth pressure theory, three states of earth pressure are often examined. They are

1. A reference state of earth pressure, which is normally taken to be the earth

pressure of the “undisturbed” soil mass (i.e. prior to construction work). This is

often known as the at-rest earth pressure.

2. Two limiting states of earth pressure, which defines two extreme states of failure

of a soil mass. These are known as the active and passive earth pressures.

2.1 At-rest earth pressure

The at-rest earth pressure is actually quite difficult to quantify accurately

because it is heavily dependent upon the soil type as well as the geological and

stress histories of the soil. For this reason, there remains a high degree of

empiricism in the determination of the at-rest earth pressure of a soil. Consider an

element of soil located at a depth z below a level ground surface in uniform soil

conditions, as shown in Figure 2.1.

γd = σ’v + u

σh = σ’h + u

Figure 2.1 State of Stress Below a Level Ground


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As Figure 2.1 shows, the total vertical stress is computed from the self-weight

of the overburden soil, plus any pre-existing ground surface loading. If the soil is fully consolidated and the groundwater table is not changing, then there is no

excess pore pressure and the pore pressure is often computed as the hydrostatic

pore pressure, that is

u = γw dw (2.1)

in which γw is the unit weight of water and dw the depth below groundwater table.

In a non-hydrostatic situation, the pore pressure has to be separately

determined. Whatever the case, the pore pressure is often determinable from the

pore pressure status or groundwater conditions. Once this is done, σv’ is known, the

main issue that remains is the determination of σh’. In other words, the main problem in at-rest earth pressure is often the determination of the effective lateral stress. Because of this, the effective lateral stress σh’ is often expressed as a

proportion of the effective vertical stress σv’ through an at-rest earth pressure

coefficient Ko, which is defined as

Ko = σh' / σv’ (2.2)

It should be emphasized that Ko is almost invariably defined as a ratio of the effective (not total) stresses.

We mentioned earlier that Ko is a function of the soil type and geologic or

stress history. The effects of soil type can be isolated by examining the Ko of a

normally consolidated soil (or a soil which has lost all memory of its previous stress

history), or Knc. In such a situation, Knc should be only a function of the soil type.

If we think of a soil as a elastic material, then by considering the stress state

of an elastic material subjected to 1-dimensional constrained compression, we can

show that

'1'K o ν−ν

= (2.3)

in which ν’ is the effective Poisson’s ratio and ranges from 0.25 to 0.35. However,

this tends to underestimate the Ko of real soils. This is because normally


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consolidated soils are in a plastic rather than elastic state of compression. This is

evident from comparing the typical stress-strain curve of an elasto-plastic material

and the e vs log p’ curve of a consolidating soil (Figure 2.2). Overconsolidated soils

are elastic due to unloading but they have even higher Ko than normally consolidated

soil. For this reason, Eq. 2.3 is seldom used.

The most commonly used relationship for determining Knc is that due to Jaky, which

states that

( )( )'sin1

'sin1'sin1K 3

2

nc φ+

φ+φ−= (2.4)

where φ’ is the effective (not total) angle of friction of the soil. This is very

commonly simplified into its approximate form

'sin1Knc φ−≈ (2.5)

There is evidence that this approximate form is sufficiently accurate for most

engineering purposes according to available experimental data (e.g. Wroth 1975,

Mayne & Kulhawy 1982). Alternative relationships have also been proposed by

Brooker and Ireland

'sin95.0Knc φ−≈ (2.6)

Figure 2.2 Idealized e vs log p’ curve for soils


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and Alpan (1967)

)(I log 233.019.0K pnc +≈ (2.7)

in which Ip is the plasticity index.

As mentioned earlier, Ko also depends upon geologic and stress history. In

the estimation of Ko, the geologic or stress history of a soil is normally represented

approximately through its over-consolidation ratio (OCR). The Ko-value for over-

consolidated soils is usually higher than that for normally consolidated soils.

Numerous relationships have been proposed for estimating Ko for over-consolidated

soils, most of these related the Ko of the over-consolidated soil to that of the same

soil in its normally consolidated state i.e. Knc. One of these is by Alpan (1967), who

suggested that

nnco OCRKK ≈ (2.8)

in which n is an exponent whose value depends upon the soil type. For clays, Alpan

proposes that n can be related to the plasticity index Ip via

281/-Ip10 x 54.0n = (2.9)

For sand, Alpan proposes that n can be related to the effective angle of friction via

Figure 2.3.

Figure 2.3 Relation of n with Friction Angle


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Wroth (1972) also proposed two semi-empirical relationships for estimating Ko

from Knc. For OCR up to 5, Wroth (1972) proposed that

( )1OCR'1

'K OCRK nco −ν−

ν−= (2.10)

This relationship is based on the assumption that the unloading process which

resulted in the over-consolidated state is an elastic process. For higher OCR, Wroth

(1972) proposed that

( ) ( ) ( )⎥⎦

⎤⎢⎣

⎡++

=⎟⎟⎠

⎞⎜⎜⎝

⎛+−

−+−

o

nc

o

o

nc

nc

K21K21 OCRln

K21K13

K21K13m (2.11)

in which m = 0.022875 PI + 1.22 (2.12)

Finally, Parry also proposes an empirical relationship for Ko as follows:

'nco OCRKK φ= (2.13)

in which φ’ is expressed in radians.

2.2 Active Earth Pressure

As mentioned above, the at-rest earth pressure is the earth pressure which

would exist prior to disturbance to the soil. Thus, if a wall can be constructed with no

deformation to the retained soil mass, the earth pressure which would act on the wall

is given by the at-rest earth pressure. This is clearly impossible since all retaining

structures move. As illustrated in Figure 2.4, the resulting changes to the earth

pressure depend upon how the presence and loading from the wall deforms the

retained soil. If the retaining wall allows the soil face to move away from the retained

soil mass, the resulting deformation will generate stresses which will try to hold up

the soil mass, thereby reducing the earth pressure on the retaining wall. Thus, any

movement of the soil from the retained side will result in a reduction in earth

pressure from the at-rest value, resulting in what is known as an active condition.

Conversely, any movement of the soil into the retained side will result in an increase

in earth pressure from the at-rest value, resulting in a passive condition.


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In most soils other than very dense sand, the earth pressure in an active state

(hereafter termed as the active earth pressure) tends to decrease monotonically with

movement away from the soil. In other words, up to a point, one can trade earth

pressure for deformation. By allowing deformation to occur, the earth pressure acting

on the wall can be reduced. However, this is only so up to a point, since the active

earth pressure reaches a steady value (usually minimum) beyond a certain level of

deformation. This is the fully active condition, in which the strength of the soil is fully

mobilized in helping to hold up its own self-weight and that generated by any loading

on the retained ground surface. At this stage, the soil mass is at ultimate failure and

any remaining earth pressure (also known as the active earth pressure) will have to

be resisted by the retaining structure.

In reality, the active earth pressure may not be easily estimated exactly since

real problems often have complicated geometries and soil behaviour tends to be

Figure 2.4 Development of Active and Passive Earth Pressures with Wall Movement


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highly complex. However, by making certain simplifying assumptions regarding the

behaviour of soils and the constraints on the problem, the active earth pressure can

be estimated approximately. The first assumption to make is to assume that the soil

behaves in an elastic-perfectly plastic manner, as indicated in Figure 2.5. In reality,

the stress-strain curve of soil is often a curve. However, in order to facilitate

estimation, we shall assume that the stress-strain curve is characterized by an

elastic portion (shown as a linear sloping portion) and a perfectly plastic portion,

along which no further increase in stress with further straining is possible. Defined

loosely, the term “perfectly plastic” refers to the fact that the stress-strain curve is

horizontal and no further increase in stress is possible.

We now consider an idealized scenario in which a gravity retaining wall

(shown in Figure 2.6 as a trapezoidal wedge) is progressively moved away from the

retained soil, and we seek to estimate the active earth pressure in the soil. We also

assume that the retained soil is deforming in an undrained manner and its shear

strength is characterized by the undrained shear strength cu, the undrained angle of

friction φu being 0. We realize that the stress state in the retained soil is rather

complicated since the stresses must necessarily change as we move away from the

retaining wall in the direction indicated by the arrow marked A. Indeed, if the distance

from the wall is sufficiently large, we should not feel the effect of the wall movement

at all. Furthermore, the portion of retained soil above the base of the wall must also

interact with that below the base. All these complicate the stress situation

significantly.

τ

γ

Idealized elastic- perfectly plastic behavior

typical nonlinear response

τ

γ

Idealized elastic- perfectly plastic behavior

typical nonlinear response

Figure 2.5 Elastic-perfectly Plastic Stress-Strain Curve


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In order to simplify the problem, we can introduce a working model which

approximates the actual problems, but with some constraints relaxed so as to

simplify the stress situation. One way of doing this is shown in Figure 2.7. We first

isolate a block of soil (marked X) located just behind the wall. We then assume the

following:

1. The interface between the wall and the soil is perfectly smooth, as represented

by the black rollers on the wall-soil interface. This is a conservative assumption

since the friction between the wall and soil helps to hold up the soil block X and

therefore reduce the lateral earth pressure on the wall.

2. We also assume that the far side boundary of the block X is also perfectly

smooth. This is also a conservative for the same reason as stated in (1).

3. We also assume that the base of the block X is also perfectly smooth and can

offer no resistance to the soil moving towards the wall. This is again a

conservative assumption.

Figure 2.6 Idealized Gravity Retaining Wall

Figure 2.7 Approximate Working Model to the Wall Problem


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We now allow the wall to move forward and develop the fully active condition,

and we try to calculate the active earth pressure arising from block X. This is an

approximate but conservative (because of the above assumptions) estimate of the

actual active earth pressure. We note that every vertical section within the block X is

identical to one another. Furthermore, the two vertical boundaries of block X is

perfectly smooth and can therefore develop no shear stress. For this reason, shear

stresses cannot develop along any vertical section of the block, that is τhv = 0. As

shown on Figure 2.8, complementarity of shear stresses must necessarily require

that τvh = 0. In other words, the vertical and horizontal normal stresses at all points

within the block are also the principal stresses. Since τhv = 0, the vertical normal

stresses at all points within the block can be computed based on the weight of the

overburden soil, that is

σv = γ d (2.14)

The horizontal normal stress is the lateral earth pressure that we are seeking to

estimate.

As shown on Figure 2.8, the relationship between the lateral earth pressure σh

and the vertical overburden stress σv can be expressed as

σh = σv – 2 cu = γ d – 2 cu (2.15)

Figure 2.8 Mohr Circle of a Cohesive Soil at Active Failure


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From Eq. 2.15, if we plot σh as a function of d, we obtain the lateral earth pressure

profile acting on the retaining structure as shown on Figure 2.9.

From Eq. 2.15, we see that when the depth d is less than a value dc , given by

γ= u

cc2d (2.16)

the lateral earth pressure becomes negative. As this is clearly impossible, what it

implies is that, above this depth, a soil with undrained shear strength cu will support

a vertical face. A vertical cut with a height less than dc will require no lateral support.

Furthermore, if a retaining wall supporting a vertical of height greater than dc is

allowed to move outwards so as to develop full active condition, the soil above a

depth dc will not follow the movement of the wall, so that a gap will develop in such a

situation. This gap is commonly termed a “tension crack”. This is a misnomer

because there is no tension. Moreover, in virtually all working load conditions, the

wall will not be allowed to move out infinitely, so that the gap will not develop to the

depth given by Eq. 2.16. The above solution constitutes part of what is commonly

known as Rankine’s Earth Pressure Theory.

We note that we have not taken into consideration the requirement that there

must be compatibility between points inside the block X. This is unnecessary since at

ultimate failure of a perfectly plastic material, all points will ultimate reach the same

state, regardless of strain. Eq. 2.15 represents an approximate solution to the

problem involving a gravity retaining wall with vertical back face retaining a uniform

Figure 2.9 Lateral Earth Pressure for Approximate Wall Problem


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cohesive soil, which was obtained by solving an approximate model of the actual

problem. Furthermore, the model was obtained by relaxing certain constraints on the

real problem, in this case the shear stresses at the boundaries of the soil block X is

set to zero. Thus, the model is weaker than the actual problem and our solution in

Eq. 2.15 is a “lower bound” solution to the actual problem. The concept of bounds to the actual solution is an important concept of plasticity theory and may be stated

simply as follows (without proof):

1. If a solution to the plastic failure or collapse load of a problem involving a perfectly

plastic material is sought which involves only consideration of equilibrium and the

failure criterion, then the solution will represent a lower bound of the actual

solution. In other words, if we represent the actual problem by a model which does

not require points to deform in a compatible fashion, the coupling between points

are weaken and we end up with a weaker model, so that the solution will be a

lower bound. This is a loose way of stating the Lower Bound Theorem.

2. If a solution to the collapse load of a problem involving a perfectly plastic material

is sought which involves a assumed failure mechanism and a “weak” equilibrium

of the failure mechanism (in the form of plastic work done and energy release in

collapse), then the solution will represent an upper bound of the actual solution. In

layman terms, if we specify or assume a failure mechanism, then our assumed

failure mechanism will be stronger and will lead us to a higher failure load than the

critical failure mechanism (which is the actual failure mechanism). This is a loose

way of stating the Upper Bound Theorem.

These two Extremum Principles play an important role in plasticity

computations since the actual failure load is often not easily determined. In such a

case, being able to bound the actual failure load gives us an idea of the actual

solution. In particular, the Lower Bound Theorem is useful in design since it

invariably yields a conservative failure load.

We can also consider variations to the case shown in Figure 2.8 by assuming

that the soil has no cohesion i.e. c’ = 0 but has a finite effective angle of friction φ’.

As shown in Figure 2.10, consideration of the Mohr’s Circle geometry leads to

'sin''''

hv

hv φ=σ+σσ−σ

(2.17)


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Rearranging the terms in Eq. 2.17 leads to

av

h K'sin1'sin1

''

=φ+φ−

=σσ

(2.18)

in which Ka is termed as the coefficient of active earth pressure.

A third variation can also be consider, in which the soil has finite c’ and φ’. As

shown in Figure 2.11, consideration of Mohr’s circle geometry leads to

'sin2

''' cos c2

'' hvhv φσ+σ

+φ=σ−σ

(2.19)

which leads to

'sin1'cos'c2

'sin1'sin1'' vh φ+

φ−

φ+φ−

σ=σ

'sin1'cos'c2'K va φ+φ

−σ= (2.20)

Figure 2.10 Mohr Circle of a Cohesionless Soil at Active Failure


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We note that

( ) a2

22

K'sin1'sin1

'sin1'sin1

'sin1'cos

=φ+φ−

=φ+φ−

=⎥⎦

⎤⎢⎣

⎡φ+

φ (2.21)

Therefore, Eq. 2.20 can be re-written as

avah K'c2'K' −σ=σ (2.23)

which is the standard form of active earth pressure equation (see e.g. BS8002).

2.3 Passive Earth Pressure

The passive earth pressure is developed when the wall, or part of it, is being

pushed into the retained soil. In a fully passive condition, the passive earth pressure

pushing the soil backwards has to overcome both the self-weight and shear strength

of the soil. For this reason, the passive earth pressure is often higher than the at-rest

earth pressure. While it is possible to repeat the three cases analysed above for the

active condition, this is unnecessary, since the last case is, in fact, the general case,

and encompasses the other two cases.

Figure 2.11 Mohr Circle of a c-φ’ Soil at Active Failure


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The governing equation for passive failure is also easily obtained by simply

swapping the terms σv’ and σh’ in Eqs. 2.19 and 2.20, which leads to

'sin1'cos'c2

'sin1'sin1'' hv φ+

φ−

φ+φ−

σ=σ (2.24)

so that

'cos'sin1'c2

'sin1'sin1'' vh φ

φ++

φ−φ+

σ=σ (2.25)

which can be expressed in its common form, as

pvph K'c2'K' +σ=σ (2.26)

where 'sin1'sin1Kp φ−φ+

= (2.27)

and is termed as the coefficient of passive earth pressure.

Graphically, the Mohr circle representation of the passive failure condition is

shown on Figure 2.12.

Figure 2.12 Mohr Circle of a c-φ’ Soil at Passive Failure


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In the special case where c’ = 0, Eq. 2.26 reduces to

'K' vph σ=σ (2.28)

Similarly, in the undrained case where φu = 0, we replace c’ by cu and φ’ by φu = 0, so

that Eq. 2.26 becomes

uvh c2+σ=σ (2.29)


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3 EARTH PRESSURE THEORIES PART II 3.1 Shortcomings of Rankine’s Earth Pressure Theory

The Rankine’s earth pressure theory discussed in the previous part has the

advantage that, under conditions of perfect plasticity, it computes a lower bound of

the collapse load. In practice, soil is not a perfectly plastic material, but we can relax

the requirements for this bounds property to hold to an ultimate failure condition

wherein the shear stresses in the soil neither increase nor decrease with shear

strain, as illustrated by points X in Figure 3.1. These states are known as the critical

states of the soil (e.g. BS8002: 1994).

The other shortcoming of the Rankine earth pressure theory is that such lower

bound solutions are only known for a very few number of highly idealized

geometries. For more complex geometries and problems where wall friction is

significant, the solutions are not easily found. Thus Rankine’s earth pressure theory

is, in practice, rather restricted in its usage.

3.2 Coulomb’s Earth Pressure Theory – Historical Perspective

An alternative approach to evaluating limiting earth pressures in a soil mass at

failure was proposed by Coulomb (1776). Charles Augustine de Coulomb (Figure

Figure 17 Mohr Circle of a c-φ’ Soil at Passive Failure


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3.2) was born in Angouleme, France on 14 June 1736. He graduated from the Ecole

du Genie (French for “School of Engineers”) at Mezieres in 1761 and then joined the

Corps du Genie (or Corps of Engineers) of the French Army with a rank of lieutenant

[Notice that, in those days, there was no distinct between military (now combat) and

civil engineers, as there is now]. In 1764, he was sent to Martinique in the West

Indies to bolster its defence (Martinique was under the sovereignty of France) by

building new fortifications at Fort Bourbon. These fortifications consisted of massive

gravity retaining walls with moat and obstacles in front (Figure 3.3). It was during this

period (1764 to 1772) that he did his seminal research work on shear strength of

soils and limit equilibrium method of retaining wall design.

Figure 3.2 Charles Augustine de Coulomb

Figure 3.3 Early modern fortifications


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Coulomb’s work on geostatics was written in his highly acclaimed essay Sur

une application des regles, de maximis a quelque problemes de statique, relatifs a

l’architecture, that was presented to the Academie des Sciences (Academy of

Science) in Paris in 1773, the objective of which, in Coulomb’s words is “…. To

determine, as far as a combination of mathematics and physics will permit, the

influence of friction and cohesion in some problems of statics…”. With this one single

article, Coulomb created the science of friction.

Coulomb had observed from his research, that failure in a soil mass tends to

occur along well-defined failure planes, as illustrated in Figure 3.4. Thus, the failure

of the soil structure as a whole can be analysed as a series of blocks sliding over

one another along the slip planes. Coulomb had two problems to solve. The first is:

what is the failure criterion along these slip planes? In other words, what contribute

to the resistance along these slip planes and how can these contributory factors be

accounted for mathematically.

Coulomb had learned from his experiments that the shear resistance of rock

consists essentially of two components. The first derives from the cementation and

bonding within the rock structure and is termed cohesion. The second derives from

the friction that exists between blocks of the rock structure. In the figure above, if the

stresses acting on a slip plane are resolved into a normal component σf and a shear

component τf, then at the point of failure, τf must be resisted by two components of

resistance. The first is a constant cohesive component, denoted by c whilst the

Figure 3.4 Failure Wedge in an Experiment


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second is a frictional component that is proportional to the normal stress σf. In other

words,

τf = c + σf tan φ (3.1)

in which φ is the angle of friction and tan φ is the coefficient of friction (commonly

denoted by µ). This is the Mohr-Coulomb failure criterion.

The second problem relates to the geometry of the failure wedges. Coulomb

noticed from his work on sandy soil that most of the failure surfaces are almost

planar. For this reason he assumed that the failure surfaces are planes. However

Coulomb’s approach can, in principle, be applied to any class of failure surfaces.

3.3 Example 1 – Cohesionless Soil, Smooth Wall, Level Ground, Active State

Coulomb’s earth pressure theory can be illustrated by a simple example

involving a gravity retaining wall retaining a sandy soil with c’ = 0 and angle of friction

φ’, as shown in Figure 3.5.

In Figure 3.5, the uniform sandy soil of unit weight γ’ is retained by a gravity

wall with a smooth vertical back face. One potential failure wedge is the triangular

wedge of soil abc, with interface ab sliding along the back face of the wall and bc

sliding along the underlying soil. The forces acting on the wedge are its weight W,

the force Q resulting from the wall reaction (whose magnitude is the sum total of the

lateral earth pressure), as well as the normal and shear forces N and F on the slip

Figure 3.5


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plane bc. We shall assume that the slip plane bc is inclined to the horizontal at an

angle θ.

Considering equilibrium in the direction of the force N leads to

N = W cos θ + Q sin θ (3.2)

Considering equilibrium in the direction of the force F leads to

F = W sin θ - Q cos θ (3.3)

Dividing Eq. 3.3 by Eq. 3.2 leads to

θ+θθ−θ

=sinQcosW

cosQsinWNF (3.4)

We note that, along the slip plane bc, limiting friction must be reached so that

'tanNF

φ= (3.5)

Combining Eqs. 3.4 and 3.5 leads to

'tansinQcosW

cosQsinWφ=

θ+θθ−θ (3.6)

which can be expressed as

Q = W tan (θ – φ’) (3.7)

We note that the length ac is h cot θ, so that the weight W of the wedge is given by

θγ= coth'W 221 (3.8)

Combining Eqs. 3.7 and 3.8 leads to

( )'-tan coth'Q 221 φθθγ=

( ) ( )'-tan tanh' 22

21 φθθ−γ= π (3.9)


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We have arbitrarily assumed that the slip plane is inclined at an angle θ to the

horizontal. There may be numerous potential slip planes inclined at all angles but

only one will lead to the failure of the soil mass; this is the critical slip plane, which is

also the slip plane which gives the highest value of Q. To evaluate this value of θ, we

equate the first derivative of Q to zero, that is

( ) ( ) ( ) ( )[ ] 0'sectan'tansech'ddQ 2

2222

21 =φ−θθ−+φ−θθ−−γ=

θππ (3.10)

which, upon solving, gives

2'45

2'

4φ

+°=φ

+π

=θ (3.11)

Substituting Eq. 3.11 into Eq. 3.9 gives the critical value of Q

⎟⎠⎞

⎜⎝⎛ φ

−π

γ=2'

4tanh'Q 22

21 (3.12)

However, this only gives the total force on the back of the wall, not the earth

pressure distribution. To find the latter, we note that there is an equally likely slip

plane occurring at a depth of h+δh and inclined at the same critical value of θ. The

total force on the back of the wall arising from the slightly lower slip plane (and

slightly larger wedge) is Q + δQ, where δQ is given by

h2'

4tanh'h

dhdQQ 2 δ⎟

⎠⎞

⎜⎝⎛ φ

−π

γ=δ=δ (3.13)

The pressure across the small increment in depth δh is thus given by

⎟⎠⎞

⎜⎝⎛ φ

−π

γ=σ=δδ

2'

4tanh''

hQ 2

ha (3.14)

We note that

( )'sin1'sin1

'sin1'sin1

'2

sin

'2

cos1

2'

4tan 2

2

2

2

2

φ+φ−

=φ−φ−

=⎟⎠⎞

⎜⎝⎛ φ−π

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ φ−π

−=⎟

⎠⎞

⎜⎝⎛ φ

−π (3.15)


1-29

Thus

vavha 'K'sin1'sin1'

'sin1'sin1h'' σ=

φ+φ−

σ=φ+φ−

γ=σ (3.16)

Hence, Coulomb’s earth pressure theory with the prescribed failure wedge gives the

same solution for active earth pressure as the Rankine’s earth pressure theory.

3.4 Coulomb’s Solution as an Optimistic (or Unsafe) Solution (?)

We have discussed earlier that Rankine’s earth pressure theory gives a lower

bound to the correct failure load. In other words, Rankine’s earth pressure theory

generally gives what can be considered as a safe or pessimistic solution. It is

tempting to draw a parallel between Coulomb earth pressure calculation and the

upper bound solution in plasticity. Firstly, the upper bound solution in plasticity

requires the prescription of a kinematically admissible failure mechanism, which in

the above calculations, can be considered to be potential failure wedge. Secondly,

an upper bound solution involves an energy balance calculation, which, in many

cases, can be equated to the consideration of force equilibrium. For instance, in the

example above, consideration of energy balance would lead to

F∆ = W∆ sin θ - Q∆ cos θ (3.17)

in which ∆ is the displacement along the slip plane. Cancelling ∆ out on both sides

leads to

F = W sin θ - Q cos θ (3.18)

which is identical with the equilibrium equation (Eq. 3.3) along the slip plane.

The situation is actually somewhat more complicated. Firstly, there is no

mathematical requirement that the failure wedge in Coulomb’s calculation must be

kinematically admissible. All that is needed is a close enough profile to the actual

slip plane. Secondly, in cases involving curvilinear slip planes, the equivalence

between force equilibrium and energy balance is often less obvious. Thirdly,

mathematically upper bound solution in plasticity requires that the material obeys the

“associated flow rule”, which, in the case of the Mohr-Coulomb criterion, would

require that the angle of friction is equal to the angle of dilation. This is not so in

most, if not all soils.


1-30

Coulomb’s calculations, on the other hand, have no such requirements.

Hence, Coulomb’s earth pressure calculations do not actually satisfy all the

mathematical premises needed to qualify it as a strict upper bound calculation.

Nonetheless, we can think of Coulomb’s calculation loosely as an optimistic or

unsafe solution. This is because in Coulomb’s calculation, the slip or rupture plane is

limited to a certain class of geometry, in the above case a plane surface, even

though we do optimize it with respective to the inclination θ. The actual slip surface

may not be plane. In that situation, our optimization will not produce the true slip

surface and the true critical failure conditions. In this sense, Coulomb’s calculation is

really a constrained optimization. Thus, if we have a soil which behaves in an

idealized elastic perfectly plastic manner, we would expect that, unless we have prior

knowledge of the geometry of the slip surface, Coulomb’s calculation is more likely

than not to miss the critical slip surface. For this reason, Coulomb’s calculation is

likely to yield an optimistic or unsafe solution. For this reason, it is very important that some form of optimization be undertaken, at least within the class of chosen slip surface, such as finding the critical value of θ as performed above. This should be considered an integral part of Coulomb’s calculations.

We now re-visit the fact that the solutions which were obtained by the

Rankine’s and Coulomb’s calculations are identical. Since Rankine’s method gives a

pessimistic or safe solution whereas Coulomb’s method gives an optimistic (or

unsafe) solution, the fact that both solutions are identical implies that, in this special

case, Rankines and Coulomb calculations produce the correct solution. This is not

unexpected, in the Rankine’s method, we have specified a stress field which is

triaxial (i.e. principal stresses act on vertical and horizontal planes, Figure 3.6); in

such a case, consideration of triaxial stress state through the Mohr circle will lead us

to the conclusion that this prescribed stress field is consistent with a slip plane which

is inclined at ( )2'45 φ+° to the horizontal (Figure 3.7); this is also the critical angle of

the slip plane used in Coulomb’s calculations. In more general scenarios, we may

not be so lucky.


1-31

Figure 3.6 Slip Plane Implied by Rankine’s Stress Field

Figure 3.7 Mohr Circle with Inclination of Slip Plane for Rankine’s Stress Field

45° + φ’/2


1-32

3.5 Example 2 - Cohesionless Soil, Smooth Wall, Level Ground, Passive State

We can repeat the same Coulomb procedure for a passive case as shown in

Figure 3.8.

In this case, the critical value of θ can be shown (mathematics not presented) to be

2'45

2'

4φ

−°=φ

−π

=θ (3.19)

⎟⎠⎞

⎜⎝⎛ φ

+π

γ=2'

4tanh'Q 22

21 (3.20)

and ⎟⎠⎞

⎜⎝⎛ φ

+π

γ=σ2'

4tanh'' 2

hp (3.21)

It can be shown that

( )'sin1'sin1

'sin1'sin1

'2

sin

'2

cos1

2'

4tan 2

2

2

2

2

φ−φ+

=φ−φ+

=⎟⎠⎞

⎜⎝⎛ φ+π

⎥⎦

⎤⎢⎣

⎡⎟⎠⎞

⎜⎝⎛ φ+π

−=⎟

⎠⎞

⎜⎝⎛ φ

+π (3.22)

Thus

vpvhp 'K'sin1'sin1'

'sin1'sin1h'' σ=

φ−φ+

σ=φ−φ+

γ=σ (3.23)

Figure 3.8


1-33

Once again, for the passive case, the Rankine’s and Coulomb’s earth

pressure calculations return the same solution. Although the Rankine’s and

Coulomb’s calculations for the above two cases give essentially the same solutions,

the solutions may not necessarily agree well with reality. In many instances,

comparison with field and laboratory studies have shown that passive earth pressure

theories tend to overestimate the actual pressure in cases where φ’ is high. Over the

years, there have been a number of attempts to try to get a better match to the

experimental results by using curvilinear surfaces. One of the factors which have

been identified as being at least a partial cause is the presence of significant wall

friction, the effects of which is aggravated in a passive state (c.f. active state). This

issue will be elaborated further upon below.

3.6 Cohesionless Soil, Non-Vertical Rough Back Wall, Sloping Ground, Active State

One of the advantages of the Coulomb’s method is its flexibility in dealing with

wall as well as sloping back wall and ground. This is illustrated by the case shown in

Figure 3.9.

The solution for this is given by

σ’a = Ka σv’ = Ka γ’h (3.24)

2h'KQ

2

aγ

= (3.25)

Figure 3.9


1-34

in which

( )

( ) ( ) ( )( )

2

a

sin'sin'sinsin

'- sin sec cK

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

β+αβ−φφ+δ

+δ+α

φαα= (3.26)

Note that Q does not act horizontally. The horizontal component Qh of Q is given by

2h'KQ

2

ahhγ

= (3.27)

in which

Kah = Ka sin (α + δ) (3.28)

This solution has sometimes been attributed to Muller-Breslau

(Tschebotarioff, 1951). However, it is now widely recognized that, even if Coulomb

had expressed the solution in the form of ratios rather than trigonometric function,

the improvement taken to convert ratios to trigonometric function is really rather too

trivial to merit any special mention.

3.7 Cohesionless Soil, Non-Vertical Rough Back Wall, Sloping Ground, Passive State

The equivalence of the above problem for passive failure is shown in Figure

3.10.

Figure 3.10

δQ


1-35

The solution is given by the following relationships:

σ’p = Kp σv’ = Kp γ’h (3.29)

2h'KQ

2

pγ

= (3.30)

in which

( )

( ) ( ) ( )( )

2

p

sin'sin'sinsin

' sin sec cK

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

β−αβ+φφ+δ

−δ−α

φ+αα= (3.31)

Note that Q does not act horizontally. The horizontal component Qh of Q is given by

2h'KQ

2

phhγ

= (3.32)

in which

Kph = Kp sin (α + δ) (3.33)

In Coulomb’s original computations, the failure wedges were assumed to be

bounded by plane surfaces. This assumption is not unduly unrealistic for active

failures where the actual failure surfaces are relatively flat curves. In passive

failures, the assumption of flat plane surfaces of failure often leads to larger errors in

computations. This is aggravated if the wall is rough. Partly for this reason, the

assumption of flat surfaces has often been found to overestimate the passive

pressure developed in field and model tests for φ’ exceeding 35°. Caquot and

Kerisel (1948) produced tables of earth pressure based on non-plane failure

surfaces. These design charts are shown in Figures 3.11 to 3.14.


1-36

Figure 3.11 Caquot and Kerisel Solution for Active Pressure with Level Ground

Figure 3.12 Caquot and Kerisel Solution for Passive Pressure with Level Ground


1-37

Figure 3.13 Caquot and Kerisel Solution for Active Pressure with Sloping Ground

Figure 3.14 Caquot and Kerisel Solution for Passive Pressure with Sloping Ground


1-38

In these figures, β is the angle of inclination of the backfill. When β = 0°,

Figures 3.13 and 3.14 reduces to Figures 3.11 and 3.12. Various other researchers

including Janbu (1957), Shields & Tomay (1973), Sokolovski (1960) and Rosenfarb

and Chen (1972) have developed numerical or iterative procedures for estimating

passive earth pressure coefficient. However, most of these are not directly

applicable without some computer software and are therefore not covered here.

3.8 Cohesive Soil, Smooth Vertical Back Wall, Horizontal Backfill Surface, Active State

Coulomb’s approach is less commonly applied to cohesive or c’-φ’ soils. This

is probably Coulomb’s solution did not include cohesion. This is not entirely

surprising since Coulomb’s application of his theory is in the construction of

fortifications, in which the soil is completely remoulded. Coulomb states three times

in his famous Essai, that newly remoulded soil has no cohesion since the remoulding

process would have destroyed any bonds which might have existed in the soil.

However, one can apply Coulomb’s approach to a cohesive soil by considering a

failure wedge as is shown in Figure 3.15.

Considering equilibrium in the direction of the force N leads to

N = W cos θ + Q sin θ (3.34)

Figure 3.15


1-39

Considering equilibrium in the direction of the force F leads to

F = W sin θ - Q cos θ (3.35)

We note that

θ

=sin

hcF u (3.36)

in which cu is the undrained shear strength, and that

θγ= coth'W 221 (3.8)

Substituting Eqs. 3.36 and 3.8 into Eq. 3.35 leads to

θ−θγ=θ

cosQcosh'21

sinhc 2u (3.37)

which gives

θ

−γ=θθ

−γ==2sinhc2h'

21

cossinhch'

21Q u2u2 (3.38)

We note that, in Eq. 3.38, only one term is dependent upon θ, and the

minimum value of this term is reached when sin 2θ has a maximum value of 1, i.e.

when θ = 45°. At this point,

hc2h'21Q u

2 −γ= (3.39)

so that

uha c2h'dhdQ'

hQ

−γ==σ=δδ (3.40)

This is the same answer as that given by Rankine’s earth pressure theory.

The above computations did not take the tension crack into account. If this is

done as shown in Figure 3.16, then the weight of the sub-wedge of soil behind the

crack can be discounted, thereby reducing W to

θγ−θγ= coth'21coth'

21W 2

c2 (3.41)


1-40

At the same time, F is also reduced to

( )cu hh

sincF −θ

= (3.42)

which gives

( ) ( )θ−

−−γ=2sin

hhc2hh'

21Q cu2

c2 (3.43)

Thus, for maximum Q, θ remains unchanged at 45°, at which

( ) ( )cu2c

2 hhc2hh'21Q −−−γ= (3.44)

and σa remains unchanged.

Figure 3.16


1-41

3.9 Culmann’s Graphical Method of Solution

Culmann’s method of solution is essentially a clever graphical implementation

of the Coulomb’s wedge calculation, which allows several trial surfaces to be tried

within a reasonable time. Figure 3.17 shows a trial wedge and the force triangle

below. We note that, if the force triangle is now rotated so that the reaction R is

aligned along the trial slip plane, the self-weight vector W will be inclined at an angle

of φ’ to the horizontal. In other words, by alignment the reactions R along their

respective trial slip surfaces, the vector W for all the trial cases can be collapsed into

a single line inclined at angle φ’ to the horizontal. Furthermore, the angle between

the wall reaction Q and the self-weight W is 180° – α – δ, which is independent of the

inclination of the trial surface θ. Thus, all the wall reactions will also be aligned along

the same direction. Thus, by appropriately rotating the force triangle, the graphical

solution can be greatly speeded up.

The implementation of Culmann’s method of solution is as follows:

Figure 3.17


1-42

1. Draw the retaining wall, backfill etc.. to a convenient scale (Fig. 3.18).

2. From point A at the heel of the wall, project line AC at an angle of inclination of φ’

to the horizontal. This will be the line along which all the self-weight vectors W will

be aligned.

3. From point A, project line AD at an angle of 180°–α–δ to line AC. All the wall

reaction vectors Q will be aligned parallel to this line.

4. For each trial wedge, compute the self-weight W1, W2 etc.., and scale off these

weights on line AC using a convenient scale for the rotated force triangle.

5. Through each end point w1, w2 etc.. corresponding to each self-weight vector,

draw lines parallel to AD so as to intersect their corresponding trial slip plane.

6. Draw a smooth curve through the point of intersection.

7. Draw a line that is tangential to the Culmann’s line and parallel to AC. At this

point, the offset between the Culmann’s line and AC is the maximum, thereby

giving the maximum wall reaction.

Figure 3.18


1-43

8. Draw a line through the tangent point that is parallel to AD to intersect the line

AC. The length of this line gives the maximum wall reaction corresponding to the

critical slip plane.

Because of the efficiency of Culmann’s method, it can applied to a fairly wide range

of problems, including irregular backfill surface. Figure 3.19 shows an application of

Culmann’s method to a cantilever wall.

Figure 3.19


1-44

4 SOME FACTORS AFFECTING EARTH PRESSURES. In this section, we discuss some of the factors which affect the earth (and

water) pressure on the retaining wall. The factors which will be discussed are

1. Wall friction,

2. Surcharge loading,

3. Pore water pressure, and

4. Compaction of soil behind the wall.

4.1 Wall Friction Real retaining walls are often not smooth. If you revisit Figures 3.9 and 3.10

(for rough walls), you would appreciate that the active wedge moves down relative to

the wall (Figure 3.9) whereas the passive wedge moves up (Figure 3.10). The effect

of wall friction is to alter the potential slip surface into curves.

Since wall friction has an important effect on soil pressures, they should be

accounted for in design. In a previous section, we have seen that the classical

Rankine theory cannot take account of wall friction, so that Ka is over-estimated and

Kp is under-estimated. In a previous section, we have already seen that Rankine’s

earth pressure formulae can be extended to cover sloping ground and wall friction

angle δ by modifying the active earth pressure coefficient Ka using a Coulomb wedge

analysis (see section 3.6 and Figure 3.9, repeated here).

Figure 3.9


1-45

This leads to the following relationships for Ka:

σ’a = Ka σv’ = Ka γ’h (3.24)

2h'KQ

2

aγ

= (3.25)

in which

( )

( ) ( ) ( )( )

2

a

sin'sin'sinsin

'- sin sec cK

⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢

⎣

⎡

β+αβ−φφ+δ

+δ+α

φαα= (3.26)

The horizontal component Qh of Q is given by

2h'KQ

2

ahhγ

= (3.27)

in which

Kah = Ka sin (α + δ) (3.28)

Rankine’s earth pressure formulae can be similarly extended to include wall

adhesion cw by modifying the earth pressure coefficient by an appropriate wedge

analysis. Consider a situation with a cohesive soil with shear strength parameters c’

and φ’, wall friction δ and wall adhesion cw. As shown on Figure 4.1, the directions of

the forces are known, as are the magnitudes of W, Cw (= cw x EB) and C (= c’ x BC).

Thus, the value of P can be determined from the force diagram for the trial failure

plane. Again, a number of trial failure planes can be analysed to obtain the

maximum value of P.

Figure 4.1

δ


1-46

For the special case of a vertical wall with horizontal soil surface and undrained

condition (i.e. φu = 0), the maximum value of the wall reaction Pa is given by

( ) ( ) ⎟⎟⎠

⎞⎜⎜⎝

⎛+−−−γ=

u

wou

2o

2a c

c1zHc2zH

21P (4.1)

We can therefore modify the cohesion contribution by a coefficient Kac, such that

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

u

wac c

c12K (4.2)

For the fully drained condition with a c’-φ’ soil, Kac is approximately given by

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

u

waac c

c1K2K (4.3)

Thus, a convenient general way to express the active pressure at depth z is

σha = Ka γ’z – Kac c’ (4.4)

Ka may be estimated from Eq. 3.26 by setting β = 0° and α = 90°; this leads to

( )

2

a' sin'sin cos

' cosK⎥⎥⎦

⎤

⎢⎢⎣

⎡

φφ+δ+δ

φ= (4.5)

The depth of the tension crack zo is then given by

a

u

w

a

aco K

cc1

''c2

'K'cK

z⎟⎟⎠

⎞⎜⎜⎝

⎛+

γ=

γ= (4.6)

Note that all these relationships only hold for a vertical back wall and horizontal soil

surface (i.e. α = 90° and β = 0°).

An equivalent set of relationships exist for passive failure modes, i.e.

σhp = Kp γ’z + Kpc c’ (4.7)

in which

( )

2

p' sin'sin cos

' cosK⎥⎥⎦

⎤

⎢⎢⎣

⎡

φφ+δ−δ

φ= (4.8)


1-47

which is obtained by setting α = 90° and β = 0° in Eq. 3.31; and Kpc is given by

⎟⎟⎠

⎞⎜⎜⎝

⎛+=

u

wppc c

c1K2K (4.9)

However, it should be noted that the inclusion wall friction often changes the

critical failure planes to curved surfaces which are different from Coulomb’s

assumption of plane surfaces. For this reason, the results may be unreliable

especially for the passive mode wherein the curved surfaces have very different

shapes from the assumed plane surfaces. The effect of this is often a slight under-

estimation of Ka but a significant over-estimation of Kp. Note that, in both instances,

Coulomb’s computation would err on the unsafe side.

A much better match to experimental results is often obtained by using

Caquot and Kerisel’s (1948) design charts, a selection of which were earlier

presented on Figures 3.11 to 3.14. Figure 4.4 shows the Caquot and Kerisel’s

passive earth pressure coefficients for an inclined backfill of angle β, and various

values of wall friction δ. The value of the wall friction angle δ is often assumed to be

between φ’/3 or 2φ’/3.

Figure 4.4 Caquot and Kerisel’s Charts for Passive Earth Pressure Coefficients

for an Inclined Backfill with Wall Friction


1-48

(c) δ = ½ φ’

Figure 4.4 Caquot and Kerisel’s Charts for Passive Earth Pressure Coefficients for an Inclined Backfill with Wall Friction (continued)


1-49

Figure 4.4 Caquot and Kerisel’s Charts for Passive Earth Pressure Coefficients

for an Inclined Backfill with Wall Friction (continued)


1-50

Suggested values of angles of friction for dissimilar materials have been suggested

by the California Trenching and Shoring Manual (see Table 4.1)

Table 4.1 Values of Wall Friction Angles and Adhesion


1-51

4.2 Surcharge Loading

Surcharge loads on the retained soil, or on the surface of the excavation,

influence both the magnitude and distribution of active or passive pressures. The

effects of uniform surcharges of intensity q (Figure 4.5) can be easily computed

using the earth pressure theories developed earlier, that is

σha = Ka q (4.10)

and σhp = Kp q (4.11)

The effects of localized point and line surcharges are less well-established

since the soil is only nominally failing. One method is to use the formulae and charts

presented in Figure 4.6, which is recommended by NAVFAC (1971) and is based on

Terzaghi’s work on anchored bulkhead. These formulae are based on the

assumption of a rigid unyielding wall and gives values which are roughly twice that

from elastic equations. Since the assumption of a rigid, unyielding wall is

conservative, its applicability to specific situations should be assessed carefully.

Figure 4.5


1-52

Figure 4.6


1-53

4.3 Compaction-induced Earth Pressures

Compaction of backfill behind the retaining wall will often increase the earth

pressure beyond that represented by the active or passive conditions. As shown in

Figure 4.7, this increase is normally limited to soil above a certain depth. Below that,

the effect of compaction is not felt. Other empirical earth pressure diagrams for

compaction earth pressure are also available, but most of them are characterized by

a similar trend, i.e. a linearly increasing portion with depth followed by a zone of

constant earth pressure.

There is a question on whether compaction stresses need to be applied. This

depends upon what the earth pressure is used for. For the assessment of stability of

gravity retaining structure, it is often unnecessary to consider compaction stresses

since a higher earth pressure will cause the gravity retaining structure to move

forward (slightly) and, in so doing, relieve the stresses. However, they are necessary

for the assessment of structural integrity of the retaining structure e.g. bridge

abutments.

Figure 4.7


1-54

4.4 Pore Water Pressure

The theories of earth pressure developed earlier can be applied to total and

effective stress analysis. In total stress analysis of saturated soils, the shear strength

parameter which is commonly used is the undrained shear strength cu (or su), with φu

= 0. In such cases, the pore pressure need not be additionally accounted for.

However, in excavation situations, the long-term stability of the structure is often

more critical than the short-term stability. This can be easily illustrated by a simple

stress path analysis. Very often, we want to express or study how the stress

changes over the course of the event in question. This can be done by plotting stress

paths, which are plots of one stress component or parameter against another. A

very simple way of visualising stress path is to consider the changes to a Mohr-Circle

in the course of a typical triaxial test as shown in Figure 4.8 below:

At the start of the test, σ1 = σ3 and the Mohr-circle is a point on the σ-axis. As

the test progresses, σ1 increases while σ3 remains constant; thus the Mohr circle

moves to the right as shown. From the start to the end of the test, the sample

progresses through an infinite number of states, each of which can be represented

by a Mohr-Circle. If we are going to sketch all these circles, the stress space will be

so congested that the trend cannot be detected. One way to solve this problem is to

just plot the path traced out by the topmost point of the circles as they change. This

Figure 4.8


1-55

path can be considered as a stress path (see Figure 4.8 above). In a triaxial test, σ1

= σv and σ3 = σh; thus the co-ordinates of the topmost point of a Mohr-circle can be

expressed as follows:

x-coordinate s = (σ1 + σ3)/2 = (σv + σh)/2 (4.12)

y-coordinate t = (σ1 – σ3)/2 = (σv – σh)/2 (4.13)

The co-ordinates (s,t) can therefore be regarded as stress path parameters. [In fact,

their use was first suggested by MIT researchers Lambe, Whitman, Ladd et al., and

thus they are commonly known as the MIT stress path parameters]. The effective

stress path equivalences of these two parameters can also be defined similarly i.e.

x-coordinate s’ = (σ’1 + σ’3)/2 = (σ’v + σ’h)/2 (4.14)

y-coordinate t’ = (σ’1 – σ’3)/2 = (σ’v – σ’h)/2 (4.15)

The relation between (s,t) and (s’,t’) can be easily ascertain as follows :

s’ = (σv’ + σh’)/2 = (σv + σh – 2u)/2 = (σv + σh)/2 –u = s – u (4.16)

t’ = (σv’ – σh’)/2 = (σv – u – σh + u)/2 = (σv – σh)/2 = t (4.17)

The increment in s, s’ and t can be similarly defined as

∆s = (∆σv + ∆σh)/2 (4.18) ∆s’ = (∆σ’v + ∆σ’h)/2 (4.19) ∆t = ∆t’ = (∆σv –∆σh)/2 (4.20)

We now want to superimpose the Mohr-Coulomb failure criterion into the stress path

plot. For a cohesionless soil, i.e. c = 0, this is easily achieved since we know that, in

the active state,

'sin1'sin1

v

h

φ+φ−

=σσ

(4.21)

Combining Eqs. 4.12 & 4.13 leads to

'sin1

1

st

v

h

v

h

hv

hv φ=

σσ

+

σσ

−=

σ+σσ−σ

= (4.22)

in view of Eq. 4.14.


1-56

Similarly, for the passive case,

'sin1'sin1

v

h

φ−φ+

=σσ

(4.23)

This leads to

t/s = – sin φ’ (4.24)

Hence, the slope of the failure stress path line, α, is related to the slope of the Mohr

Coulomb line (friction angle), φ, by

tan α = ± sin φ’ (4.25)

Graphically, this is illustrated in Figure 4.9

With a cohesive soil, failure occurs when

t = (σv – σh)/2 = cu in compressive failure (4.26)

t = (σv – σh)/2 = – cu in extension failure (4.27)

α

αts

t/s =

Figure 4.9


1-57

Thus, in t-s stress space the failure envelope and its complement (in extension

failure) are two horizontal straight lines with t-intercepts of cu and –cu, respectively.

With a c’-φ’ soil, the failure envelope can be similarly derived. As shown in Figure

4.10, the abscissa of the centre of the Mohr Circle is s while the radius of the circle is

t, thus

t = s sin φ’ + c’ cos φ’ (4.28)

and the failure envelopes are also straight lines in t-s space. In summary, the failure

envelopes in τ-σ (Figure 4.10) and t-s stress space (Figure 4.11) are very similar in

shape.

Figure 4.10

α

α

Figure 4.11


1-58

Consider now a situation in which a soil with K0 = 1 is excavated on one side as

illustrated in Figure 4.12. In this situation, σv is likely to remain nearly constant

whereas σh decrease; in other words

∆σv = 0 and ∆σh < 0 (4.29)

At the initial state, σh = σv and σh’ = σv’, so that

so = σv , s’o = σ’v and to = 0

so that the initial stress states can be represented by points on the s-axis.

Δs = Δσh < 0 (4.30) Δt = -Δσh > 0 (4.31) Δt/Δs = –1 (4.32)

We can now plot the total (s, t) stress path as shown in Figure 4.13.

Figure 4.12

Figure 4.13


1-59

For the effective stress path, s’ cannot change in the short term if the soil has a low

permeability and cannot easily drain, so that

∆s’ = 0 (4.33)

In the long term, ∆u = 0 so that ∆s’ = ∆s.

As shown in Figure 4.13, the total stress path heads towards the t-axis

indicating an immediate decrease in s but the effective stress path moves vertically,

indicating no immediate decrease in s’.

In other words, there is no decrease in the effective stress and therefore the

strength of the soil. In the long term, the effective stress path moves towards the

total stress path as pore pressure equilibrates, thereby leading to a decrease in

effective stress and strength in the long term. Thus, the critical scenario for most

excavation retaining structure is the long-term.

CE5108 Part 1 - Introduction and Earth Pressure Theories

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