2. Lateral Earth Pressure Theories

2. Lateral Earth Pressure Theories

Significance of Lateral Pressure Why study lateral earth pressure? Retaining walls are designed and constructed essentially to resist lateral earth pressures.

What is lateral earth pressure? Lateral earth pressure is the pressure that soil exerts against a structure in a sideways, mainly horizontal _________ direction.

How to calculate lateral earth pressure? Not Easy!! Unlike vertical pressure, which can be calculated quite reliably based on the soil unit weight and applied surcharge.

Significance of Lateral Pressure Depending on the ground conditions (e (e.g. g stress history of the soil) and the wall conditions (movements embedment) (movements, embedment), different states of lateral earth pressures may be mobilized. This section focuses on Lateral Earth Pressures, and how you can estimate these pressures for use in engineering design and analysis.

Learning Objectives Different States of Lateral Earth Stresses Insitu, Active, Passive

Methods M h d off calculating l l i active i and d passive i earth pressures Rankine, Coulomb, Caquot & Kerisel

Key features/limitations K f t /li it ti off th the diff differentt limiting earth pressure methods Effect of surcharge, compaction, etc

Stresses in the g ground – At Rest Condition Vertical stresses (v,  v) Horizontal stresses (h, h) Ground surface

v = dry(d-dw) + bulkdw

d = dry unit weight bulk = bulk unit weight

v = v - u v = dry(d-dw) + bulkdw h = ?? h = ’h + u

d

dw

u = wdw v = v - u h = ?? h =  ’h + u

Stresses in the g ground – At Rest Condition y

y , y

At-rest condition,

x 0z = 0

x

z

x , x that is,, strains in the horizontal directions are zero.

z , z

It is reasonable to treat the horizontal soil strains to be zero for at-rest condition, especially p y if the ground is very large (in plan area).

Stresses in the g ground – At Rest Condition For at at-rest rest condition, If the soil is a linear elastic material (which it is not!) we can evaluate the effective horizontal not!), stress ’h (in terms of the effective vertical stress ’v) using just one elastic parameter. parameter What elastic parameter is that?? Poisson’s ratio:



What is the ratio ’h/’v (or ’x/’y ) ?

Stresses in the g ground – At Rest Condition For an elastic soil in at at-rest rest condition,

  'h  ' x   1   'v  ' y The ratio ’h/’v is called the lateral earth pressure coefficient K. The ratio ’h/’v for the at-rest condition is denoted as Ko.

 'h  Ko    'v 1 

(elastic material onl only!) !)

Some information on Poisson’s Poisson s Ratio Typical Poisson Poisson’s s Ratio for Soils 0.25 to 0.35 for soils (effective stress or  = ___________ drained analysis) 0.5  = ____ 0.49 to 0.495 = _____________

(undrained = incompressible) (to avoid numerical instability when modeling undrained prob.)

What are the physical limits of Poisson’s ratio? –1 1    ____ 05 0.5 ___ What is the behavior associated with negative ?

Some information on Poisson’s Poisson s Ratio

Some information on Poisson’s Poisson s Ratio

Earth Pressure at Rest If we consider soil as a elastic material, then

' Ko  1  '

' h  v 1  '

For 0.25  ’  0.35

 0.35  Ko  0.5 Generally on the low side T d to Tend t underestimate d ti t reall soils il Ko

How do we obtain more representative values of Ko for real soils?

Earth Pressure at Rest Ko is a function of the  the soil type  geologic or stress history. The soil type refers to whether the soil is sand (loose, dense) or clay (soft, stiff), and how it is characterized using i strength t th parameters t such h as the th friction f i ti angle l  and the plasticity index PI The geologic or stress history usually refers to whether the soil is normally or overconsolidated. (See next slide)

Normally Consolidated vs O Overconsolidated lid t d Soils S il Void ratio,, e

Overconsolidation ratio

 v (max prev) OCR   v (current) Normal (Virgin) Compression Line

v (current)

v (max prev)

v (log (l scale) l )

Earth Pressure at Rest – NC Soils Let us first consider Ko for a normally y consolidated soil. (Also referred to as Knc or Konc).

Estimating Knc  The most commonly used relationship for determining Knc is th t due that d to t Jâky Jâk (1944) (1944), which hi h states t t th thatt

K nc

 1  32 sin  '   (1  sin  ' ) (1  sin  ' )

where   is the effective (not total) angle of friction of the soil. This is very commonly simplified into its approximate form

K nc  (1  sin ' )  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) See Figure on next page.

Estimating Knc

Estimating Knc Variation of Knc with Friction Angle

08 0.8 0.7 06 0.6

Simplified Jaky

Kncc

0.5 Original Jaky

0.4 0.3 0.2 0.1 0 0

10

20

30

Friction Angle

40

50

60

Alternative Relationships for Knc Brooker and Ireland (1965) ( ) Knc = 0.95 – sin  Variation of Ko with Friction Angle 0.8 0.7 0.6 Simplified Jaky

Kncc

0.5 Original Jaky

0.4 0.3 02 0.2

Brooker and Ireland

0.1 0 0

10

20

30

Friction Angle

40

50

60

Alternative Relationships for Knc  Alpan (1967) Knc = 0.19 + 0.233 log (PI) where PI is the plasticity index Variation of Ko with PI Ip 0.8 0.7 0.6 Alpan

Knc

0.5 0.4 0.3 0.2 0.1 0 0

20

40

60

80

Plasticity Index (PI)

100

120

Knc for Singapore Marine Clay Using Effective Stress Parameter:  = 22 0.63 (Jaky, simplified)  Knc = ____ 0.58 (Jaky, original) = ____ 0.58 (Brooker and Ireland) = ____ U i Atterberg Using Att b li limits: it PI = 55  Knc = _____ 0.60 (Alpan)

Other Relationships for Knc

Earth Pressure at Rest – OC Soils

Over-consolidated soil are elastic due to unloading but they have even higher Ko than normally y consolidated soil.

Ko for overconsolidated soils  There are many proposed relationships. One of these is by Alpan (1967) (1967), who suggested that

Ko = Knc OCRn where n is an exponent whose value depends upon the soil type. For clays, Alpan proposed that n can be related to the plasticity index PI PI/281 n = 0.54 0 54 x 10-PI/281

 Kulhawy and Mayne (1982) proposed that n can be related to the effective angle of friction ’:

n = sin

Ko for Overconsolidated Marine Clay Using Alpan’s PI method: PI = 55 Knc = 0.60 0 60 n = 0.54 x 10-PI/281 = 0.344 0 344 Ko = Knc OCRn = 0.6 0 6 OCR0.344

Using Jaky’s Jaky s approach:  = 22 Knc = 0.63 0 63 n = sin  = sin 22 = 0.37

(Kulhawy and Mayne)

Ko = Knc OCRn = 0.63 OCR0.37

Ko for Overconsolidated Marine Clay Variation of Ko with OCR

2 16 1.6

Jaky with Kulhawy and Mayne

1.2 Ko

Alpan 0.8 0.4 0 0

2

4

6

OCR

8

10

12

Ko for overconsolidated soils ((continued))  Wroth (1972) also proposed two semi-empirical relationships for estimating Ko from Knc. For OCR up to 5, 5 Wroth (1972) proposed that

K o  OCR K nc -

' (OCR  1) 1 ' 1

This relationship is based on the assumption that the unloading process which resulted in the over-consolidated over consolidated state is an elastic process.  For higher OCR, OCR Wroth (1972) proposed that

 3(1  K nc ) 3(1  K 0 )  OCR (1  2K nc )    ln m     1  2 K 1 2 K 1 2 K    nc 0  0 in which m = 0.022875 Ip + 1.22

Typical yp Ko for Different Soil Types yp

Ko for overconsolidated soils ((continued)) Variation of Ko with OCR

2 16 1.6

Wroth

1.2 Ko

Jaky with Kulhawy and Mayne

0.8 0.4

Based on  ’ = 22 22 0 0

2

4

6

OCR

8

10

12

Earth Pressure at Rest – Summary Now, you h N have the th basic b i tools t l for f estimating ti ti the th lateral earth pressure at rest Ko. What about the ‘non at-rest’ states of the soil? These states are usually mobilized when the soil is disturbed, as for example, during an excavation. What are the other possible states of lateral earth pressure?

Lateral Earth Pressure Coefficient For Soils Lateral earth pressure coefficient

 h' K ' v  Lateral earth pressure (at rest) Ko  Lateral earth pressure (active)

KA

 Lateral L t l earth th pressure (passive) ( i ) KP Wh are th When the active ti and d passive i states t t mobilized? bili d?

What happens pp to Ko when soil is disturbed • Consider a mass of undisturbed ground Initial conditions ’v , ’h = Ko’v

What happens pp to Ko when soil is disturbed • Consider a mass of undisturbed ground Initial conditions ’v , ’h = Ko’v • A block of soil is removed as shown • If the remaining g soil is supported by a very rigid wall which prevents any l t l deformation lateral d f ti off the th remaining soil, then the stress state in the remaining soil (on the right) will remain unchanged, g that is, it remains in Ko condition.

removed block remaining soil

What happens pp to Ko when soil is disturbed • However, it is unrealistic to expect that the remaining soil will not deform at all. • Usually the retaining wall is not that rigid, rigid so that it will deform outward by some extent, and the base will heave. • We will focus only on the changes in the stress state caused by the lateral deformation of the unsupported side.

remaining soil

What happens pp to Ko when soil is disturbed • Due to the outward deformation, the lateral stress at point A will decrease • It will decrease to a limiting level such that ’h = Ka’v where Ka < Ko • When that happens happens, we say that ’h has reached the active pressure p condition.

A

’h = Ka’v

remaining soil

What happens pp to Ko when soil is disturbed • What happens when we push the unsupported side inwards into the remaining soil? • At point A A, the stress will increase to a limiting level such that ’h = Kp’v where Kp > Ko • When that happens, we say that ’h has reached the passive pressure condition.

A

’h = Kp’v

remaining soil

Earth p pressure related to wall movement Earth Pressure Coefficient, K

KP

ACTIVE STATE

PASSIVE STATE

06 0.6

KA

Ko

0.4

Wall movement

1.5


H

ACTIVE STATE

KA

Ko

Wall movement

PASSIVE STATE


H

ACTIVE STATE

Ko

KA

Wall movement to mobilize full active pressure

PASSIVE STATE

Wall movement to mobilize full passive pressure

How to Calculate Active and Passive W ll P Wall Pressures The active and p passive pressures p are known as the limiting earth pressures. Their magnitudes depend on the ____________. soil strength Hence, how we characterize the soil strength is very important important. 1. For clay, silts – undrained condition Undrained shear strength – su 2 F 2. For clay, l silts ilt – drained d i d condition diti For sands Effective Stress Strength Parameters – c’-’

How to Calculate Active and Passive W ll P Wall Pressures Methods of Evaluating Active and Passive Pressures 1. Strictly speaking, they should be evaluated taking into consideration the stress-strain relationship of the soil. (See previous figure where Ka and Kp depend on wall movement, or more accurately, soil deformation) Finite Element Analysis, Beam-Spring Analysis 2 Limit Equilibrium Method 2. Lower Bound, Upper Bound Methods

How to Calculate Active and Passive W ll P Wall Pressures Limit Equilibrium q Method  For solving various soil stability problems  Weakness – do not consider stress-strain relationship of soil Hence cannot yield deformation  Uses yield strength criterion Mohr-Coulomb Yield Criterion  Upper and Lower Bound Theorems

Lower and Upper Bounds  A “lower bound” solution means finding a set of stresses that nowhere exceed the collapse condition diti (i.e. (i it will ill always l be b on the “safe” side as the collapse condition is not exceeded)  An “upper pp bound” solution means finding a kinematically feasible mechanism of collapse (i.e. it will always l be b on the th “unsafe” “ f ” side id as a collapse mechanism has formed)

Failure load

Upper bound Correct Solution Lower bound

How to Calculate Active and Passive W ll P Wall Pressures In this module, we will focus on the classical approach of evaluating the active and passive earth pressures using limit equilibrium methods. There are three broad classes of limit equilibrium methods that are commonly adopted in practice. These are: 1. The Rankine Method 2. The Coulomb Method 3. The Caquot-Kerisel Log-Spiral Method We will cover all three methods in this module.

How to Calculate Active and Passive W ll P Wall Pressures Historically, Historically 1. the Coulomb method is probably the earliest known analytical approach to deal with the issue of lateral forces on retaining walls. It was developed in the 1700s by y Coulomb. It is an upper bound method. 2. the Rankine method was developed in the 1800s. It is a lower bound method. 3. the log-spiral method was proposed in the 1900s. It seeks k tto overcome some off th the k key li limitations it ti associated with the Coulomb and Rankine method. Of course, other methods were proposed along the way, but the above three are probably the most commonly used.

How to Calculate Active and Passive W ll P Wall Pressures For learning purposes purposes, 1. We will start with the Rankine method, as the resulting equations are quite simple, and can be quite readily derived using basic soil mechanics concepts. 2. We will then follow up with the Coulomb approach which overcomes some of the limitations associated with the Rankine method. 3. Lastly, y, we will discuss the log-spiral g p method which is generally accepted as providing the most realistic solution.

Rankine Method: Active Earth Pressure

William Rankine  William John Maquorn Rankine (1820-1872) was the Chair of Engineering of Glasgow University. These days, the annual Rankine Lecture organized by the British Geotechnical Society is named after him him.  He H studied t di d the th state t t off stress t att a point i t att failure in a semi-infinite, cohesionless medium, and related the principal stresses at failure (1857) (lower bound approach).

Rankine Method – Active Condition

Rankine Method – Active Condition Rankine used a highly theoretical approach, involving the consideration of stresses at a point point, to arrive at the active earth pressure coefficient for calculating the active stress parallel to the sloping backfill:

K A  cos 



cos   cos   cos  2

2

cos   cos 2   cos 2 

a = KAv

where  is the angle of the sloping backfill (wrt the horizontal) and  is the friction angle of the soil. The horizontal component of the active earth pressure is obtained using

K AH  cos 2 

cos   cos 2   cos 2 

cos   cos 2   cos 2 

Rankine Method – Active Condition For horizontal backfill,  = 0, and the coefficient reduces to :

K A  cos 0



cos 0  cos 0  cos  2

2

cos 0  cos 0  cos  2

2

1  sin  KA  1  sin 

Very simple and commonly cited formula for active earth pressure.

Rankine Method – Active Condition In this lecture, we will not derive Rankine’s original g formula (which is a highly complex mathematical process):

K A  cos 

cos   cos 2   cos 2  cos   cos 2   cos 2 

Instead, we will use basic soil mechanics concepts p to show how we can arrive at the special case of the Rankine formula for a horizontal backfill:

1  sin  KA  1  sin 

Rankine Method – Active Condition Assume no friction along the sides of the soil mass

smooth

v

Initially

 h = Ko  v

h

Outward Movement of Wall due to Soil Removal (Ka condition) What happens when h the th wall ll moves outward?

v reduces

h

remains constant

Rankine Method – Active Condition Is the soil strength characteri ed using characterized sing (i) su (undrained shear strength) (ii) c’-’ (effective stress strength parameters)

Rankine Method – Cases Considered Active Case (i) Effective stress purely frictional response, Drained (’) (ii) Effective stress frictional response with cohesion cohesion, Drained (c’ - ’) (iii) Total stress response response, undrained (undrained shear strength su )

Passive Case (i) Effective stress purely frictional response, Drained (’) (ii) Effective stress frictional response with cohesion, Drained (c’ - ’) (iii) Total stress response, undrained (undrained shear strength su ) Earth Pressures - 56



Consider first a frictional  ’ material.

Initial Stress State of a ’ material 

Mohr-Coulomb Failure Envelope

In terms of friction parameter ’

Mohr circle for initial stress state

’

 h

v

’h Complementary p y Failure envelope

Stable Condition (Not Yielded Yet)

’v

As wall moves outwards - ’h reduces 

Failure envelope  

 h

Complementary p y Failure envelope p

v


Failure envelope  

 h


v



 h


v

KA for a  ’ material 


Effective stress circle at active failure

sin    (v-h)/2

max





h

(v+h)/2


v

 v ' h ' / 2  v ' h ' / 2

 h 1  sin  '   v 1  sin  '

 1  sin  ' KA  1  sin '

h = KA v

Distribution of earth p pressure behind wall

If wall doesn’t move

h = Kov

Distribution of active earth p pressure behind wall

If wall moves outward

ah = KAv

Effect of Surcharge g Loading g

Effect of surcharge loading

q

ah = KAv qh = KAq

Mohr Circle Failure Planes for ’ material (Ka condition)   f  45 

f  f

Failure envelope

'

 

2

v h

90

max

2f



h v h

(v-h)/2

v

h

2f = 90 +  ’

 v acting on

(v+h)/2

v

h

horizontal plane

f = 45 + ’/2

The failure plane is Effective oriented at f to the stress circle at ’ plane, where v active failure


'  f  45   2

KA Slip p Planes in Soil Mass ( (’ material))

Active Condition  f  45 

' 2


Passive Case (i) Effective stress purely frictional response, Drained (’) (ii) Effective stress frictional response with cohesion, Drained (c’ - ’) (iii) Total stress response, undrained (undrained shear strength su )

Frictional material with cohesion,, c’-’ 

Active Condition for c-  soil  Effective stress circle at active failure

Failure envelope 

(v-h)/2

max

 cc a

 h

(v+h)/2

v

 v ' h ' sin   

2  ' h ' a v 2

Complementary Failure envelope

tan   

c a

Active Condition for c-  soil c tan    a



a  c

cos   sin  





a sin   



c  cos   



h'



  1  sin   1  sin   h'  v '2c 

a sin    c  cos  

 v ' h ' 2 sin     ' h ' a v 2 N t th Note thatt

 cos  '     1  sin  ' 

2

 1  sin 2  '  =  2     1 sin  '  

=

1  sin  '  1  sin  ' 

 v ' h ' sin    v ' h ' 2

2

 v ' h ' sin    v ' h ' 2

1  sin  

(1  sin  )

2

v '2c 

(1  sin  )

  h '  K A v '2c  K A

cos   (1  sin  ) (1  sin  )

where

 1  sin '  KA  1  sin ' 

Rankine Method – Active Condition -  ’ material We have derived an expression for the active earth pressure coefficient of a purely frictional material (commonly called the Rankine earth pressure coefficient):

1  sin  KA  1  sin 

For a purely frictional material, material

h = KA v For a frictional material with cohesion c’,

 h'  K A v' 2c K A You may come across notes and books that attribute the last equation t Rankine. to R ki St i tl speaking, Strictly ki that th t is i nott correct, t as Rankine R ki did nott consider a c’-’ material in his paper. That equation represents an extension of Rankine’s method to a c’-’ material.



Consider next a material exhibiting undrained response response, su.

Earth Pressures - 75

Initial Stress State of Soil Mass 

In terms of undrained strength: cu or su Apparent Failure envelope

cu


or

su

 h

v

h A Apparent F Failure il envelope l


v

Stress state changes for su material due to outward wall movement  Apparent Failure envelope Initial condition

su

 h

v

A Apparent F Failure il envelope l


Stress state changes for su material due to outward wall movement  Apparent Failure envelope su  h

v


Mohr M h Circles Ci l expand d by b reducing d i h , while hil v remains constant

Stress state changes for su material due to outward wall movement  Apparent Failure envelope su  h

v


Mohr M h Circles Ci l expand d by b reducing d i h , while hil v remains constant

Stress state changes for su material due to outward wall movement  Apparent Failure envelope Total stress circle at active failure

su

 h

2su

v


At failure

h = v– 2s 2 u = d d – 2s 2u

Mohr Circle Failure Planes for su material (Ka condition) 

Stress State on Failure Plane Apparent Failure envelope

v

(f , f) h

su

2su

v

h

Stress State on Horizontal Plane

2f = 90 90 h

f = 45

f  f

v



Stress State on Vertical Plane

v h

v

At failure h

h = v– 2su = d – 2su Note that the failure plane is oriented at ___ 45 to the horizontal.

KA Slip p Planes in Soil Mass ((su material)) v h

f = 45

h v


h v

f = 45


h v

f = 45

Active earth pressure distribution f a su material for t i l

h = d d – 2s 2u If we plot h vs d, h < 0 when

d – 2su < 0 d < 2su d < 2su /

Active earth pressure distribution f a su material for t i l dc  Direction of wall movement

2 su



h = d – 2su

Rankine Method – Active Condition – su material We have derived an expression relating the horizontal stress to the vertical stress under the active condition for undrained response involving g the undrained shear strength su

h = v– 2s 2 u = d d – 2s 2u Note that, again, this equation cannot be strictly attributed to Rankine, as he only considered a frictional ’ material. It may be more appropriate to call this a lower bound solution for the active condition under undrained condition.

Rankine Method: Passive Earth Pressure

Consider Soil Mass behind Wall

v increases h

remains constant



Consider first a frictional  ’ material.

KP for a  ’ material 

 



initial

h

Failure envelope l

v



 

Failure envelope l

 v


Mohr Circles first reduce until h = v.


 

Failure envelope l

 v




 

Failure envelope l

 v




 

Failure envelope l

 v




 

Failure envelope l




Mohr Circles then expand when h > v .


 

Failure envelope l






 

Failure envelope l






 

Failure envelope l






 

Failure envelope l






 

Failure envelope l






 

Failure envelope l





 

Failure envelope l sin   

(h-v)/2

max



initial

v

(h+v)/2


h

 h ' v ' / 2  h ' v ' / 2

 h 1  sin  '   v 1  sin  ' KP 

1  sin ' 1  sin i  '

Effective stress circle at passive failure

h = KP   v

Distribution of earth p pressure behind wall

If wall doesn’t move

h = Kov

Distribution of p passive earth pressure p behind wall

If wall is pushed against soil

ph = KPv

Effect of Surcharge g Loading g


q

ph = Kpv

qh = Kpq

Mohr Circle Failure Planes for ’ material (Kp condition) 

f f 45 

 f  45 

'

Failure envelope l

 

2

' 2

(v-h)/2

2f

initial v

(v+h)/2

max

The failure plane is oriented at f to the  ’h plane (vertical), ___ h acting on where vertical plane


 f  45 

Effective stress circle at active failure

or

 f  45 

' 2

' 2

to the horizontal plane

Kp Slip p Planes in Soil Mass ( (’ material))

Passive Condition  f  45 

' 2

Comparison of Active and Passive Slip Planes (for  ’ material)

Active Condition  f  45 

' 2

Passive Condition  f  45 

' 2



Frictional material with cohesion,, c’-’ 


Passive Earth Condition for c-  soil  Effective stress circle at passive failure

Failure envelope 

(h-v)/2

max

cc a

 v

(h+v)/2

h

 h ' v ' sin    Complementary Failure envelope

2  ' ' a h v 2

tan 

c a

Passive Earth Condition for c-  soil v ' 

1  sin  

(1  sin  )

h '2c 

 1  sin   h'  v '2c  (1  sin  )

cos   (1  sin  )

cos   (1  sin  )

Note that

 cos  '     1  sin  ' 

2

 h'  K P v' 2c K P

 1  sin 2  '  =  2     1 sin  '  

=

1  sin  '  1  sin  ' 

=

Kp

For the special case where c’ = 0 this reduces to

 h'  K P v' Similarly, for the undrained case where u = 0 we replace c’ c by su and  ’ by u = 0, 0 so we get

h = v + 2su

Rankine Method – Passive Condition - ’ material We have derived an expression for the Rankine passive earth pressure coefficient of a purely frictional material:

1  sin  KP  1  sin i  For a purely frictional material, material

h = KP v For a frictional material with cohesion c’,

 h'  K P v' 2c K P Again the last equation is not strictly attributable to Rankine. Again, Rankine



Consider next a material exhibiting undrained response response, su.




Mohr Circle progression from I iti l tto Passive Initial P i St State t Apparent Failure envelope

su  h


v




su  h

v






su  h

v






su  v = h






su  v

h


Mohr Circles first reduce until h = v. Mohr Circles then expand when h > v .




su  v

h





Mohr Circle progression from I iti l tto Passive Initial P i St State t Total stress circle at passive failure

Apparent Failure envelope su  h


Mohr Circles then expand p when h>v until the Mohr circle touches the Apparent Failure Envelope.

v

2su

h

At failure

h = v+2su = d +2su

Mohr Circle Failure Planes for cu material (Kp condition) f  f



Stress State on Failure F il Pl Plane Apparent Failure envelope

( f ,  f )

v

f = 45 h

su

Stress State on Vertical Plane

2 f

v

2su

h

v

h



Stress State on Horizontal Plane

At failure

v h

v

h

h = v+ 2su = d + 2su

Note that the failure plane is oriented at 45 to the horizontal.

Kp Slip p Planes in Soil Mass ((cu material)) v h

h v

Identical to Active Condition Slip Planes f = 45

Passive earth pressure distribution f a su material for t i l 2su

Direction of wall movement

h = d + 2su

Rankine Method – Passive Condition – su material We have derived an expression relating the horizontal stress to the vertical stress under the passive condition for undrained response involving g the undrained shear strength su

h = v+ 2s 2 u = d d + 2s 2u Note that, again, this equation cannot be strictly attributed to Rankine, as he only considered a frictional ’ material. It may be more appropriate to call this a lower bound solution for the passive condition under undrained condition.

Summary up to this point  We’ve adopted the Rankine approach for evaluating the l t l earth lateral th pressures.  Consider only stress states of the soil (h , v) and the yield i ld limits li it in i deriving d i i the th active ti and d passive i failure f il conditions.  Failure plane is obtained from consideration of stresses using Mohr circle concepts.  No wall friction considered.  No assumption of failure mechanism (e.g. failure wedge behind the wall) is made. Hence, no force equilibrium is considered.  Extended the Rankine approach to materials with cohesion c’ or undrained shear strength su.

Generalized Case for Rankine Earth Pressure of a Granular G a u a So Soil (o (only y friction ct o a angle ge ’)) with Non-vertical Wall Face and Sl i B Sloping Backfill kfill

A Generalized Case for Rankine Active and Passive Pressure – Granular Backfill ( (’ material) Chu (1991) “Rankine’s Analysis of Active and Passive Pressures in Dry Sands” Soils and Foundations, Vol 31, Vol. 31 No. No 4 4, 115–120

  

 



Frictionless wall

Pa’ or Pp’

  ’

A Generalized Case for Rankine Active and Passive Pressure – Granular Backfill ( (’ material) For Rankine Active Case The active force PA per unit length of the wall:

1 2 PA  H K a 2 where

KA 

cos(    ) 1  sin ' 2 sin ' cos  a 2



cos 2  cos   sin 2 '  sin 2   sin       2  a  sin i   sin '  1

 sin i ' sin i a    a  tan   1  sin ' cos  a  1



A Generalized Case for Rankine Active and Passive Pressure – Granular Backfill ( (’ material) For Rankine Passive Case The passive force PP per unit length of the wall:

1 PP  H 2 K P 2

where

KP 

cos(    ) 1  sin 2 ' 2 sin ' cos  p



cos  cos   sin '  sin  2

2

 sin       2  p  sin i   sin '  1

 sin i ' sin i p     p  tan    1  sin ' cos  p  1

2



Generalized Case for Rankine Earth Pressure of a Cohesive Co es e So Soil (c (c’ -  ’)) with Vertical Wall Face and Sl i B Sloping Backfill kfill

Rankine Active and Passive Pressure for a Cohesive Soil (c (c’-’  material) – Inclined Backfill  

Mazindrani and Ganjali (1997) “Lateral Earth Pressure Problem of Cohesive Backfill with Inclined Surface” z

Journal of Geotechnical and Geoenvironmental Engineering, g g ASCE, Vol. 123, No. 2, 110–112

pa or pp

’c’

Frictionless wall

Rankine Active Pressure for a Cohesive Soil (c’-’ material) – Inclined Backfill For Rankine Active Case The active pressure pa per unit length of the wall: pa = zka cos  

cos  2  2  z  cos   2c' cos ' sin ' 2 cos ' 

4 cos cos 2

2

  cos 2 ' 2 z 2  4c 2  cos 2 '



 8c' z  cos 2   sin ' cos '  z  cos 

and  c'  1  2 ka   2 cos   2   cos ' sin ' 2 cos '   z  

2     c'    c 2 2 2 2 2  4 cos  cos   cos '  4   cos '  8  cos   sin ' cos '   1  z     z  

Rankine Active Pressure for a Cohesive Soil (c’-’ material) – Inclined Backfill Variation of ka with , c/z and  

c/z




Rankine Passive Pressure for a Cohesive Soil (c’-’ material) – Inclined Backfill For Rankine Passive Case The passive pressure pa per unit length of the wall: pp = zkp cos  

cos  2  2  z  cos   2c' cos ' sin ' 2 cos ' 

4 cos cos 2

2

  cos 2 ' 2 z 2  4c 2  cos 2 '



 8c' z  cos 2   sin ' cos '  z  cos 

and  c'  1  2 kp   2 cos   2   cos ' sin ' 2 cos '   z  

2     c'    c 2 2 2 2 2  4 cos  cos   cos '  4   cos '  8  cos   sin ' cos '   1  z     z  

Rankine Active Pressure for a Cohesive Soil (c’-’ material) – Inclined Backfill Variation of kp with , c/z and  



c/z

Limitation of the Rankine method  A shortcoming of the Rankine earth pressure theory is that h such h lower l b bound d solutions l i are only l known k f a for very few number of highly idealized geometries, and a smooth wall is assumed. assumed For more complex geometries and problems where wall friction is significant, the solutions are not easily y found.  Thus Rankine’s earth pressure theory is, in practice, rather restricted in its usage.  In fact, a more general method was proposed even earlier in time.  Recall it was mentioned earlier that Coulomb had proposed a method for calculating lateral earth pressures in the 1700s.  We will study Coulomb’s method next.

Charles Coulomb  Born 14 June 1736.  Graduated from Ecole du Genie in 1761.  Joined Corps of Engineers of the French Army with a rank of lieutenant  Posted to Martinique q in the Caribbean to bolster its defence by building new fortifications at Fort Bourbon.  These fortifications consisted of massive gravity retaining walls with moat and obstacles in front.  1764 to 1772 – famous research work on shear strength of soils and limit equilibrium method of retaining wall design. g  Upper Bound Approach.

Coulomb’s Coulomb s failure mechanism

 Coulomb observed that for sandy soil that most of the failure surfaces were almost _______. planar  For this reason he assumed that the failure surfaces are ______. However Coulomb’s planes approach h can, in i principle, i i l be b applied li d to t any class of failure surface.

Coulomb’s Coulomb s failure mechanism Failure wedge in an experiment

Upper pp Bound Approach pp  Coulomb observed that for sandy soil that most of the failure surfaces were almost _______. planar  For this reason he assumed that the failure surfaces are ______. However Coulomb’s planes approach h can, in i principle, i i l be b applied li d to t any class of failure surface.

Coulomb’s Coulomb s failure mechanism In Coulomb’s method, you extract the wedge and consider id the th forces f acting ti on the th wedge. d

PV F

PH

N W You fill in the forces that you know (W, (W F, F N) and work out the unknowns PH and Pv.

Coulomb – Active State

Example 1 – Cohesionless Soil (), Smooth Wall, Level Ground Ground, Active State h cot Choose a Wedge: decide the values of h and 


W Wcos  P

Psin 

Wsin  Pcos 



F

h

N

N = W cos + P sin F = W sin - P cos

W  1   h 2 cot  2

Example p 1: Coulomb’s Ka Derivation 1 N = W cos + P sin F = W sin - P cos F W sin   P cos   N W cos   P sin 

We note that,, along g the slip pp plane,, limiting g friction must be reached so that F  tan t  N 

W sin   P cos   tan t  W cos   P sin 

Example p 1: Coulomb’s Ka Derivation 2 W sin   P cos   tan t  W cos   P sin 

tan   tan    W tan(     ) P W 1  tan  tan  

W  1 2  h cot  2

P  1 2  h 2 cot  tan(     )

 2 1  P  h tan(   ) tan(     ) 2 2

tan( A  B ) 

tan A  tan B 1  tan A tan B

Example p 1: Coulomb’s Ka Derivation 3 • We have arbitrarily assumed that the slip plane is inclined at an angle  to the horizontal horizontal. • There may be numerous potential slip planes inclined at all angles but only y 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 P. • To evaluate this value of c, we equate the first derivative of P to zero, that is   dP 1      ' h 2   sec 2     tan  '   tan    sec 2   '   0  2 d 2  2    which, upon solving gives

c 

 4



 2

 45  

 2

Substituting this critical value of  gives

   P  1 2  ' h 2 tan 2    4 2

Example p 1: Coulomb’s Ka Derivation 4 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+h and inclined at the same critical value of c. h cotc W P h P + P c

h

Example p 1: Coulomb’s Ka Derivation 5 The total force on the back of the wall arising from the slightly l lower slip li plane l (and ( d slightly li htl larger l wedge) d ) is i Q + Q, Q where h Q is i given by

  dP 2  P  h   h tan   h 4 2  dh

Recall

   P  1  ' h 2 tan 2    2 4 2 dP   2    h tan    dh 4 2 

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

P   2    h tan      ha h 4 2 

Example p 1: Coulomb’s Ka Derivation 6 2



sin      '   4 2 tan 2      4 2  cos 2         4 2  



Thus

ha

1  sin  ' 1  sin  '

     2  4

=  h tan 2 

  1  cos         2  sin    2 4 2

cos 2 A  1  2 sin 2 A cos 2 A  2 cos 2 A  1

  1  cos             2  cos2     1  sin i 2    2 4 2 4 2   cos      sin   2 

1  sin  ' = ’h 1  sin  '

= v

1  sin  ' 1  sin  '

= Kav

Hence, Coulomb’s earth pressure theory with the prescribed failure solution for active earth p wedge g g gives the same ____________ pressure as the Rankine’s earth pressure theory.

Coulomb – Passive State

Example 2 – Cohesionless Soil, Smooth Wall, Level Ground Ground, Passive State h cot

W Wcos 


P Psin 

Wsin  Pcos 



F

h

N

N = W cos + P sin F = P cos - W sin

W  1   h 2 cot  2

Example p 2: Coulomb’s Kp Derivation 1 N = W cos + P sin F = P cos -W sin

F P cos   W sin   N W cos   P sin  We note that, along the slip plane, limiting friction must be reached h d so that th t

F  tan   N

P cos   W sin    tan   W cos   P sin 

Example p 2: Coulomb’s Kp Derivation 2 P cos   W sin   tan   W cos   P sin  ttan   tan t  P W  W tan(     ) 1  tan  tan  

W  1 2  h cot  2

P  1 2  h 2 cot  tan(     )



P  1  h 2 tan(   ) tan(     ) 2 2

tan( A  B ) 

tan A  tan B 1  tan A tan B

Example p 2: Coulomb’s Kp Derivation 3 Differentiating and solving gives

 This gives

 4



 2

 45  

 2

 2 2 1 P  2  ' h tan    4 2     sin2      '   4 2   1  sin  ' tan2      4 2  cos2       1  sin  '   4 2

  1  cos         2  sin    2 4 2

  1  cos             2  cos2     1  sin2     2 4 2 4 2   cos       sin((   )   sin((   ) 2 

= KP

dP 1  sin  ' 1  sin  ' Thus hp = = ’h = v = KP v dh 1  sin  ' 1  sin  '

Once again, for the passive case, the Rankine’s and Coulomb’s earth pressure calculations return the same solution.

Examples 1 & 2: Coulomb’s Ka and Kp Derivation (Summary) Note that Coulomb’s method is based on the assumption of a ____________ ) failure wedge ((or failure mechanism). It is very important that some form of ___________ optimization be undertaken as part of Coulomb’s method, at least within the choice of slip surface, such as finding the critical value of . This should be considered an integral part of Coulomb’s calculations. calculations We can think of Coulomb’s calculation loosely as an upper unsafe solution optimistic or ______ bound. It will tend to give an _________ (since we are obtaining our solution from a scenario that already involves a failure mechanism). smooth wall Interestingly, we note that, with the ___________ assumption, Coulomb’s earth pressure theory with the prescribed failure wedge gives the same solution for active and passive earth pressure as the Rankine’s earth pressure theory.

Examples 1 & 2: Coulomb’s Ka and Kp Derivation (Summary) Recall that, Rankine’s earth pressure theory gives a lower bound to the correct failure load. In other words, Rankine’s earth p pressure theory y g generally y g gives what can be considered as a safe or pessimistic solution. Since Rankine’s method gives a pessimistic or safe f solution (lower ( bound)) whereas Coulomb’s method gives an optimistic or unsafe solution (upper bound), the fact that both solutions are identical implies p that,, in this special p case,, Rankine’s and Coulomb calculations correct solution. produce the _______

Failure load

Upper bound Correct Solution Lower bound

Examples 1 & 2: Coulomb’s Ka and Kp Derivation (Summary) This is not unexpected, as in the Rankine’s method, we h have specified ifi d a stress field fi ld which hi h is i triaxial i i l (i.e. (i principal stresses act on vertical and horizontal planes); in such a case, case consideration of triaxial stress state through the Mohr circle will lead us to the conclusion that this p prescribed stress field is consistent with a slip p plane which is inclined at (45 + ’/2 ) to the horizontal for the active case (see next slide) and (45 - ’/2 ) to the h i horizontal t l for f the th active ti case. These are also the critical angle of the slip plane used in Coulomb’s calculations. In more general scenarios, i we may nott be b so lucky. l k

Consideration of Wall Friction in Coulomb’s Method  Note that most walls are not smooth, and there will likely be wall friction between the back of the wall and the retained soil.  The presence of wall friction will introduce shear stresses along the back of the wall. Hence, Rankine’s theoretical approach pp is no longer g applicable. pp  Coulomb’s method of considering failure wedges can incorporate wall friction. friction  However, the approach just discussed in Examples 1 and 2 using geometry/trigonometry can be quite cumbersome when wall friction is considered.

Consideration of Wall Friction in Coulomb’s Method  An alternative way of evaluating the active/passive force on the wall is via the use of force polygons. A force polygon is a closed polygon whose sides taken in order represent in magnitude and direction a system of forces in equilibrium.  In the Coulomb method, method the force polygon is obtained by considering the forces acting on the assumed failure wedge g behind the wall.  As the failure wedge is in equilibrium under the action of the forces, forces these forces have to form a closed polygon.  The method is illustrated in the next few slides. slides

Coulomb Active Force via Force Polygon: Non-vertical rough wall and sloping ground surface (cohesionless soil soil, soil friction angle =  ’, wall friction angle = )

Forces Acting on the Failure Wedge as per Coulomb (for cohesionless soil with friction angle  and wall friction )



W N R 





assumed

P 

F

Using the Force Polygon Approach to solve C l b’ W Coulomb’s Wedge d

(for cohesionless soil with friction angle  and wall friction )

    

P

180 180- 

W P = force on the soil wedge due to retaining wall = force f acting ti on the wall

R

  





Coulomb Cou o b Active ct e Force o ce via a Force o ce Polygon: o ygo Non-vertical rough wall and sloping ground surface (soil with friction angle = ,  cohesion = c; wall with friction angle = , adhesion = cw ) This is a more general condition where (a) the soil strength is characterized using friction angle  and cohesion c (b) the shear resistance along the wall is characterized using wall friction angle  and wall adhesion cw .

Forces Acting on the Failure Wedge as per Coulomb (for soil with strength parameters cc,  and wall friction cw, )

Soil: soil = c +  tan  Wall: wall = cw +  tan 



Lb

Fc = c La

W

Fcw = cw Lb

Fc N

Fcw



 P

La

assumed



R  F

Using the Force Polygon Approach to solve C l b’ W Coulomb’s Wedge d

(for soil with strength parameters c,  and wall friction cw, )    

P

R W P = force on the soil wedge due to retaining wall = force f acting ti on the wall

Fcw Fc

Coulomb Active Force via Force Polygon: Non-vertical rough wall and sloping ground surface (cohesive soil with undrained strength = su ; wall with adhesion = cw ) This is for the analysis of a cohesive soil under undrained condition where (a) the soil strength is characterized using the undrained shear strength su (b) the shear resistance along the wall is characterized c a acte ed us using g wall a ad adhesion es o cw .

Forces Acting on the Failure Wedge as per Coulomb (for cohesive cohesi e soil with ith undrained ndrained strength su and wall all adhesion cw )

Soil: soil = su Wall: wall = cw

La



Lb

Fc = su La

W

Fcw = cw Lb Fcw

 assumed

P 

Fc

R

Using the Force Polygon Approach to solve C l b’ W Coulomb’s Wedge d (for cohesive soil with undrained strength su and wall adhesion cw )   

P

R W P = force on the soil wedge due to retaining wall = force f acting ti on the wall

Fcw Fc

Use of Force Polygon in Coulomb’s Method  Note that the force polygons shown on the previous few slides are obtained for an arbitrary or assumed slip plane failure fail re angle .   The force polygons have to be repeated for different trial angles of ,  until the critical value of the active/passive force (P) is obtained.  Again Again, this can be quite cumbersome as it requires repeated graphical construction of the force polygon to obtain the critical value of P.  There is a ‘clever’ graphical technique called the Culmann method which can help p reduce the effort,, which we will cover shortly.  In addition, some closed closed-form form solutions of Coulomb Coulomb’s s active/passive forces are available for special cases.

Closed Form Coulomb Co lomb Solutions Sol tions for Special Cases

Pa , Ka for non-vertical rough wall and sloping ground surface (f cohesionless (for h i l soilil with ith ffriction i ti angle l  and d wallll ffriction i ti )

Active Condition



Soil  Wall 


H  Pa 

Earth Pressure 2-174

Pa , Ka for non-vertical rough wall and sloping ground surface (f cohesionless (for h i l soilil with ith ffriction i ti angle l  and d wallll ffriction i ti )

1 The solution for this is given by Pa  Ka H2 2     sin   '  / sin    where Ka =  sin   '  sin '     sin      sin      

2

Note that Pa does not act horizontally. The horizontal component Pah of Pa is given by

1 Pah  KahH2 where 2 Earth Pressure 2-175

Kahh = Ka sin (+)

Pa , Ka for non-vertical rough wall and sloping ground surface (for soil with strength parameters cc,  and wall adhesion cw, wall friction )

Active Condition



Soil c ,  Wall cw , 


H  Pa 


Pa , Ka for non-vertical rough wall and sloping ground surface (for soil with strength parameters cc,  and wall adhesion cw, wall friction )

1 The solution for this is given by Pa  Ka H2  KaccH 2     sin   '  / sin    where Ka =  sin   '  sin '     sin      sin      

2

cw   and Kac = 2 K a 1  c   Note that Pa does not act horizontally. The horizontal component Pah of given by y Pa is g

Pah = Pa sin (+)


Pp , Kp for non-vertical rough wall and sloping ground surface (f cohesionless (for h i l soilil with ith ffriction i ti angle l  and d wallll ffriction i ti )

Passive Condition



Soil  W ll  Wall


H Pp  


Pp , Kp for non-vertical rough wall and sloping ground surface (f cohesionless (for h i l soilil with ith ffriction i ti angle l  and d wallll ffriction i ti )

1 The solution for the passive case is given by Pp  Kp H2 2     sin    '  / sin   in which Kp =  sin    '  sin  '       sin       sin      

2

Note that Pp does not act horizontally. The horizontal component Pph of P is given by

1 Pph  KphH2 where Kph = Kp sin (+) 2 Earth Pressure 2-179

Pp , Kp for non-vertical rough wall and sloping ground surface (for soil with strength parameters cc,  and wall adhesion cw, wall friction )

Passive Condition



Soil c ,  Wall cw , 

Direction of wall movement Pp



 


H

Pp , Kp for non-vertical rough wall and sloping ground surface (for soil with strength parameters cc,  and wall adhesion cw, wall friction )

1 2 The solution for this is given by Pp  Kp H  KpccH 2       sin    ' / sin  where Kp =   sin    '  sin  '     i        sin sin      

2

 cw  and Kpc = 2 Kp 1  c   Note that Pp does not act horizontally. The horizontal component Pph of Pp is given by

Pph = Pp sin ((+) +)


Coulomb’s Summary Method y of forCoulomb’s More Complex Method p Geometries 

 For F problems bl involving i l i non-vertical ti l walls and sloping backfills like those shown on the right, g , we can use the analytical (closed-form) solutions provided in the preceding slides slides.

wall

 For more geometrically complex problems with non-even non even backfill like those shown on the right, closed-form solutions are not readily available.

wall

soil



soil



Coulomb’s Summary Method y of forCoulomb’s More Complex Method p Geometries  For geometrically more complex problems, problems Coulomb Coulomb’s s method using the force polygon method is rational, but is quite cumbersome.  Remember that it has to be repeated for different trial failure surfaces to obtain the wedge corresponding to the largest Pa or smallest Pp .  A ‘‘simpler’, i l ’ more efficient ffi i t method th d (without ( ith t resorting ti to t computer) is the Culmann’s method.

Culmann’s Culmann s method graphical  Culmann’s method of solution is essentially a clever _________ implementation of the Coulomb Coulomb’s s wedge calculation, calculation which allows several trial surfaces to be tried within a reasonable time.  If we rotate the force triangle g for the failure wedge g so that the reaction R is aligned along the trial slip plane, the self-weight vector W will be inclined at an angle of ’ to the horizontal.

Culmann’s Graphical p Method of Solution Note that, in order for the angle between R and N to be ,  it is necessary for cc’ = 0. 0



Therefore, Fc = 0 (in the earlier slide).



W

Direction of wall movement P

180-

 P

 

W R 

F



R

 N F

Culmann’s Culmann s method graphical  Culmann’s method of solution is essentially a clever _________ implementation of the Coulomb Coulomb’s s wedge calculation, calculation which allows several trial surfaces to be tried within a reasonable time.  If we rotate the force triangle g for the failure wedge g so that the reaction R is aligned along the trial slip plane, the self-weight vector W will be inclined at an angle of ’ to the horizontal.  In other words, words by aligning 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 ’ to the horizontal.  Furthermore, the angle between the wall reaction P and the selfweight W is 180--, which is independent of the inclination of the trial surface .  Thus, all the wall reactions will also be aligned along the same direction.  Hence, by appropriately rotating the force triangle, the graphical solution can be greatly speeded up.

Using Culmann’s Method For Active Pressure Condition


W1

R1

Weight Line W

P1 W1





P Line


W2

Weight Line W

P2 R2 W2





P Line

Using Culmann’s Method For Active Pressure Condition Failure Plane

R4

R3 R2 R1

Culmann’s Line Weight Line W

P2 P1 W2 W1





P Line

Applying pp y g Culmann’s method to a cantilever wall

W

W1

1. 2.

3. 4.

5.

6. 6 7.

8.

scale Draw the retaining wall, backfill etc.. to a convenient _______________. From point A at the heel of the wall, project line AC at an angle of inclination of ’ to the horizontal. This will be the line along which all the self-weight vectors W will be aligned. From point A, project line AD at an angle of 180 180--   to line AC. All the wall reaction vectors P will be aligned parallel to this line. For each trial wedge, compute the self-weight W1, W2 etc.., and scale off these weights on line AC using a _______________ i t scale l for convenient the rotated force triangle. Through each end point w1, w2 etc.. corresponding to each selfweight vector, draw lines parallel to AD so as to intersect their corresponding trial slip planes. Draw a _____________ intersection smooth curve through the points of intersection. 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 i the is h maximum, i thereby h b giving i i the h maximum i wall ll reaction. i Draw a line through the tangent point that is parallel to AD to intersect the line AC. The length g of this line g gives the maximum ________ wall reaction corresponding to the critical slip plane ___________

Summaryy of Culmann’s Method N Note t th thatt C Culmann’s l ’ method th d is i a graphical hi l approach h to t solve for the forces acting on Coulomb’s wedge. Hence it is still based on Coulomb Coulomb’s s method method.  Culmann’s method is quite versatile in that it can solve for different problem geometries (sloping backfill, sloping backface of wall, and also consider wall friction).

Summaryy of Coulomb’s Method  Note that Coulomb’s method for calculating the active and passive forces accounts for wall friction, friction which is present in almost all practical cases. Rankine’s method, though more elegant mathematically, assumes a smooth wall.  Coulomb’s method is usually associated with a linear failure plane behind the wall. Strictly speaking, the method can be applied to non-linear failure planes, but historically Coulomb assumed a linear failure plane. historically, plane  Note that Coulomb’s (and Rankine’s) solutions are approximate solutions solutions, whose accuracy depends on how closely the assumed linear failure plane matches the failure surface in reality.  How do realistic failure surfaces look like?

Realistic Failure Surfaces The failure surfaces behind the walls are affected by: 1. Direction of Wall Movement 2 Di 2. Direction ti off Soil S il Movement M t

Realistic failure surfaces Active Condition • • • •

Wall moves away from soil mass down relative to wall Soil moves ___________ upwards PA inclined ________ +ve  is ____

Passive Condition • • • •

Wall moves towards soil mass moves up relative to wall Soil _________ downwards PA inclined ___________  is +ve ___

Realistic failure surfaces Active Condition • • • •

Wall moves away from soil mass up relative to wall Soil moves _________ downwards PA inclined ___________ -ve  is ___

Passive Condition • • • •

Wall moves towards soil mass moves down relative to wall Soil ___________ upwards PA inclined ________  is -ve ___

Planar vs Non-planar p Failure Surfaces  In Coulomb Coulomb’s s original computations, computations the failure wedges were assumed to be bounded by plane surfaces. This assumption is not unduly unrealistic active i for ______ failures where the actual failure surfaces are relatively flat ___ curves.  In passive failures, the assumption of flat plane g errors in surfaces of failure often leads to large __________ 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 ’ 35 exceeding ___.

How to obtain Ka and Kp for More Realistic Non-Planar Non Planar Failure Surfaces

Ka and Kp for More Realistic Non-Planar Failure Surfaces  Terzaghi (1943) extended the Coulomb earth pressure theory to accommodate a failure surface geometry consisting of log log-spiral spiral and linear segments. segments

Ka and Kp for More Realistic Non-Planar Failure Surfaces C Caquott and d Kerisel K i l (1948) made d further f th developments d l t to the non-linear failure plane theory, and produced charts/tables for Ka and Kp values based on the log logspiral + linear failure surface.

Jean Kerisel Albert Caquot

Non-planar p Failure Surfaces  Caquot and Kerisel (1948) produced tables of earth pressure based on non-planar failure surfaces. spiral to represent the  They used a logarithmic _______________ rupture surface instead. i  Thi This modification difi ti i extremely is t l important i t t for f passive ______ earth pressure where there is soil-wall friction. E Eurocode d 7 Annex A C provides id charts h t based b d on Caquot and Kerisel’s work. These are “informative” – you are not required to use them. them

Eurocode 7 Annex C ((KA for  =0,, =90º))

Eurocode 7 Annex C ((KP for  =0,, =90º))

Eurocode 7 Annex C ((KA for =0,, =90º))

Eurocode 7 Annex C ((KA for =,  , =90º))

Summary of the Log-Spiral + Linear Segments Failure Surface Method  The log g spiral p shape p of the failure p plane has generally been borne out by experiments, and hence the Terzaghi/ Caquot & Kerisel solution is generally ll preferred f d over that th t off Coulomb. C l b  For the active pressure coefficient, the results using a logarithmic spiral rupture surface f provides negligible difference to those obtained using a planar surface. surface  For passive conditions, does the Caquot & Kerisel or the Coulomb method yield a higher value of KP ?  Where is the line of action of the resulting active / passive force based on the log-spiral log spiral failure surface?

Summary y Timeline of Earth Pressure Research

1700s

- Coulomb Method

1800s

- Rankine method

1900s

- Caquot and Kerisel (log-spiral failure surface)

Which Earth Pressure Theory?

What we have covered up to now  Basic Soil Mechanics: Effective stress, NC vs OC, Mohr Circle  Different states of lateral stresses: At-rest, t est, act active ea and d pass passive e  Different methods of evaluating active and passive stresses: Rankine Coulomb Rankine, Coulomb, Log-Spiral and their key features (similarities and differences) of these methods.

Other Loads Acting g on Retaining g Walls

Surcharge Loads Next to Retaining Structures

Effect of Uniform Surcharge g Loading g


q

ah = Kav ph = Kpv

ah = Kaq qh = Kpq

Uniform Surcharge g Loading g

Uniform surcharge loading is relatively straightforward t i htf d to t deal d l with. ith The ‘harder’ part is to deal with point, line or strip surcharge loadings.

Point Surcharge Loads (based on Terzaghi, 1954)

Point Load ((Side View))

Lateral Pressure due to Point Load (Terzaghi, 1954)

Point Surcharge Loads (based on Terzaghi, 1954)

Point Load (Plan View)

’H = H cos2 ((1.1))

Lateral Pressure due to Point Load, based on Boussinesq equation modified by experiment (Terzaghi, 1954)

Line Surcharge Loads (based on Terzaghi, 1954)

Line Load

Lateral Pressure due to Line Load, based on Boussinesq equation modified by experiment (Terzaghi, 1954)

Strip Surcharge Loads (based on Teng, 1962)

Strip Load

in radians

Lateral Pressure due to Strip Load, based on Boussinesq equation modified by experiment (Teng, 1962)

Line and Point loads (NAVFAC 7.02) (based on Terzaghi, Terzaghi 1954)

See also Gaba et al (2003) (CIRIA) Fig. 4.6. They say “the use of elastic (Boussinesq) lateral stress distribution to model a strip surcharge is not recommended unless the wall is rigid with no deformation. deformation ”

Strip Surcharge Loads

Pappin, Simpson, Felton and Raison (1985) “N “Numerical i lA Analysis l i off Flexible Retaining Walls” Proceedings of the NUMETA ‘85 Conference, Swansea, 7-11 January 1985

Strip Surcharge Loads

Georgiadis and Anagnostopoulos (1998) “Lateral Lateral Pressure on sheet pile walls due to Strip load” Journal of Geotechnical and Geoenvironmental Engineering, ASCE, Vol. 124, No. 1, 95-98

Effect of Compaction Stresses on Lateral Earth Pressures

Compaction p stresses There is a question on whether compaction stresses t need d to t be b applied. li d Thi This d depends d upon what the earth pressure is used for. F the For th assessmentt off stability t bilit off gravity it retaining structure, it is often unnecessary t consider to id compaction ti stresses t since i a higher earth pressure will cause the gravity retaining str structure ct re to mo move e for forward ard (slightl (slightly)) and, in so doing, relieve the stresses. However they are necessary for the However, assessment of structural integrity of the retaining structure e e.g. g bridge abutments abutments.

Compaction p stresses Eurocode 7 says “Measurements indicate that the additional pressures depend on the applied pp compactive p energy, gy the thickness of the compacted layers and the travel pattern of the compaction plant. Horizontal pressure normal to the wall in a layer may reduce when the next layer is placed and compacted. When backfilling is complete, the additional pressure normally acts only on the upper part of the wall.”

Compaction stress (NAVFAC 7.02) (based on Ingold Ingold, 1979)

Water Pressures due to Seepage behind the Wall

Simplified Water Pressure Distribution for Seepage Flow across a Gravity Retaining Structure

Total flow path = hw + b hydraulic h d li  hw hw  b gradient

hw

h

Pw

b

A uA

Pu

uA

  hw  u A   hw  hw  w hw  b  

 hw2  hw b  hw2   hw  b 



hw b w hw  b

  w 

Seepage p g Flow – Embedded Walls

Fl Flow Net N t

Head drop=h+i-j

Total flow path along wall =2d+h-i-j

hydraulic gradient

(h  i  j )  ( 2d  h  i  j )

Simplified Water Pressure Distribution for Seepage Flow across an Embedded Wall

Head drop=h+i drop=h+i-jj

Net pore water pressure distribution um

d-i

d-i ut

ut

Total flow path =2d+h-i-j hydraulic  i  ( h  i  j ) gradient ( 2d  h  i  j )

  (h  i  j ) ( h  i  j )  w um  ( h  i  j )  ( 2d  h  i  j )    um 

2( d  i )( h  i  j ) w ( 2d  h  i  j )

More on Water Pressures Behind Retaining Walls

Example p on Simplified p Seepage p g Calculations 1.5 m

4.5 m 7.5 m

Impermeable Layer

Before Excavation Problem Geometry Key Equation for Seepage Flow: TTotal head at any point X = Elevation Head at X + Pressure Head at X lh d i X El i H d X P H d X THx = EHx + PHx


4.5 m

3 m A

B C

Elevation Datum EH =0 EH =0

Impermeable Layer

Before Excavation Consider points A, B and C Set elevation datum l d

7.5 m


4.5 m

3 m A

B

7.5 m

C Elevation Datum EH =0 EH =0

Impermeable Layer

Before Excavation Total head at A = Elevation Head at A + Pressure Head at A + Velocity Head at A THA = EHA + PHA 9 4 5 4 5 = 9 m THA = 4.5 + 4.5

usually negligible for usually negligible for seepage problems


4.5 m

3 m A

B C


Impermeable Layer

Before Excavation Total head at B = Elevation Head at B + Pressure Head at B THB = EHB + PHB 9 4 5 4 5 = 9 m THB = 4.5 + 4.5

7.5 m


4.5 m

3 m A

B C


Impermeable Layer

Before Excavation Total head at C = Elevation Head at C + Pressure Head at C THC = EHC + PHC 9 3 6 = 9 m THC = 3 + 6

7.5 m

Example p on Simplified p Seepage p g Calculations

3.0 m 1.5 m 4.5 m

3 m A

3.0 m

B

6.0 m

C

Note change in N h i water table level After Excavation in front of wall in front of wall At Point C Impermeable Layer

At Point A

At Point B

THA = EHA + PHA

THB = EHB + PHB

THC = EHC + PHC

THA = 4.5 + PH 4 5 PHA

THB = 4.5 + PH 4 5 PHB

THC = 3 + PH 3 + PHC


3.0 m 1.5 m 3 m

4.5 m

A

3.0 m

B

6.0 m

C

Impermeable Layer

After Excavation At Point A

At Point B

At Point C

THA = EHA + PHA

THB = EHB + PHB

THC = EHC + PHC

THA = 4.5 + PH 4 5 PHA

THB = 4.5 + PH 4 5 PHB


??

??

??

??

??

??


3.0 m 1.5 m 4.5 m

3 m A

B

3.0 m 6.0 m

C

Impermeable Layer

Unknowns are THA , PHA , THB , PHB , THC , PHC In fact, if we can work out THA , THB , THC , the pressure heads PHA , PHB , PHC can be obtained. The values of TH The values of THA , TH THB , TH THC can be estimated by assuming a uniform can be estimated by assuming a uniform head loss with length along the retaining wall.

Example p on Simplified p Seepage p g Calculations X

4.5 m

3.0 m 1.5 m Y

3 m A

B

3.0 m 6.0 m

C

Impermeable Layer

What is the flow length of a particle along the wall (from Pt X to Pt Y)? 45 3 75 Flow path length = l hl h 4.5 + 3 = 7.5 m What is the head loss as the particle flows from point X to point Y? THX = EHX + PHX = 7.5 + 1.5 = 9 m THY = EHY + PHY = 6 + 0 = 6 m

THX‐Y = THX – THY = 9 ‐ 6 = 3 m


4.5 m

3.0 m 1.5 m Y

3 m A

B

3.0 m

C

Impermeable Layer

Hence, the head loss gradient (assuming linear) : Head loss gradient = dl di Head loss / flow path length dl / fl hl h = THX‐Y / 7.5 m = 3 / 7.5 = 0.4

6.0 m


4.5 m

3.0 m 1.5 m Y

3 m A

B

3.0 m 6.0 m

C

Impermeable Layer

Hence, total head loss from X to A = 3 x 0.4 = 1.2 m total head loss from X to C = 4.5 x 0.4 = 1.8 m total head loss from X to C = 45x04=18m total head loss from X to B = 6 x 0.4 =2.4 m total head at X – head loss X to A = 9 – 1.2 = 7.8 m , Hence, new total head at A = new total head at C = total head at X – head loss X to C = 9 – 1.8 = 7.2 m new total head at B = total head at X – head loss X to B = 9 – 2.4 = 6.6 m


3.0 m 1.5 m 3 m

4.5 m

A

3.0 m

B

6.0 m

C

Impermeable Layer

At Point A

At Point B

At Point C

THA = EHA + PHA

THB = EHB + PHB

THC = EHC + PHC

THA = 4.5 + PH 4 5 PHA

THB = 4.5 + PH 4 5 PHB


?? 7.8

?? 3.3

?? 6.6

?? 2.1

?? 7.2

?? 4.2


3.0 m 1.5 m 4.5 m

3 m A

B

3.0 m 6.0 m

C

Impermeable Layer

Hence, pore pressure at A due to seepage = PHA x w = 3.3 x 10 = 33 kN/m2 pore pressure at B due to seepage = PHB x w = 2.1 x 10 = 21 kN/m2 pore pressure at C due to seepage = PHC x w = 4.2 x 10 = 42 kN/m2 Compare with hydrostatic values (if ignore seepage) 60 kN/m2 2 2 uC = uB = 15 kN/m uA = 45 kN/m 30 kN/m2

Simplified Water Pressure Distribution for Seepage Flow across an Embedded Wall

Head drop=h+i drop=h+i-jj

Net pore water pressure distribution um

d-i

d-i ut

ut

Total flow path =2d+h-i-j hydraulic  i  ( h  i  j ) gradient ( 2d  h  i  j )

  (h  i  j ) ( h  i  j )  w um  ( h  i  j )  ( 2d  h  i  j )    um 

2( d  i )( h  i  j ) w ( 2d  h  i  j )

General Expression for umax behind the wall for the condition shown below X b

a

Y

Z Impermeable Layer Impermeable Layer

hydraulic gradient 

b change in head between back and front of wall  total flow path length 2a  b

Take Elevation Head (EH) to be zero at Point Z At Point X, EH(X) = a + b

PH(X) = 0

TH(X) = EH(X) + PH(X) = a + b

At Point Y, EH(Y) = a

PH(Y) = ?

TH(Y) = EH(Y) + PH(Y) = a + ?

At Point Z, EH(Z) = 0

PH(Z) = ?

TH(Z) = EH(Z) + PH(Z) = ?

General Expression for umax behind the wall for the condition shown below At Point Y:

X

H dl Head loss = HXY b

= hydraulic gradient x distance XY b b2  .b  2a  b 2a  b

a

TH(Y) = TH(X) – HXY b2  a  b   2a  b 2a 2  3ab  b 2  b 2 2a 2  3ab   2a  b 2a  b

Z Impermeable Layer p y

Compare with earlier figure:

PH(Y) = TH(Y) – EH(Y) 2a 2  3ab  a 2a  b

2a 2  3ab  2a 2  ab   2a  b 2ab  2a  b

Y

a = d – i b h i j b = h + i – Substitute into equation for PH(Y) PH(Y) =

2ab 2a  b



2d  ih  i  j 2d  h  i  j

Multiply PH(Y) by w to obtain the pressure.

Some common pore pressure distributions related to earth retaining walls

Pore Water Pressure Distribution 1

• Water table levels same on both sides of wall • Hydrostatic pressures • Balanced on both sides • Do not have to be considered in calculations

Pore Water Pressure Distribution 2 • Water table levels different on both sides of wall • Distribution unbalanced • Can be rigorously determined from flow nets

uc

• Approximate solution based on assumption ti that th t total t t l head h d is dissipated uniformly along back and front wall surfaces between two water table levels. • Maximum net pressure occurs opposite the lower water table level

2ba b uc  w 2b  a

Pore Water Pressure Distribution 3

• A depth of water (c) in front of wall • Approximate distribution DEFG may be used water

uG 

2b  c a 2b  c  a

w

Pore Water Pressure Distribution 4 • Wall constructed mainly in a soil of high g permeability y

high permeability

• Lower part of wall penetrates a layer of clay of low permeability • Assume undrained conditions in the clay • Pore pressures in the overlying soil would be hydrostatic

low permeability

• Net pressure distribution would be HJKL

Pore Water Pressure Distribution 5 • Wall constructed in a clay which contains thin layers of fine sand or silt

clay sand / ilt /silt

• Assume that sand or silt clay allows water at hydrostatic sand pressure to reach the back /silt surface of wall • Implies pressure in excess of hydrostatic in front of wall • Consequent upward seepage in front of the wall

These simplified water pressure diagrams will be useful later on, especially when we consider embedded flexible walls.

Soil Parameters for Earth Pressure Calculations

Soil Parameters for Coulomb’s Method and Caquot & Kerisel’s Method

Undrained Shear Strength cu or su Soil friction angle ’ Wall friction  Wall adhesion ca

Interpretation p of Undrained Shear Strength g • Using SPT data

(cu or su  5N)

• Using CPT data • Using field vane shear test • Lab Test (UU or CU)

Mohr Circles from UU tests

Highly empirical!! i i l!!

Interpretation p of Undrained Shear Strength g • Using SPT data

(cu or su  5N)

• Using CPT data • Using field vane shear test • Lab Test (UU or CU)

Mohr Circles from UU tests

average? ?

Highly empirical!! i i l!!



Friction Angle g ’ from CU Tests Mohr Circles

Mohr Circles from CU tests using different initial effective stresses  (kPa)

’

i1

f1 i2

i3

f2

f3  (kPa)

Shear Strength of Soils : Angle of Internal Friction

Typical Values of the Friction Angle ’ Gravel G Gravel (sandy) ( )

Friction Angle ’ 40 – 55 35 – 50

Sand Loose dry Loose saturated D Dense d dry Dense saturated

28 – 34 28 – 34 36 – 45 45 36 – 45

Silt or silty sand Loose Dense

27 – 30 30 – 35

Clay General Singapore marine clay

19 – 27 ~22

CD - 7



Wall friction  California Trenching and Shoring Manual (http://www dot ca gov/hq/esc/construction/manuals/TrenchingandShoring/) (http://www.dot.ca.gov/hq/esc/construction/manuals/TrenchingandShoring/)

Wall friction  (continued)

Wall Friction (Eurocode 7)  A concrete wall or steel sheet pile wall supporting sand or gravel may be assumed to have a design wall ground interface parameter d = kk. cv;d.  For precast concrete or steel sheet piling k should not exceed 2/3 2/3.  For concrete cast against soil, a value of k = 1.0 may be assu ed assumed.  For a steel sheet pile in clay under undrained conditions immediatelyy after driving, g no adhesive or frictional resistance should be assumed. (Increases in these values may take place over a period of time.)

Wall friction  (continued) Note that, in Plaxis: W can model We d l interface i t f friction f i ti using i a parameter t Rinter

Rinter = tan  / tan 

 /

H Hence Rinter is i nott equivalent i l t to t k. k



Wall Adhesion ca

Gaba et al. (2003) (CIRIA) suggest ca=  su (0.5)

Wall Adhesion ca

R Reported t d adhesion dh i ffactors t  from f different diff t studies t di

Taken from Cherubini and Vessia (2007)

Wall Adhesion ca

How to estimate adhesion factor 

Alternatively

  'vo     0.5    su 

0.45

from Sladen (1992)

Taken from Cherubini and Vessia (2007)

What we have covered up to now  Basic Soil Mechanics: Effective stress, NC vs OC, Mohr Circle  Different states of lateral stresses: At-rest, active and passive and how they are affected by wall ll movements t  Different methods of evaluating active and passive stresses: Rankine, Coulomb, Log-Spiral and their key features (similarities and differences) of these methods.  Additional stress components acting on retaining walls Surcharge, line and point loads, compaction stresses, water pressures  Choice of Soil Parameters for Lateral Stress Evaluation F i ti Friction angle, l undrained d i d shear h strength, t th wall ll friction, f i ti wall ll adhesion

Learning Objectives Different States of Lateral Earth Stresses Insitu, Active, Passive

Methods M h d off calculating l l i active i and d passive i earth pressures Rankine, Coulomb, Caquot & Kerisel

Key features/limitations K f t /li it ti off th the diff differentt limiting earth pressure methods Effect of surcharge, compaction, etc

Short-Term and Long-Term g Response of Lateral Earth Pressures

Introduction • The short-term and long-term response of the earth g y influenced by y the presence pressures are significantly of water. • This is because the shear strength g of soils is a function of the effective stresses in the soil Effective stress = Total Stress – Pore Pressure • Hence, if the pore pressures in the short term and long term are different (say, due to consolidation), then the soil strengths in the short term and long term are also different different. • These different soil strengths will affect the ShortT Term and d Long-Term L T Stability St bilit off Retaining R t i i Wall W ll Systems.

Methods of Analysis for Short-term and Long-term Responses Undrained • Short Term • Total Stress or Effective Stress

D i d Drained • Long Term • Effective Stress

Total vs Effective Stress Analysis y  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 u = 0. In such cases the pore pressure need not be additionally cases, accounted for. This is appropriate for short-term conditions (e.g. temporary works) in low permeability soils (k < 10-8 m/s – see Gaba et al, 2003) • However However, in excavation situations, situations the long long-term 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.

Introduction to Stress Paths Recall our earlier discussion on the active pressure condition, condition and how it is related to changes in the Mohr circles

Initial Stress State of a ’ material 

Mohr-Coulomb Failure Envelope

In terms of friction parameter ’


’

 h

v

’h Complementary p y Failure envelope


’v



 h


v



 h


v



 h


v

Drawing g the stress path from the Mohr cirlces 


Draw a line through g the top p point of the Mohr Circles  h


v

Drawing g the stress path from the Mohr cirlces 


Stress Path

 h


v

Kf -line : Failure envelope (slope ) associated with the Stress Path 

 Kf-line


Kf-line is the failure line based on the top point of the Mohr Circle

Stress Path

 h


v

Introduction to Stress Paths • Previously, we have represented states of stress by a Mohr circle in a - coordinate system. • Sometimes it is convenient to represent that state of stress by a stress point, which has the (1 + 3)/2 and _________. (1 - 3)/2 coordinates _________ • F For many situations it ti in i geotechnical t h i l engineering, i i we assume 1 and 3 act on vertical and horizontal planes, so the coordinates of the stress point become (v - h)/2 ( ________ v + h)/2 and _________. • For simplicity, define

1   3 s 2

t is considered positive when v > h

1   3 t 2

Introduction to Stress Paths •

Often, we want to show successive states of stress which a test specimen (lab) or a typical element in the field undergoes during loading or unloading. unloading

•

One way is to use a diagram showing the successive states with a series of Mohr circles.

•

For example, in a triaxial test with constant 3 and increasing 1



(a) Successive Mohr circles

t

(b) Stress Path

45

D C B A

3

1



s

Relationship between soil strength parameters c’-’ and the stress path parameters c For a general failure envelope based on the c’ and ’ parameters, the equivalent stress path parameters a’ a and  ’ parameters are given as

sin ’ = tan ’ c’ = a’ / cos ’

 (kPa)

’  ’

Kf - line

c’

a’ 200

400

600

800

1000

 (kPa)

Effective vs Total Stress Paths • Stress Paths can be plotted using either effective stresses or total stresses. 1  3 1  3 t • Total stress path: s  2 2 • Effective stress path: 1  3 ( 1  u)  ( 3  u)  1   3  2u  1   3    u  su s  2 2 2 2 1  3 ( 1  u)  ( 3  u)  1   3 t    t 2 2 2

• Hence the effective stress path (ESP) is displaced laterally from the total stress path (TSP)by the pore pressure u. • Or, the horizontal distance between the ESP and TSP is the pore pressure u.

Use of Stress Paths to Illustrate the Soil Response in the Short-Term and L Long-Term T due d to t An A Excavation E ti

Stresses behind a retaining g wall

’v Excavation

Retaining wall

’h

Figure 4.12

Stress Changes due to Excavation in Clay (Total Stress)

 At the initial state,, h′ = Kov ′ (assuming installation of the wall has had no effect)

 Changes induced by excavation  v = 0 and    h < 0 (for the active case)  v   h  h s   2 2

(negative)

 v   h  h t   2 2

(positive)

t / s  1

In terms of total stress  total stress path Doesn’t say anything about pore pressure

Stress Paths due to Excavation in Clay y t Failure envelope

'

Ko

a' s,s'


'

Ko

a'

uo initial pore pressure

s,s'


Total stress 1 path -1

a'

'

Ko

uo initial pore pressure

s,s'

Stress Changes due to Excavation in Clay (Effective Stress)

 What happens to the effective stress? short term  undrained  no volume change  Volume change is related to mean effective stress s’  No volume change  no change in s’  s’ = 0  t/s’ = 

Slope of the stress path

1  3 2   3 t'  t  1 2 s' 

 effective stress path is a vertical line Strictly true only for elastic behaviour behaviour.


'

Short term - Undrained Total stress 1 path -1

a'

Effective stress path

Ko

uo

s,s'

Stress Paths due to Excavation in Clay y t '

Failure envelope

Note: N t u = pore pressure immediately after the excav uo = initial hydrostatic pp

ue uo

u

ue = excess pore pressure u

u = ue + uo

1

(-ve) (-ve) (+ve)

-1

a'

Ko uo

s,s'

Stress Changes due to Excavation in Clay (Effective Stress)  What happens to the effective stress in the long term?  Long term  drainage occurs

 excess pore pressure ue dissipates until only hydrostatic pressure uo remains  effective stress = total stress - hydrostatic pore pressure uo

Stress Changes due to Excavation in Clay (Effective Stress)  What happens to the effective stress in the long term?  Long term  drainage occurs

 excess pore pressure ue dissipates until only hydrostatic pressure uo remains  effective stress = total stress - hydrostatic pore pressure uo

In order that effective ff stress = total stress - uo , Either (a) total stress must increase or (b) effective stress must decrease

or (c) both change so that they meet halfway

???

Stress Paths due to Excavation in Clay y t

Failure envelope

' 

Short term - Undrained TS

ES

Ko

a'

uo

s,s'


Failure envelope

' 

With time TS

ES

? a'

Ko

uo

s,s'


Failure envelope

' 

With time TS

ES

? a'

Ko

uo

s,s'


Failure envelope

' 

With time TS

ES

? a'

Ko

uo

s,s'


Failure envelope

' 

With time TS

ES

Ko

a'

uo

s,s'


Failure envelope

' 

With time with time Total T t l stress path

a'

Effective stress t path

Ko

uo

s,s'


Failure envelope

long term B

A uo Total stress path

a'

short term Effective stress path

' 

Long Term - Drained Ko

uo

s,s'


 Recall that shear strength is dependent upon effective stress, not total stress.  In the short term, term due to the excavation, excavation the effective stress state is at A.  In the long term, after the pore pressure has dissipated, the effective stress state is at B.  Consider stress states A and B: Which point is closer to the failure envelope?


Failure envelope

long term B

a'

A uo

' 

excess strength capacity

short term Effective stress path

Ko

uo

s,s'


 Recall that shear strength is dependent upon effective stress, not total stress.  In the short term, term due to the excavation, excavation the effective stress state is at A.  In the long term, after the pore pressure has dissipated, the effective stress state is at B.  Consider stress states A and B: Which point is closer to the failure envelope? Hence, point B (long term) is actually closer to failure! In some cases, it may actually reach the failure line.

Short term vs Long term response d to due t Excavation E ti  Short term – undrained Reduction in h produces reduction of pore water pressure (may induce suctions). No immediate change in effective stress, so design can be based on undrained strength based on initial conditions

 Long term – drained Pore water P t pressure changes h i d induced db by excavation ti dissipate – water is drawn in and pore water pressure rises Effective stress state moves closer to failure rises. failure. Design has to be based on effective stress parameters using p g an estimation of the long g term effective stress state and pore water pressure regime

Idealized vs Realistic Effective Stress Path for Undrained Shearing t Failure envelope

'

A true only for elastic behaviour

more realistic for real soils

Ko

a' s,s'

That concludes our brief discussion on Stress Paths. If you are still interested in learning more about stress paths, paths you can sign up for CE6101 – Geotechnical Constitutive Modelling!

What we have covered up to now  Basic Soil Mechanics: Effective stress, NC vs OC, Mohr Circle  Different states of lateral stresses: At-rest, active and passive  Different methods of evaluating active and passive stresses: Rankine, Coulomb, Log-Spiral and their key features (similarities and differences)) of these methods.  Additional stress components acting on retaining walls Surcharge, line and point loads, compaction stresses, water pressures  Choice of Soil Parameters for Lateral Stress Evaluation Friction angle angle, undrained ndrained shear strength, strength wall all friction, friction wall all adhesion  Short-term Short term (undrained) vs Long-term Long term (Drained) Response Stress path concepts

2. Lateral Earth Pressure Theories

Recommend Documents