CIVE1129 Revision 2013 Part 2

5/22/2018 CIVE1129 Revision 2013 Part 2

1/93

Earth Pressures and Retaining Walls

Muni Budhu - Soil Mechanics and Foundations

2nd

edition

Dr Jie Li

School of Civil, Environmental and Chemical EngineeringRMIT University ( )

Melbourne, Australia

CIVE1129 Geotechnical Engineering 2

Revision Part 2


2/93

Importantly, at K0 state, there are no lateral strains

GL

A

v= 1

h= 3

Lateral Earth Pressure At Rest

Soil is usually deposited in the horizontal layers and consolidation

occurs solely in the vertical direction with no lateral deformation. Such

a soil mass is said to be in the at-rest condition. The ratio between the

horizontal and vertical effective stresses is a constant known as the Ko

coefficient.

' ' '

3' ' '

1

h xo

v z

K

In a homogeneous

natural soil deposit

Lateral Earth PressureCoefficient at Rest:


3/93

smooth wall

Wall movesaway from soil

Wall moves

towards soil

A

B

Lets look at the soil elements A and B during the wall

movement.

Active Earth Pressure in granular soils


4/93

A

z

xz

As the wall moves away from the soil,

Initially, there is no lateral movement.

z= v= z

x= h= K0 z= K0 z

zremains the same; and

xdecreases till failure occurs.

Active state



5/93


z

decreasing x

Initially (K0 state)

Failure (Active state)


active earth pressure

A

z

x

zWJM Rankine

(1820-1872)

' '( )x za a

K Rankines coefficient of active earth pressure

Can be derived from Geometry of Mohrs circles


6/93

21 sin tan (45 / 2)

1 sin

aK

z(x)a

Az

x45 + /2

90+

Failure plane is at

45 + /2 to horizontal

' '( )x za a

K



7/93

A

zx

z


xdecreases until failure occurs.

wall movement

x

Active

state

K0 state



8/93

1 and 3 Relation at Failure

1 3

' ' 2tan (45 '/ 2) 2 ' tan(45 '/ 2)c

3 1

' ' 2

tan (45 '/ 2) 2 ' tan(45 '/ 2)c

1 3

1 3

OF 2sinOA

cot '2

c

The failure criterion can also be expressed in terms of principal stresses

3 1

2O

F

A

c

1 3

2

1 3

2

ccot

' ' tan 'f

c

= 45o + /2


9/93

zx

245

active wedge

Active Earth Pressure

2

3 1tan (45 '/ 2) 2x z ztan (45 )

2 a

K

2 'tan (45 )

2

aK

Rankine Theory of Earth Pressure in Sand

Mohr-Coulomb failure criterion:

For sand, c=0. This leads to:

effective active earth pressure

Coefficient of active earth pressure

3 1

' ' 2tan (45 '/ 2) 2 ' tan(45 '/ 2)c


10/93

B

zx

Initially, soil is in K0 state.

As the wall moves towards the soil,

zremains the same, and

xincreases until failure occurs.

Passive state

Passive Earth Pressure in granular soils


11/93

z

Initially (K0 state)

Failure (passive state)


increasing x

passive earth

pressure


( ') 'x p P z

K

Rankines coefficient ofpassive earth pressureB

z

x

Again, from Geometry of Mohrs circles


12/93

z (x)p

Azx

90+

Failure plane is at

45 -/2 to horizontal45 -/2

)2/45(tansin1

sin1 2

PK


( ') 'x p P zK


13/93

B

zx


xincreases till failure occurs.

wall movement

x

K0 state

Passive state



14/93

1 3

' ' 2tan (45 '/ 2)

z x

2

x z ztan (45 '/ 2) pK

2tan (45 '/ 2)

pK

245

passive wedge

Passive Earth Pressure


For sand, c=0. This leads to:1 3

' ' 2tan (45 '/ 2) 2 ' tan(45 '/ 2)c

Coefficient of passive earth pressure

effective passive earth pressure



15/93

' '

1 3' '

1 3

' ' ' '

1 3 1 3

2sin '

2

f f

f f

f f f f

OBOA

Mohr-Coulomb Failure Criterion

'

12

'

3

1 sin ' 'tan 45

1 sin ' 2

f

p

f

k

'

32

'

1

1 sin ' 'tan 45

1 sin ' 2

f

a

f

k

kp = passive earth pressure coefficient ka = active earth pressure coefficient

1 3(1 sin ) (1 sin )


16/93

Normal stress

Shear stress

zoActive state Passive state

The 3 States:

At Rest


17/93

x zx

z

At-Rest

xz

Active pressure

x z

z x

xz

Passive pressure


From the at-rest

condition and with thevertical effective stress

kept constant, the soil

can fail in two ways.

First, the horizontal

effective stress may

decrease when the soil

is allowed to have

lateral extension. Thiscreates the active earth

pressure.

Second, the horizontal

stress may increasewhen the soil is

compressed laterally.

This creates the passive

earth pressure.


18/93

2

a x z a

1

2a a aP dz K dz z K dz K H

2

p x z p

1

2p p p

P dz K dz z K dz K H

x

aP

pP

x

PA and PP are the resultant active

and passive thrusts on the wall

Ha

Hp


Active thrust:

Passive thrust:

Ka HaKp Hp


19/93

Hv

uaP

h

For the case shown above:z

( )

z z h

h z h z h

)(w hzu

a x

0 0

2

( ( ))

1 1( ) (2 ( ))

2 2

H h H

a a

h

a a

P dz z K dz h z h K dz

K h K H h h H h

x zaK

2w )(

2

1hHdzuU

H

h

a aP P U

1. Calculate the vertical effective stressdistribution.

2. Calculate the effective earth pressure.

3. Calculate the effective earth thrust.

4. Calculate the pore pressure u.

5. Calculate the total force U due to u.

6. Calculate the total earth thrust Pa = Pa + U

Presence of pore pressure:


20/93

The general relation: Kp > Ko > Ka

Kp

Ka

Ko

Kp KaKo> >


21/93

x z 2a a

K c K

zxc

2tan (45 '/ 2)

aK


active wedge

245

cz

Rankine Theory for Clay


S i n c e x= 3a n d z = 1a t f a i l u r e , t h e p a s s i v e e a r t hp r e s s u r e :

3 1

' ' 2tan (45 '/ 2) 2 ' tan(45 '/ 2)c


22/93

Can soil undergo tension?

If z = 0, thenx 0Now if z = z,

At what depth willx = 0?

x

= Ka

z- 2cK

a

This depth is called the depth of cracking, zc, &

defines the potential tension zone



23/93

Depth of Cracking

By definition:

At zc, x = 0

Therefore,

x = 0 = Ka zc - 2cKa

Therefore,

zc = [2cKa][ Ka ]Or

c

a

2cz

K



24/93

x z 2a aK c K

zxc

ac

2K

cz

2tan (45 '/ 2)aK


active wedge

245

cz


The depth of tensile cracking:,

c

a x

1 2

2

H

a

z a

cP dz K H

K

Active thrust:


25/93

Active earth pressure

aP

x 3 z 1

Passive earth pressure

pP

x 1 z 3

Advantage: Simple and explicit

expressions for earth pressures and

earth thrust.

Solution Steps:

1. Find vertical stress z2. Find horizontal stress x3. Integrate x to find the effective

earth trust.

4. If a water table is present,estimate the forces due to pore

pressures.

5. Estimate the total earth thrust.

Assumptions:

Rankine Earth Pressure: Summary

Horizontal ground surface Vertical retaining wall

Initially at an at-rest state

No friction between the wall and soil

Failure occurs throughout the soil mass behind the wall


26/93

Orientation of Failure PlanesFrom Mohrs circles

Active state:

(45 + /2) to horizontal

Passive state:

(45 -/2) to horizontal

Rankine Earth Pressure: Summary

[ '] ' 2x p P z PK c K

[ '] ' 2x a a z aK c K


27/93

Quiz

1. Which requires more retaining wall rotation to be mobilized?A) Neutral resistance

B) Passive resistance

C) Active resistance

D) None of above

2. For the same soil, what is usually true?

A) Passive pressure > at-rest stress > active pressure

B) Passive pressure > active pressure > at-rest stress

C) Active pressure > passive pressure

D) At-rest stress > passive pressure


28/93

Rankine Active Earth Pressure

Determine the active lateral earth pressure on the frictionlesswall shown below. Calculate the resultant force and its location

from the base of the wall. Neglect seepage effects.

1. The active lateral earth coefficient

2 2tan (45 '/ 2) tan (45 30 / 2) 0.333o o oaK

2. The effective stresses and pore pressure

E l 12 1 R ki A ti E th P ( 495)


29/93

Example 12.1 Rankine Active Earth Pressure (p495)

3. These stresses are sketched as below:

4. The lateral force

Pa = Pa +U = 165 kN/m

Pa = x 17 x 5 = 42.5 kN/m U = Pw = x 49 x 5 = 122.5 kN/m

Since both the distributions of the effective lateral earth

pressure x and the pore pressure u are triangular overthe whole depth, the resultant is at the centroid of the

triangle, that is H/3=1.67m from the base.

5. Location of resultant

S R ki E th P


30/93

x 3 z 1 x 1 z 3

Solution Steps:

1. Find vertical stress z

2. Find horizontal stress x3. Integrate x to find the effective earth trust.

4. If a water table is present, estimate the forces due to pore pressures.

5. Estimate the total earth thrust.

Summary - Rankine Earth Pressure

pPaP

Active earth pressure Passive earth pressure


31/93

1. Vertical backs of walls only

2. Backfill surface must be regular

a solution exists for a sloping backfil l, provided slope

angle, < 3. Backfill loads / surcharge effects approximated

4. Wall friction ignored!

a beneficial effect

Limitations of Rankines earth pressure theory


32/93

Example: Rankine Earth Pressure

A 6 m high smooth retaining wall is to support a soil with unit

weight =17.4 kN/m3, friction angle = 26o and cohesion c =

14.36 kN/m2. The groundwater table is below the base of the wall.

Determine the Rankine active force per unit length of the wall, and

determine the line of action of the resultant.

(1) Active lateral earth pressure coefficient:

2

tan (45 '/ 2) 0.39

o

aK

' ' 2 ' ) 17.95x a z a

K c K kPa

2 '2.64

c

a

cz m

K

At surface: z = 0 kPa

(2) The depth of tensile cracks :

Soil cannot undergo tension!

The depth of tensile cracks


33/93

' ' 2 ' ) 22.77x a z a

K c K kPa At base: z = 17.4 x 6 =104.4 kPa

(3) Active lateral earth pressure:

The active force is the positive area of active earth pressure distribution:' 1 1

( )( 2 ' ) 22.8(6 2.64) 38.3 /2 2

a c a aP H z HK c K kN m

Location of resultant active earth pressure is (H-zc)/3 = (6-2.64)/3 =1.12m

z= 104.4 kPa

2 '2.64

c

a

cz m

K

x= 22.8 kPa

x= -18 kPa

1.12 m


34/93

For the frictionless wall shown below, determine:

1. Active lateral earth pressure distribution

2. Passive lateral earth pressure distribution

3. Magnitudes and locations of active and

passive forces

4. Resultant force and its location

5. Ratio of passive moment to active moment

Earth Pressure in Layered Soils

(1) Lateral earth pressure coefficients

2tan (45 25 / 2) 0.41o oaK

2tan (45 30 / 2) 0.33o oaK

Top layer:

Bottom layer:

2tan (45 30 / 2) 1/ 3o op aK K Passive coefficient:


35/93

(2)Active lateral earth pressure (at the back of the wall)

At surface:

x = Kaz = 0.41 x 20 kPa = 8.2 kPa, u = 0 kPa

At 2- m,z = 20+H = 20+19 x 2 = 58 kPa

z = qs = 20 kPa

x = Kaz = 0.41 x 58 = 23.8 kPa u = 0 kPa

At 2+ m,z = 20+H = 20+19 x 2 = 58 kPa

x = Kaz = 0.33 x 58 = 19.33 kPa u = 0 kPa

At base, z = 58 + H = 58 + (20 9.8) x 4 = 98.8 kPa

x = Kaz = 0.33 x 98.8 = 32.93 kPa

u = w H = 9.8 x 4 = 39.2 kPa


36/93

(3) Passive lateral earth pressure (in the front of the wall)

At surface:

x = Kpz = 3 x 0 = 0 kPa, u = 0 kPa

z = 0 kPa

At base:

z = H2 = (20 9.8) x 4 = 40.8 kPa

x = Kpz = 3 x 40.8 = 122.4 kPa

u = w H = 9.8 x 4 = 39.2 kPa

x = 8.2 kPa


37/93

(4) Active Force

x= 122.4 kPa

U = x 39.2 x 4 = 78.4 kN/m

u = w H = 9.8 x 4 = 39.2 kPa

x

x = 23.8 kPa

x = 19.3 kPa

x= 32.9 kPau = 39.2 kPa u = 39.2 kPa

' 8.2 23.8 19.3 32.92 4 136.4kN/m2 2

aP

The effective active force acting on the back of the wall is the total

area of the active lateral earth pressure distribution:

Pa = Pa + U = 214.8 kN/mTotal

4 m

2 m

x = 8.2 kPa


38/93

(4) Total moment about the base line:

x= 122.4 kPa

Mu = x (39.2 x 4) x 1/3 x 4 = 78.4 kN

u = w H = 9.8 x 4 = 39.2 kPa

x

x = 23.8 kPa

x = 19.3 kPa

x= 32.9 kPau = 39.2 kPa u = 39.2 kPa

' 1 2 1 48.2 2 (1 4) 15.6 2( 4) 19.3 4 2 13.6 4 345.6kN2 3 2 3

aM

Ma = Ma + Mu = 214.8 kNTotal

32.9-19.3 =13.6

Location of active lateral force = Ma/Pa = 450.1/214.8 = 2.1m from the base

4 m

2 m

(5) Passive Force


39/93

Total moment about the base line: Mp

= Pp

x 4/3 = 430.8 kN

Rx = Pp - Pa = 323.2 214.8 =108.4 kN/m

The negative sign indicates the location is below the base.

Ratio of moment = Mp/Ma = 430.8/450.1 = 0.96 < 1

( )

The effective passive force acting on the front face of the wall

is the total area of the passive lateral earth pressure distribution:

Pp = x 122.4 x 4 = 244.8 kN/m

Pp = Pp + U = 323.2 kN/m

U = x 39.2 x 4 = 78.4 kN/m

Location of passive lateral force = 4/3 = 1.33m from the base

Its location is:

The resultant lateral force

(Mp - Ma)/R = (430.8 - 450.1)/108.4 = - 0.18 m

(6) The moment ratio

Wall wil l rotate


40/93

TWO METHODS

to estimate the

earth pressures on structures

1. Rankine

Plausible stress states

2. Coulomb

Plausible failure mechanisms

Relative merits of approaches?

COULOMB APPROACH


41/93

HANDLES: Irregular backfill surfaces

Sloping backs of walls

Sloping backfill Surface surcharges

Wall friction

Interface friction angle, Although failure surfaces were known to be

curved, a planar approximation was adopted

(OK for Active pressures)

Coulombs Earth Pressure Theory


42/93

At failure, the stress on the slip surface through the soil obey the Coulomb

failure criterion Stresses:

ntanc

Slip surface

tanT C N

nT ds N ds C c ds

Passive wedge failure

Slip surface

Active wedge failure

Coulomb s Earth Pressure Theory

Coulomb wedge analysis is a limited equilibrium method. A collapse

mechanism is assumed, which is usually a straight-line slip surface, and theresulting forces are deducted graphically from the principle of equilibrium.

Integrating we obtain the corresponding forces according to

where

n



43/93

H

Failed soil wedge

T ds

Free-body diagram of failed soil wedge

Pa

W

T

NnN ds

H cot

sand

W = weight of soil wedge - needs support

N = force normal to sliding plane from underlying supporting soil

T = tangential force along sliding plane = N tanPa = Max reaction from wall required for equilibrium for critical

wedge angle,




44/93

T ds H

Failed soil wedgeFree-body diagram of failed soil wedge

Pa

W

T

NnN ds

H cot

Fx = Pa + T cos- N sin= 0

Fy = W - T sin- N cos= 0W = H2 cot

T = N tan

sand

Pa = H2

cot tan(-)Solving for Pa, The critical value of which makes the wall force Pa a maximum is

given by solving Pa/ = 0. This leads to


= 45

o

+ /2

Influence of Wall Friction


45/93

Influence of Wall Friction

PaActive

Pp

Passive

angle of frictionbetween wall and soil

For rough wall, = . Forsmooth wall, = 2/3



46/93

Active Earth Pressure: Sand

H

q

N

T Sand

, ',

tanNT

w wtanT N

21 cot2

W H

cotq

Q qB qH

Bq = H cot

W

Nw

Tw

Consider both Wall friction + Surface surcharges q

Pa

Nw

Tw

Pa



47/93

For sand, the analysis is done in terms of effective stress using and and c =

cw= 0. Consider a vertical wall of height of H retaining a sand backfill in level with

its coping. A surcharge q is applied to the ground surface. The water table is

assumed to below the base of the wall.

xF =0

H

q

N

T Sand

, ',

tanNT w wtanT N

21 cot2

W H cotQ qH

Two unknowns: ,wN N

w sin cos sin tan 'cos cos (tan tan ')N N T N N N

y wF =0 cos sin cos (1 tan tan ')Q W T N T N

H cot

W

Nw

Tw

Dividing 1st by 2nd equation, we have tan( ')w

w

N

Q W T

The two equations can be used to solveN and Nw, in terms of the angle .



48/93


The critical value of which makes the wall force Nw a maximum is

given by solving Nw/= 0. This leads to

2a

12

aP qH H K

2

cos 'tan1 sin ' 1

tan '

aK

2( ) tan( ') 1 cot tan( ')

1 tan tan( ') 2 1 tan tan( ')w

Q WN qH H

tan( ')( ) tan( ')( tan )w w wN Q W T Q W N

with

2

2 ' cos 'tan 452 1 sin '

o

aK

Comparing with Rankine result derived previously

when = 0 (smooth wall) we recover the Rankine expression for Ka



49/93

Ka and Kp for retaining wall

with sloping back, wall friction

and sloping soil surface.

(12.16)

(12.17)

Page 500

The above type of analysis can

also be extended for the caseof an inclined wall and an

inclined ground surface.

Rankine Theory vs Coulomb Wedge Method


50/93

Rankine Theory

At no point in the soil mass did the stress state exceed the failure

stress state.

In reality, a more efficient distribution of stress could exist

This is a lower bound solution (lower than the true solution)

Coulomb Wedge Method

A planar slip surface is assumed and the solution is obtained by

requiring force equilibrium on the soil mass In reality, a more efficient failure mechanism may be possible

This is upper bound solution (greater than the true solution)

Rankine Theory vs Coulomb Wedge Method

Attention Quiz Time!


51/93

1. Coulombs approach assumes that (click all apply)

A) The soil is rigid-plastic

B) The soil is elastic-plasticC) A slip plane occurs through the soil mass

D) The soil is homogeneous

Attention Quiz Time!

2. Coulombs approach for earth pressure is based on

A) Limit analysis

B) Finite element method

C) Limit equilibrium method

D) Lower bound solution


52/93

1) Erosion

2) Rainfall

3) Earthquakes

4) Geological Features

5) External Loading

6) Construction Activities

Summary - Causes of Slope Failure

SLOPE STABILITY

SUMMARY: KEY POINTS


53/93

Angle of repose for dry granular soils

Influence of seepage on granular soils

Slope stability for homogeneous slopes in saturated clay

- Taylors method/charts

Frictional soils more difficult

- Method of slices

Bishops method for circular slips

Slope stability programs search for failure surface with lowest FS

- Circular or non-circular slips?

Importance of shear strength parameters

- Drained and/or undrained?

- Peak, ultimate or critical state?

Slope Stability Analysis


54/93


When the ground surface is sloping, forces exist which tend to cause

the soil to move from high points to low points. The most important of

these forces are the force of gravity and the force of seeping water.

Failure will occur when the resultant shear forces exceeds the shearstrength of the soil mass.

Factor of Safety is the ratio of the available shear strength of the soil to

the minimum shear strength required to maintain equilibrium.

m

fFoS

Where

f is the shear strength of the soil

m

is the minimum shear strength required to maintain stability

Sl St bilit A l i


55/93


In stability analysis Factor of Safety is defined as the

ratio of the resisting stress to the disturbing stress

resisting stress

disturbing stressFoS

ESA Effective Stress Analysis

TSA Total Stress Analysis

Infinite slopes


56/93

h

b

Vertical

slice

W

s

Force equilibrium the slice


57/93

W

N = WN = W coss

l = length of sliding surface

s

s

WT = W sins


58/93

Stability IF, WT = T = cl + N tan

where cl = resistance due to cohesion (kN)

and N tan' = resistance due to friction

Factor of Safety (FoS) = restoring force

disturbing force

STABILITY

Restoring forceFoS N = WN = W coss


59/93

CASE 1: c = c = 0 (clean sand)

T

FoSDisturbing force

cL N tan

W

cL cos tan sin

s

s

W aW a

The natural angle of repose

W

s

s s

WT = W sins

cos tan ' tan 'FoS= FS = F

sin tan

s

s s

W a

W a a

At limit equilibrium, FS = 1

s =

CASE 1: c = c = 0 (clean sand)


60/93

The natural angle of repose

s =

the maximum angle of slope (measured fromhorizontal plane) at which loose sand will

come to rest on a pile of sand.

bCase 2: c = 0, Seepage down the slope

Pore force U on sliding base due to pore


61/93

N = WN = b h coss

' cos tan ' ' tan 'FS =

sin tan

f s

m sat s sat s

b h a

bh a a

W

s

Pore force U

h

Seepage force parallel to slope

Js = w b h i = w b h sins

(i =h/l = s)

Available shear strength:

f= N tan = b h coss tan

m = Wsins + Js = b h sins + w b h sins = ( +w)bh sins

sat

Pore force, U, on sliding base due to pore

water pressure

Case 2: c = 0,

Seepage down the slope


62/93

almost only half the FS!

' cos tan ' ' tan 'FS =

sin tan

f s

m sat s sat s

b h a

bh a a

Seepage down the slope

Effective normal force reduced less friction!


' tan ' 1tan = tan '

2s

sat

a

b

Sh t t l t bilit fi d i d

Case 3: TSA, c 0,


63/93

l = b / coss

sin ( )sin= sin cos

/ cos

s ss s

s

W a bh aTh a a

l l b a

2h =z =

sin(2 )u

s

s

a

Ws

h

Short term slope stability, find-grained

soils, Total Stress Analysis (TSA)

Shear strength on l for TSA:

s = 1/2 sin-1(2 su /h )

l


2

FS = =sin cos sin(2 )

u u u

m s s s

s s

h a a h a

zcr= hcr= 2 su /s = 45o min. value of z (or h)

zcr(or hcr) = max. slope under short term loading

Example 13.1 Infinite slope stability considering

seepage


64/93

seepage

s = ?

Example Infinite slope stability considering seepage


65/93

s = cs = 30oStep 2: Determine max. slope under dry condition

For FS =1.25, the safe slope is s

= 24.8o

Step 3: Determine max. slope under saturated condition

For sand, cs not significantly affected by the wet or dry

condition same s

Example Infinite slope stability considering seepage


66/93

' tan ' tan =

s

sat

a

s = 14o


67/93

Q1: A sand has a fiction angle of 24 . Determine the

maximum slope angle impending failure (assuming infiniteslop failure). What is the safe slope for FS =1.25?

s = = 24o s = ?(1)

o

-1 -1 os 24 =tan tan 19 6

FS 1 25tan tan .

.

(2)

Q2: Discharge from a dam cause tidal variation in the

downstream river For a sand of fiction angle of 25o and


68/93

downstream river. For a sand of fiction angle of 25 and

saturated unit weight of 18.8 kN/m3 that is subjected to thistidal variation (rapid drawdown at low tide), determine the

maximum slope angle (infinite slop failure).

' tan ' tan =s

sat

a

318.8 9.8 9 kN m

-1 os

9 =tan tan25 12 618 8

..

Method of Slices


69/93

Bishops Method (1955):

Assume circular slip plane

Consider only moment equilibrium

Neglect seepage forces

Circular Failure Surface

FBD of a slice

Bishops Method (p578)


70/93

FBD of a slice

Fy = 0

Introduce porewater

pressure ratio

Mo = 0

eq (13.18)


' tan( ')T N


71/93

FBD of a slice

Substitute

tan( ')FS =

ff j j j

m j j

T N

T T

' tan( ')

FS

j j

j

NT

Into eq 13.18

Solving for Nj

take eq (13.26)

Bishops Method (p578) FBD of a slice


72/93

Ignoring (Xj Xj+1)

1% error, gives

Bishops equation for ESA

(Effective Stress Analysis):

(13.30)

Groundwater below

surface ru = 0(13.31)


Total Stress Analysis (TSA)


73/93

FBD of a slice

cosFS =

sin

j

u jf j

m j j

b

s

W

(13.34)

f= (su)j x lj

Bishops equation for TSA:

y ( )

bj = lj cos j

FS =

sin

u jf j

m j j

s l

W

Effect of tension crack:

Modify the failure surface


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Effect of tension crack on

the slip surface

Tension crack may be filled with water (reduce FS) Provide a channel for water to reach underlying soil layers

Bishops method: Moment caused by

hydrostatic pressure

Method of Slices

Procedures to determine FS of a slope using Bishop method


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Procedures to determine FS of a slope using Bishop method.

(1) Draw the slope to scale including the soil layers

R

O

Phreatic

surface

(2) Draw a trial slip surface and identify its point of rotation (locate the

phreatic surface).

(4) Divide soil mass above the slip

surface into a number of slices. Forease of calculations, try to make as

many as possible of the same width.

About 10 slides are satisfactory for

most hand calculations

1 23

45

6

7

8

(3) If soil is clay, calculate the tension crack depth and sketch in a

possible location of tension crack.

Method of Slices



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(5) Measure b, z and etc. Make a table as shown and record b, z,

zw, and for each slice

R

O

Phreatic

surface

1 23

45

6

7

8

b

equipotential

line for slice 4

zW

z

-+

B width,

z mean height,

unit weight,zw vertical projection of the

equipotential line (porewater

pressure head)

Method of Slices



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RO

Phreatic

surface

1 23

45

6

7

8

b

equipotential

line for slice 4

zW

z

-+

(7) Divide the sum of last column by thesum of column 9 to get FS. If FS is not

equal to the assumed value, reiterate

until FS calculated and FS assumed

are same or within a small tolerance(=0.01)

(6) Calculate W = bz, ru = zw w/ z , assume a value of FS anddetermine mj from following equation or the chart shown in Fig 13.13.complete the calculations for the next 3 columns.

Example 13.3 (p585) - Bishops Method


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Homogenous soil

su 30 kPa

' 33 deg.w 9.8 kN/m3

sat 18 kN/m3

zcr 3.33 m

zs 4 m

FS 1.06 assumed

Example 13.3 (p585) - Bishops Method

(1) Draw the slope to scale


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(1) Draw the slope to scale

(2) Find depth of the tension crack

(3) Divide the slide mass into 9 slices

(4) Set up a spreadsheet

bR

(4) Set up a spreadsheet


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ESA TSA

Slice b z W=bz zw ru mj Wsin W (1 - ru)tan' mj sub/cos

m m kN m deg

1 4.9 1 88.2 1 0.54 -23 1.47 -34.5 38.3 159.7

2 2.5 3.6 162.0 3.6 0.54 -10 1.14 -28.1 54.6 76.2

3 2 4.6 165.6 4.6 0.54 0 1.00 0.0 49.0 60.0

4 2 5.6 201.6 5 0.49 9 0.92 31.5 62.1 60.7

5 2 6.5 234.0 5.5 0.46 17 0.88 68.4 72.2 62.7

6 2 6.9 248.4 5.3 0.42 29 0.85 120.4 80.1 68.6

7 2 6.8 244.8 4.5 0.36 39.5 0.86 155.7 87.6 77.88 2.5 5.3 238.5 2.9 0.30 49.5 0.90 181.4 97.5 115.5

9 1.6 1.6 46.1 0.1 0.03 65 1.02 41.8 29.6 113.6

Sum 536.6 570.9 794.8

FS 1.06 1.48

zzw

For ESA, assume a value of FS first

Taylors Method slope stability for undrained shear strength, su (or cu)


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Simple slopes

Homogeneous

- slopes are rarely homogeneous

Failure occurring by rotation

No surcharge/external loading

No open water outside the slope

Relative depth, nd

Stability number, N0

cos

FS =sin

j

u jf j

m j j

bs

W

(13.34)

Based on TSA and the following equation:

Taylors Charts (1) Calculate nd = D0 / H0


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0ou

FSN H

s

Bedrock

D0

H0

sUnit weight of soil =

Shear strength = cu

FS = FoS(2) Calculate

(3) Read s from the Charts

Example H0 = 10 m, D0 = 3 m, F = 1.25

= 18 kN/m3, su = 30 kPa(1)


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nd = D0/H0 = 3/10 = 0.3

0

1.25 (18 10)30

7.5

o

u

FSN Hs

20

(2)

(3) Read s value at theintersection of nd and No

Which of the following is not true for Taylor s method,


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a) Slope failure depends on soil typeb) Slope failure depends on soil colour

c) Slope failure depends on seepage

d) Slope failure depends on slope geometry

In Bishops method, the failure surface is

a) Any shape

b) Planar

c) Circular d) Non-circular

Which of the following is not true for Taylor s method,


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a) used for Homogeneous soilsb) Not consider surcharge

c) Rotation Failure

d) ESA (Effective Stress Analysis)

Which change can lead to slope failure?

a) addition of vegetation cover on slope surface

b) removal of material from the top of a slope

c) addition of material to the base of a sloped) rise of water table in slope material

Bearing Capacity of Soils and


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Settlement of Shallow Foundations

Muni Budhu - Soil Mechanics and Foundations

2nd edition

Dr Jie Li

VERTICAL CENTRIC LOADS


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Short-Term (TSA): qu = 5.14suscdc

Long Term (ESA):

qu = Df(Nq 1) sq dq + 0.5 B'N s d

INCLINED LOADS

Short-Term (TSA): qu = 5.14suic

Long Term (ESA):qu = Df(Nq 1) iq + 0.5 B' N i

Nq

and Nrare all functions of (effective!)

GEOMETRIC FACTORS


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Choice of Strengths


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SAND

Clean sand

c = 0 = c

, ,

CLAY

Saturated, NC

undrained (Short-Term) loading

u = 0

usually more critical

drained (Long-Term) loading

c,

Problem Solving 9.6 A square footing, 3m wide, is located 1.5mbelow the surface of a stiff clay. Determine the allowable bearing

capacity for short term condition if su 100 kPa and sat 20 kN/m3

If


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capacity for short term condition if su= 100 kPa and sat = 20 kN/m Ifthe footing were located on the surface, what would be the allowablebearing capacity? Use FS = 3.

c

B

s 1 0 2 1 0 2 1 2L. . . dc = 1+0.33 tan-1

(Df/B'), take as 1

u u c cq 5 14 s s d 5 14 100 1 2 1 617kPa. . .

Position 1 Df= 1.5 m:

u

a f

q 617q D 20 1 5 236kPa

FS 3.

Position 2 Df= 0 m:

u

a f

q 617q D 20 0 206kPa

FS 3

Problem Solving 9.7 A column carrying a load of 750 kN is to befounded on a square footing at a depth of 2 m below the ground surface

in deep clay stratum. What will be the size of the footing for FS = 3 for

ESA and TSA? The soil properties are = 28o = 18 5 kN/m3 and


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ESA and TSA? The soil properties are p = 28o, sat = 18.5 kN/m3, andsu= 55 kPa. The ground water level is at the base of the footing but it is

expected to rise to the ground surface during rainy seasons.

c

B 2 4s 1 0 2 1 0 2 1 2

L 2 4

.. . .

. dc = 1

Check both short and long-term cases

(a) Short-Term (TSA)

Assume B = 2.4 m

u u c cq 5 14 s s d 5 14 55 1 2 1 339kPa. . .

,

Applied stress = a750

130 2 kPa2 4 2 4 .. .

u

a f

q 339FS 3 0

D 130 2 8 7 2.

' . .

tan 28 2 'tan (45 / 2) 14.7q pN e

(b) Long Term (ESA ) Problem Solving 9.7


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N = 0.1054 exp(9.6p) = 11.5

u 1q 8 7 2 0 13 7 1 53 1 21 8 7 2 4 11 5 0 6 1 0 513kPa2

. . . . . . . . . .

u

a f

q 513FS 4 6

D 130 2 8 7 2

.

' . .

TSA governs, use B = 2.4 m

sq = 1 + B/Ltan(p)=1+2.4/2.4 tan (28) =1.53

Use Davis & Booker expression (p351):

dr= 1

sr= 1 0.4 B/L= 1 0.4 x 2.4/2.4 = 0.6

dq = 1 + 2tan(p)(1-sin(p))2 tan-1(Df/B) =1.21

(Table 9.1 p351)

2nd edition

1) Which analysis should be used to calculate the bearing capacity of

a fine-grained soil?,

A) TSA

B) ESA


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B) ESA

C) Both TSA and ESA

D) Finite Element Method

2) Which analysis should be used to calculate the bearing capacity of

a coarse-grained soil?,

A) TSA

B) ESAC) Both TSA and ESA

D) Finite Element Method

TSA: qu = 5.14 suscdc

ESA: qu = Df(Nq 1) sq dq + 0.5 BN s dN

qand N

rare all functions of (effective!)

Documents

CIVE1129 Revision 2013 Part 2