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Flow Nets for Anisotropic Materials

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Flow nets for anisotropic soil

k hx

k hz

H V

22

2

2 0+ =

(4)

Governing Equation

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k hx

k hz

H V

22

2

2 0+ =(4)

Governing Equation

(5b)k

k

h

x

h

z

H

V

2

2

2

2

2 0+ =

+


(5a)x x

and

z z

=

=

Transformation

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(5b)k

k

h

x

h

z

H

V

2

2

2

2

2

0+ =

= kkHV (5c)

22

2

2 0

h

x

h

z+ = (5d)

+


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x

z

Impermeable bedrock

L

H1H2

Example: Flow net for anisotropic soil

Fig. 4 Shows the dam drawn at its natural scale

Impermeable

dam

Soil laer!

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"et us assume that the soil has different hori#ontal

and $ertical permeabilities such that %&' 4 %

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Transformation

(a)




= =4 2* kk

V

V

k

k

H

V

=

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Transformation

(a)




= =

= =

=

42

22

* k

k

so

x x or x x

z z

V

V

k

k

H

V

=

(b)

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z

Impermeable bedrock

L/2

H1H2

x


Fig. 5 Shows the dam drawn to its transformed scale

Soil laer

!

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+he purpose of drawing the flow net is to determine

, -ore pressures and hence effecti$e stresses

these are determined as for isotropic soil

remember to allow for the scale in calculating forces from

the pressures

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+he purpose of drawing the flow net is to determine

, -ore pressures and hence effecti$e stresses

these are determined as for isotropic soil

remember to allow for the scale in calculating forces from

the pressures

, +he /uantit of flow

calculated using 0 ' % h as before

with % ' %e/' (1)k kH V

E i l t bilit f

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Equivalent permeability foranisotropic flow

xx

Natural scaletransformed

scale

0t

h h 2 h h h 2 h

E i l t bilit f

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Q k t h

xH=

(1a)

3onsidering hori#ontal flow we ha$e

(a) Natural scale

xx


scale

0t

h h 2 h h h 2 h

Equi alent permeabilit for

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Q k t h

xH=

Q k t h

xk t

h

x

k

keq eq

H

V

= =

(1a)

(1b)


(a) Natural scale

(b) +ransformed scale

xx


scale

0t

h h 2 h h h 2 h

Equivalent permeability for

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Q k t h

xH=

Q k t h

xk t

h

x

k

keq eq

H

V

= =

(1a)

(1b)

/uating 1a and 1b gi$es k k keq H V=


(a) Natural scale

(b) +ransformed scale

xx


scale

0t

h h 2 h h h 2 h

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Example: Seepae !nder a dam

&

' 6.7 m

&8 ' 8.5 m

% ' 72 m9s

%& ' 4 *72 m9s

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h

' 6.7 m

h8 ' 8.5 m

% ' 72 m9s

%& ' 4 *72 m9s

k meq = = " # " # $ sec4 %0 %0 2 %0

& & &(:a)

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h

' 6.7 m

h8 ' 8.5 m

% ' 72 m9s

%& ' 4 *72 m9s

k meq = = " # " # $ sec4 %0 %0 2 %0

& & &

h m=

=

" ' #

'

%( 2 )

%4 0 *)

(:a)

(:b)

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h

' 6.7 m

h8 ' 8.5 m

% ' 72 m9s

%& ' 4 *72 m9s

k meq = = " # " # $ sec4 %0 %0 2 %0

& & &

h m=

=

" ' #

'

%( 2 )

%4 0 *)

(:a)

(:b)

Q +=

" # " ' # ' $ $2 %0 0 *) % ) %0& & (

m s m

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h

' 6.7 m

h8 ' 8.5 m

% ' 72 m9s

%& ' 4 *72 m9s

k meq = = " # " # $ sec4 %0 %0 2 %0

& & &

h m=

=

" ' #

'

%( 2 )

%4 0 *)

(:a)

(:b)

Q +=

= =

" # " ' # ' $ $

' $ $ $ $

2 %0 0 *) % ) %0

& % ) , %0

& & (

( & (

m s m

th!s

Q m s m m s m

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0;I3

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Area =A

(z=z2 , h=h2, u=u2)

(z=z1 , h=h1,u=u1)

Ele-ation

.lan

u2

u1

.ipin "Q!icksand#

Soil element e*periencing upward flow of water

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!2

!%

(>a)

.ipin

/plift Force ! !

Force d!e to weiht z zsat

=

=

" #

" #

% 2

2 %

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!2

!%

(>a)

! h z

and

! h z

w

w

2 2 2

% % %

=

=

" #" #

(>b)

From the definition of head

.ipin

/plift Force ! !

Force d!e to weiht z zsat

=

=

" #

" #

% 2

2 %

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For piping uplift ? weight

.ipin

/plift Force ! != " #% 2

Force d!e to weiht z zsat= " #2 %

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(>c)


.ipin



! ! z zsat" # " #% 2 2 % >

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(>c)


.ipin



! ! z zsat" # " #2 % 2 % >

h h z z z zw w sat" # " # " #% 2 % 2 2 % >

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(>c)


.ipin



! ! z zsat" # " #2 % 2 % >

h h z z z zw w sat" # " # " #% 2 % 2 2 % >

h h z z z zw sat w" # " # " #% 2 2 % 2 % >

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(>c)


.ipin



! ! z zsat" # " #2 % 2 % >

h h z z z zw w sat" # " # " #% 2 % 2 2 % >

h h z z z zw sat w" # " # " #% 2 2 % 2 % >

h h

z z

sat w

w

" #

" #

% 2

2 %

>

(>d)

i i

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.ipin

(z=z2

, h=h2

)

(z=z1 , h=h1)

u2

u1

h h

z z

" #

" #

% 2

2 %

>

sat w

w

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E l

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Example

Suppose the dam in Figure is 64 metres wide and the

water heights are as before &' 6 m &8' 8.5 m.

-iping is most li%el to occur at the toe of the dam which

has the smallest element in the flow net and upward flow.

Now h2 h8' h ' 7.15 m ( as before)

#82 #' .85 m (scaled from Fig. )

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E ample

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Example







thus i ' ' 7.1

0 *)

%%2)

'

'

Example

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Example







thus i ' ' 7.1

and icrit' ' 7.:6

+he safet factor against piping (icrit9 i) is thus .85 which is

probabl not ade/uate considering the serious

0 *)

%%2)

'

'

%1 , 1%, 1% ''

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Total Stress, Effective Stress, PorePressure

The vertical normal stresses induced in soil duto self weight or overburden

where v = vertical geostatic stress

3 + unit weight of soil

z = depth under consideration

he -ertical eostatic stress5 th!s5 -aries linearl6 with depth in

this case

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otal stress7 he stresses ind!ced either d!e to self8weiht of the

soil or d!e to external applied forces or d!e to 9oth5 at an6 point

inside a soil mass is resisted 96 the soil rains as also 96 water

present in the pores or -oid spaces in the case of a sat!rated soil '

e!tral stress 8 he stress carried 96 the pore water' his is also

called ;pore water press!re< and is desinated 96 u.

u = w'z

Total Stress, Effective Stress, NeutralStress

+ 3sat 'z

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Effecti-e stress8 =efined as the difference 9etween the total

stress and the ne!tral stress > this is also referred to as the

interran!lar press!re and is denoted 96

= u

Total Stress, Effective Stress, NeutralStress

+ " 7 u) = sat'z w'z = z(sat w)

+ 3 ?'z

St d t Fl

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Stresses due to Flow

X

soil

hw

@

Static Situation (o !lo")

z

-+ whwA satz

! + w "hwA z#

-B+ Bz

t C5

St d t Fl

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Stresses due to Flow#o"n"ard $lo"

hw

@

flow

X

soil

z

-+ whwA satz

w hwA w"@8h@#"z$@#

-B+ Bz % "iz

t C5

h@ ! + w hw

! + w

"hwA@8h

@#

D as for static case

+ w hwA w"z8iz#

+ w

"hwAz# &

"iz

Reductiondue to flow

Increasedue to flow

! +

St d t Fl

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Stresses due to Flow

flow

'p"ard $lo"

hw

@X

soil

z

-+ whwA satz

w hwA w"@Ah@#"z$@#

-B+ Bz 8 wiz

t C5

h@

! + w hw

! + w "hwA@Ah@#

D as for static case

+ w hwA w"zAiz#

+ w

"hwAz# %

"iz

Increasedue to flow

Reductiondue to flow

! +

Q i ! diti i G l S il

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Quic !ondition in Granular Soils

=!rin !pward flow5 at C:

-B+ Bz 8 wiz

flow

hw

@X

soil

z

h@

= izw

w

B

3ritical hdraulic gradient (ic)

f i + ic5 the effecti-e stresses is zero'

i'e'5 no inter8ran!lar contact th!s fail!re'

& uick condition

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Pi i F il

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Piping Failures

+eton =am ;SA (>1)

Piping Failures

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Piping Failures

+eton =am remains after failure ;SA (>1)

Filt " i t t ! t l

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Filter "equirements to !ontrolPiping

, Filter drains are required on the downstreamsides of hydraulic structures and arounddrainage pipes# $ properly graded filterprevents the erosion of soil in contact with itdue to seepage forces# To prevent themovement of erodible soils into or throughfilters, the pore spaces between the filterparticles should be small enough to hold someof the protected materials in place#

Filt " i t t ! t l

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Filt " i t t ! t l

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Grain size distri9!tion c!r-es for raded filter and protected materials

S th h E th %

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Seepage through an Earth %am onan &mpervious 'ase

S th h E th %

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he rate of seepae per !nit lenth of the dam "at riht anles to

the cross section shown in Fi!re

S th h E th %

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