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Soil Permeability&

Seepage

Soil Permeability- Definition

٣

What is Permeability?

• Permeability is the measure of the soil’s ability to permit water to flow through its pores or voids

water

Loose soil

- easy to flow

- high permeability

Dense soil

- difficult to flow

- low permeability

Importance of Permeability

The following applications illustrate the importance of permeability in geotechnical design:

• The design of earth dams is very much based upon the permeability of the soils used.

• The stability of slopes and retaining structures can be greatly affected by the permeability of the soils involved.

• Filters made of soils are designed based upon their permeability.

٥

Importance of Permeability

٦

Importance of Permeability

• Estimating the quantity of underground seepage (Ch. 8)

• Solving problems involving pumping seepage water from construction excavation

٧

Flow rate through soil

What is the flow rate through a soil?

SOIL

Concrete dam

Flow rate =

Q [m3/sec]

٩

١٠

tank of water…

A

B

Why does water flow?

If flow is from A to B, the energyis higher at A than at B.

١٢

water

A B

Energy is dissipated in overcoming the soil resistance and hence is the head loss.

Soil Permeability -Definition

Soils consists of solid particles with interconnected voids wherewater can flow from a point ofhigh energy to a point of low energy

١٣

water

Bernoulli’s Equation

2. Kinetic energy

١٤

datum

z

fluid particle

The energy of fluid comprise of:

1. Potential energy

- due to velocity

- due to pressure

- due to elevation(z) with respect to a datum

١٥

١٦

Soil piezometer

pw hu ⋅= γ

A

hp

Pressure head at A.

The “pore water pressure” at A is

Hydrostatic pore water pressure

pwh hu ⋅= γ

zw1

zw2

1wwh zu ⋅γ=

2wwh zu ⋅γ=

Depth, z

zw

Then at any point in the fluid, the total energy is equal to

Total Energy = Potential energy + Pressure energy + Kinetic energy= mgh + P + ½mv2

Expressing the total energy as head (units of length)

Total Head = Elevation Head + Pressure Head + Velocity Head

g

vuZh

w 2

2

++=γ

Bernoulli’s Equation

For flow through soils, velocity (and thus velocity head) is very small. Therefore, v2/2g = zero.

0

٢٠

٢١

Bernoulli’s Equation

At any point

w

uZh

γ+=

The head loss between A and B

+−

+=−=∆

w

BB

w

AABA

uZ

uZhhh

γγ

Head loss in non-dimensionalform

L

hi

∆=Hydraulic gradient

Distance between points A and B

Difference in total head

Water In

Head Loss orHead Difference or Energy Loss

hA

hB

i = Hydraulic Gradient

(q)Water

out

Datum

hA

W.T.

hB

)h = hA - hB

W.T.

Impervious Soil

Impervious Soil

ZA

Datum

ZB

Water In

Head Loss orHead Difference or Energy Loss

hA

hB

A BDatum

PorousStone

PorousStone

Seepage Through Porous Media

i = Hydraulic Gradient

Soil

Waterout

L = Drainage Path

L

Water In

)h =hA - hB

Head Loss orHead Difference or Energy Loss

ZA

hB

A B

Datum

PorousStone

PorousStone

Seepage Through Porous Media

i = Hydraulic Gradient

Soil

Waterout

L = Drainage Path

L

hA

ZB

PA/γw

PB/γw

Tricky case!!

Remember always to look at total head

Hydraulic Gradient

• In the field, the gradient of the head is the head difference over the distance separating the 2 wells.

٢٧

X

HHi

∆−= 21

Also…..

Water Movement in Soil

1. Darcy’s Law

2. Continuity Equation:mass in = mass out + change in storage

Two Principles to Remember:

Darcy’s Law

Assumptions:flow is laminarsoil properties do not ∆ with time

Darcy’s Law

kiAQ =

Hydraulic gradient

Cross-sectional area to flow

Hydraulic conductivity

“permeability” [cm/s]

Since velocity in soil is small, flow can be considered laminar

v ∝ i v = discharge velocity = i = hydraulic gradientv = k i k = coefficient of permeability

k

Units are in cm/sec

but

k = velocity

Datum

hA = total head

W.T.

h = hA - hB

W.T.

Impervious Soil

Impervious Soil hB= total head

Flow in Soil

A

B

L

h

L

hhi BA ∆=−= )(

L

Ah

kAikQ L

∆==

Solution

Q = kiA

k = 4x10-2 cm/sec

i = ∆h/L = (167.3m – 165m) / 256m = 0.009

A = (3.2 m) (1000 m) = 3200 m2

Q = kiA = 0.0115 m3/sec = 41.5 m3/hr

٣٦

٣٧

• The hydraulic conductivity k is a measure of how easy the water can flow through the soil.

• The hydraulic conductivity is expressed in the units of velocity (such as cm/sec and m/sec).

Hydraulic Conductivity

k

depends on

soil fluid

Measure of a soil-fluid system’s resistance to flow

Void size

Fabric (structure)

Void continuity

Specific surface (drag)

Viscosity

Mass density

٣٩

• Hydraulic conductivity of soils depends on several factors:

– Fluid viscosity (η): as the viscosity increases, the hydraulic conductivity decreases – Pore size distribution– Temperature– Grain size distribution– Degree of soil saturation

Hydraulic Conductivity

It is conventional to express the value of k at a temperature of 20oC.

٤٠

To determine the quantity of flow, two parameters are needed

* k = hydraulic conductivity* i = hydraulic gradient

k can be determined using 1- Laboratory Testing � [constant head test & falling head test]

2- Field Testing � [pumping from wells]

3- Empirical Equations

i can be determined 1- from the head loss and geometry2- flow net (chapter 8)

٤٢

Laboratory Testing of Hydraulic Conductivity

Two standard laboratory tests are used to determine the hydraulic conductivity of soil

• The constant-head test

• The falling-head test.

٤٣

Constant Head Test

٤٤

Constant Head Test

From Darcy’s Law

Then compute:

٤٥

Constant Head Test

٤٧

Falling Head Test

• The falling head test is mainly for fine-grained soils.Simplified Procedure:

– Record initial head difference, h1 at t1 = 0

– Allow water to flow through the soil specimen

– Record the final head difference, h2

at time t = t2

Then compute:

٤٨

Falling Head Test

Limitations of Laboratory tests for Hydraulic Conductivity

i. It is generally hard to duplicate in-situ soil conditions (such as stratification).

ii. The structure of in-situ soils may be disturbed because of sampling and test preparation.

iii. Small size of laboratory samples lead to effects of boundary conditions.

٥٠

٥١

٥٢

٥٣

٥٤

Equivalent Hydraulic Conductivity on Stratified Soils

• Horizontal flow

• Constant hydraulic gradient conditions

• Analogous to resistors in series

Equivalent Coefficient of Vertical Permeability (kv’)Equivalent Coefficient of Vertical Permeability (kv’)Equivalent Coefficient of Vertical Permeability (kv’)Equivalent Coefficient of Vertical Permeability (kv’)

ikhHvAqaverage

'*** ==

iHkiHkiHkikhH nn .........'.. 2211 +++=

H

HkHkHkkh nn

......' 2211

+++=

Equivalent Hydraulic Conductivity on Stratified Soils

• Vertical flow

• Constant velocity

• Analogous to resistors in parallel

Equivalent Coefficient of Vertical Permeability (kv’) Equivalent Coefficient of Vertical Permeability (kv’) Equivalent Coefficient of Vertical Permeability (kv’) Equivalent Coefficient of Vertical Permeability (kv’)

Basic Concept

•qin = qout

•v constantn

n

n H

hk

H

hk

H

hkikvv ......'.

2

2

2

1

1

1=====

vh

kH

.....vh

kH

;vh

kH

;vh

kH n

n

n3

3

32

2

21

1

1 ================

n

n

3

3

2

2

1

1n321

kH

...kH

kH

kH

vh

...vh

vh

vh ++++++++++++++++====++++++++++++++++

HHHHHn

=++++ ...321

n

n

kH

kH

kH

kH

Hkv

++++=

...'

3

3

2

2

1

1

EXAMPLE EXAMPLE EXAMPLE EXAMPLE 2222

q

Questions :- determine h- determine q in cc/sec

Section 1Section 2

EXAMPLE EXAMPLE EXAMPLE EXAMPLE 2222

1111 A.i.kq ====

Bina Nusantara

Section 1 Section 2

2222 A.i.kq ====

25.40

50.02.01

hq

−= 25.40

5.007.02

−= hq

21 qq ====)5.(007.0)50.(02.0 −=− hh

h = 38.33 cm

Determination of h

EXAMPLE EXAMPLE EXAMPLE EXAMPLE 2222

1111 A.i.kq ====2222 A.i.kq ====

25.40

33.3850.02.0

−=q

Bina Nusantara

Determination of water flow rate

or

q = 0.15 cc/s

Determination of Hydraulic conductivity in the Field

1. Pumping Wells with observation holes

2. Borehole test.

3. Packer Test.

Pumping Well with Observation holes

Pumping Well in an Unconfined Aquifer

)(

ln.

21

22

1

2

hh

r

rq

k−

=π

q

)(

log. 303.2

21

22

1

210

hh

r

rq

k−

⋅=

π

OR

If q, h1, h2, r1, r2 are known , k can be calculated

Stress ConceptStress ConceptStress ConceptStress Concept

Stresses in SoilsStresses in SoilsStresses in SoilsStresses in Soils

0.248811

0.241673

ASi

g

81.950 ti

0 10 20 30 40 50 60 70 80 900.4

0.2

0

0.2

0.4

1. Geostatic Stresses

Due to soil’s self weight

2. Induced Stresses

Due to added loads (structures)

3. Dynamic Stresses

e.g., earthquakes

٦٧

٦٨

٦٩

Effective Stress – Spring Analogy

∗ ∗ ∗ ∗ σσσσ’ = σσσσ – u

∗ σ∗ σ∗ σ∗ σ’ = effective stress

∗ σ∗ σ∗ σ∗ σ = total stress

* u = pore pressure

X

σσσσ’

σσσσ

u

P

Geostatic StressesGeostatic StressesGeostatic StressesGeostatic Stresses

SHEAR STRESSES

If ground surface is flat, all geostatic shear stresses = zero

٧٢

٧٣

Geostatic StressesGeostatic StressesGeostatic StressesGeostatic Stresses

AA z⋅= γσ

TOTAL VERTICAL STRESS AT A POINT

z = depth = 5 m

A

Ground surface

Soil , γ = 18 kN/m3

“total vertical stress at A”

٧٥

٧٦

Pore water pressures

he uuu +=

uhydrostatic = uh = due to hydrostatic condition only

uexcess = ue = due to additional processes

Geostatic StressesGeostatic StressesGeostatic StressesGeostatic Stresses

ApwA hu ⋅= γ

PORE WATER PRESSURE AT A POINT

z = 5 m

A

Ground surface

Soil , γ = 18 kN/m3

“pore water pressure at A”

hpA

٧٩

٨٠

٨١

٨٢

٨٣

٨٤

EXAMPLEEXAMPLEEXAMPLEEXAMPLEPlot the variation of total and effective vertical stresses, and pore water pressure with depth for the soil profile shown below in Fig.

٨٥

Solution:Within a soil layer, the unit weight is constant, and therefore the stresses vary linearly.

Therefore, it is adequate if we compute the values at the layer interfaces and water table

location, and join them by straight lines.

٨٦

Pore Water Pressure

٨٧

Effective Stress – General Expression

٨٨

Methods of Computations Effective Stress

٨٩

14 ft

3 ft

12 ft

In Flow

Out Flow 2 ft

4 ft

Datum

3 ft

3 ft

8 ft

Piezometer

A

B

C

D

u =

6 x

62.

4

u =

14

x 62

.4

No Seepage

Buoyancy

Ws

Ws

Ws

Ws

Ws

٩١

Buoyancy

Ws

Ws

Ws

Ws

Ws

3 ft

4 ft

6 ft

12 ft

γ1 =110 pcf

W.T.1

2

3

4

- =

Total Stress Pore WaterPressure

No Seepage

Buoyancy

Ws

Ws

Ws

Ws

Ws

Effective Stress

3 ft

4 ft

6 ft

12 ft

γ1 =110 pcfW.T.

σ1 =

σ2 =

σ3 =

2

3

4

5

σ4 =

- =

Total Stress Pore WaterPressure

Total Stress Pore Water Pressure

2

3

4

5

No Seepage

Buoyancy

Ws

Ws

Ws

Ws

Ws

u1 =

u2 =

u3 =

u4 =

Effective Stress

σ1 =

σ2 =

σ3 =

σ4 =

1 1

σ5 = u5 = σ5 =

Effective Stress

3 ft

4 ft

6 ft

12 ft

γ1 =110 pcf

W.T.

1

3

4

5

- =

Total Stress Pore WaterPressure

No Seepage

Buoyancy

Ws

Ws

Ws

Ws

Ws

3 ft

2

Effective Stress

10 ft

3 ft

12 ft

In Flow

Out Flow

2 ft

4 ft

Datum

3 ft

3 ft

8 ft

Piezometer

A

B

C

D

u =

6 x

62.

4 -

∆u

u =

17

x 62

.4

Downward Seepage

Buoyancy - Seepage Force

Ws

Ws

Ws

Ws

Ws

Seepage Force

٩٦

Buoyancy - Seepage Force

Ws

Ws

Ws

Ws

Ws

Seepage Force

3 ft

4 ft

6 ft

12 ft

γ1 =110 pcfW.T.

1

2

3

4

- =

1

2

3

4

Total Stress Pore WaterPressure

Total Stress Pore Water Pressure

3 ft

Downward Seepage

Buoyancy - Seepage Force

Ws

Ws

Ws

Ws

Ws

Seepage ForceEffective Stress

Effective Stress

17 ft

3 ft

12 ft

In Flow

Out Flow 2 ft

4 ft

Datum

3 ft

3 ft

8 ft

Piezometer

A

B

C

D

u =

6 x

62.

4 +

∆u

∆u

u =

17

x 62

.4

Upward Seepage

Buoyancy + Seepage Force

Ws

Ws

Ws

Ws

Ws

٩٩

Buoyancy + Seepage Force

Ws

Ws

Ws

Ws

Ws

3 ft

4 ft

6 ft

12 ft

W.T.

4

Total Stress Pore Water Pressure

Upward Seepage

Buoyancy + Seepage Force

Ws

Ws

Ws

Ws

Ws

=

Pore WaterPressure

5 ftγ1 =110 pcf

1

3

54

-

Total Stress

4

2

Effective Stress

Effective Stress

3 ft

4 ft

6 ft

12 ft

γ1 =110 pcf W.T. W.T. 3 ft

4 ft

6 ft

12 ft

SEEPAGE FORCESEEPAGE FORCESEEPAGE FORCESEEPAGE FORCE

Ahhw

ALt

F ).21

.(.. −−=∑ γγ

volume

ForceTotalFForceBody =)(

TOTAL FORCE

BODY FORCE

L Soil weight = γγγγt.L.A

γγγγw . h2 . A

γγγγw . h1 . A

H

h2

L

h1

SEEPAGE FORCESEEPAGE FORCESEEPAGE FORCESEEPAGE FORCE

wbouyancy

wtwt

wt

iF

iL

LHF

AL

AhhALF

γγ

γγγγ

γγ

.

)1(

.

)..(.. 21

−=

+−=

+−=

−−=

w.i γγγγ

e11G

i

0.i

s

w

bouyantc

wbouyant

++++−−−−====

γγγγγγγγ

====

====γγγγ−−−−γγγγ

SEEPAGE BODY FORCE (j)=

CRITICAL CONDITION

γγγγbouyancy = γγγγt - γγγγw

EXAMPLE :EXAMPLE :EXAMPLE :EXAMPLE :k = 1x10-3 cm/s

n = 0.67

Questions :

1. Water Flow Rate

2. Flow Velocity

3. Seepage Velocity

4. Seepage Force at point A

• Water Flow Rate

• Flow Velocity

A.i.kq ==== 144

LH

i ============

A10x1A.1.10x1q 55 −−−−−−−− ========

i.kv ==== s/m10x11.10x1v 55 −−−−−−−− ========

� Seepage Velocity

� Seepage Force

nv

ni.k

'v ========

smxv /105.167.010*1

' 5

5

−−

==

ws .iF γγγγ====2/10001000*1 mkgF

s==

Laplace equation of Continuity

• In reality, the flow of water through soil is not in one direction only, nor is it uniform over the entire area perpendicular to the flow.

• The flow of water in two dimensional is described using Laplace equation.

• Laplace equation is the combination of the equation of continuity and Darcy’s law.

١١٠

Laplace equation of Continuity

Flow in:

Flow out:

Flow in = Flow out (Continuity equation)

By simplification, we get

١١١

dxdydzz

vvdydzdx

x

vv z

zx

x

∂∂+

∂∂+

dxdyvdydzv zx

dxdyvdydzvdxdydzz

vvdydzdx

x

vv zx

zz

xx +=

∂∂++

∂∂+

0=∂∂+

∂∂

dxdydzz

vdxdydz

x

v zx

0=∂∂+

∂∂

z

v

x

v zx

Laplace equation of Continuity

From Darcy’s Law:

Replace in the continuity equation

If soil is isotropic (i.e. kx = kz = k)

١١٢

0 2

2

2

2

=∂∂+

∂∂

z

hk

x

hk zx

dz

dhkv

dx

dhkv zzxx

==

0 2

2

2

2

=∂∂+

∂∂

z

h

x

h

Laplace equation

This equation governs the steady flow condition for a given point in the soil mass

q = A k i = A k ∆hL

Flow Lines

EquipotentialLines

Equipotential Lines

Flow Element

Principles of the Flow Net

Piezometer

)h = head loss = one drop

Datum

1

2

3

4

5

Principles of the Flow Net

Equipotential LinesTotal heads along this line are the same

Flow Element

Flow nets

• Flow nets are a graphical solution method of Laplace equation for 2D flow in a homogeneous, isotropic aquifer.

• In an isotropic medium, the continuity equation represents two orthogonal families of curves:

1. Flow lines: the line along which a water particle will travel from upstream to the downstream side in the permeable soil medium

2. Equipotential lines: the line along which the potential (pressure) head at all points is equal.

١١٦

Datum

Flow nets• Flow nets are the combination of flow lines and equipotential lines.

• To complete the graphic construction of a flow net, one must draw the flow and equipotential lines in such away that:

1. The equipotential lines intersect the flow lines at right angles.

2. The flow elements formed are approximate squares.

١١٧

Flow channel

Flow line

Equipotential line

١١٨

Boundary Conditions

١١٩

H

H

H-3

∆h

0

١٢٠

Seepage Calculation from Flow Net

١٢١

• In a flow net, the strip between any two adjacent flow lines is called a flow channel.

• The drop in the total head between any two adjacent equipotential lines is called the potential drop.

• If the ratio of the sides of the flow element are the same along the flow channel, then:

1. Rate of flow through the flow channel per unit width perpendicular to the flow direction is the same.

∆q1 = ∆q2 = ∆q3 = ∆q

2. The potential drop is the same and equal to:

Flow element

433221dN

Hhhhhhh =−=−=−

Where H: head difference between the upstream and downstream sides.Nd: number of potential drops.

Seepage Calculation from Flow Net

١٢٢

From Darcy’s Equation, the rate of flow is equal to:

• If the number of flow channels in a flow net is equal to Nf, the total rate of

flow through all the channels per unit length can be given by:

f

=∆

dN

NHkq

1

...3

3

43

2

2

32

1

1

21

=∆

=

−=

−=

−=∆

dN

Hkq

ll

hhkl

l

hhkl

l

hhkq

Example on estimating the total flow under dam

Example: if k = 10-7 m/sec, what would be the flow per day over a 100 m length of wall?

١٢٣

Dam

cutoff

Low permeability rock

50 m of water

5 m of water

Dam

cutoff

Low permeability rock

50 m of water

5 m of water

Calculations

Nf = 5

Nd = 14

∆h = 45 m

k = 10-7 m/sec

١٢٤

Answer:

= 10-7(5/14) 45 x 100 m length= 0.000161 m3/sec= 13.9 m3/day

Pressure head at any point

impervious strata

concrete damdatum

X

z

hL

HT = hL HT = 0

Total head = hL - # of drops from upstream x ∆h

∆h

Elevation head = -z

Pressure head = Total head – Elevation head d

L

N

h=

+ve

-ve

8

2

7

6

5

3

4

1

2

)h

u = [14 - (3. )h)].(water

14 in

Feff = *(soil + * (water - ( - )h) * (water

)h)h)h)h)h)h)h

3 in

2 in

Buoyancy + Seepage Force

Ws

Ws

Ws

Ws

Ws

In Flow

Out Flow

Uplift Pressure under Hydraulic structures

Datum

At point a:

ua/γw = (7- 1x1) - (-2) = 8 kN/m2

At point f:

ua/γw = (7- 6x1) - (-2) = 3 kN/m2

At point b:

ua/γw = (7- 2x1) - (-2) = 7 kN/m2

∆h = hL/Nd = 7/7=1

Example of Dam Failure

١٢٨

Around 7:00 am on June 5, 1976 a leak about 30 m from the top of Teton dam was observed.

Example of Dam Failure

١٢٩

The Dam Broke at 11:59 AM

١٣٠

١٣١

١٣٢

١٣٣

Flow Nets: an example

134

Posit ion: A B C D E F G H I JDist ancef romf ront t oe( f t)

0 3 22 37.5 50 62.5 75 86 94 100

n 16.5 9 8 7 6 5 4 3 2 1. 2

The flow net is drawn with: m = 5 n = 17

Flow Nets: the solution

•Solve for the flow per unit width:

q = (m/n) K h

= (5/17)(150)(35)

= 1544 ft3/day per ft

135

Flow Nets: An Example

•There is an earthen dam 13 meters across and 7.5 meters high.The Impounded water is 6.2 meters deep, while the tail water is 2.2 meters deep. The dam is 72 meters long. If the hydraulic conductivity is 6.1 x 10-4 centimeter per second, what is the seepage through the dam if n = 21

K = 6.1 x 104cm/sec = 0.527 m/day

136

Flow Nets: the solution

•From the flow net, the total head loss, H, is 6.2 -2.2 = 4.0 meters.

•There are 6 flow channels (m) and 21 head drops along each flow path (n): Q = (KmH/n) x dam length = (0.527 m/day x 6 x 4m / 21) x (dam length) = 0.60 m3/day per m of dam

•= 43.4 m3/day for the entire 72-meterlength of the dam

137

Example of Dam Failure

١٣٨

Post Failure Investigation

• Seepage piping and internal erosion

• Seepage through rock openings

• Hydraulic fracture

• Differential settlement and cracking

• Settlement in bedrock

Piping in Granular Soils

concrete dam

impervious strata

H

At the downstream, near the dam,

∆h = total head drop∆l

l

hiexit ∆

∆=the exit hydraulic gradient

Piping in Granular Soils

concrete dam

impervious strata

H

If iexit exceeds the critical hydraulic gradient (ic), first

the soil grains at exit get washed away.

no soil; all water

This phenomenon progresses towards the upstream, forming a free passage of water (“pipe”).

Critical hydraulic gradient, ic

The critical hydraulic gradient (ic),

Consequences:

no stresses to hold granular soils together

∴ soil may flow ⇒

“boiling” or “piping” = EROSION

١٤١

wsat γγγ −=′

Piping in Granular Soils

Piping is a very serious problem. It leads to downstream flooding which can result in loss of lives.

concrete dam

impervious strata

Therefore, provide adequate safety factor against piping.

3>=exit

cpiping i

iF

Filters

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Used for:

� facilitating drainage

� preventing fines from being washed away

Used in:

� earth dams

� retaining walls

Filter Materials:

� granular soils

� geotextiless

Granular Filter Design

The proper design of filters should satisfy two conditions:

Condition 1:

The size of the voids in the filter material should be small enough to hold the larger particles of the protected material in place

The filter material should have a high hydraulic conductivity to prevent buildup of large seepage forces and hydrostatic pressures in the filters.

granular filter

Condition 2:

Granular Filter Design

Condition 1:

after Terzaghi & Peck (1967)

Condition 2:

after US Navy (1971)

GSD Curves for the soil and filter must be parallel

5 to4)(85

)(15 ≤soil

filter

D

D5 to4

)(15

)(15 ≥soil

filter

D

D

5 )(85

)(15 <soil

filter

D

D

25 )(50

)(50 <soil

filter

D

D

20 )(15

)(15 <soil

filter

D

D

4 )(15

)(15 >soil

filter

D

D

Granular Filter Design

GSD Curves for the soil and filter must be parallel

5 to4)(85

)(15 ≤soil

filter

D

D

5 to4)(15

)(15 ≥soil

filter

D

D

GSD of Soil

EXAMPLE

• A stiff clay layer underlies a 12 m thick silty sand deposit. A sheet pile is driven into the sand to a depth of 7 m, and the upstream and downstream water levels are as shown in the figure. Permeability of the silty sand is 8.6 × 10-4

cm/s. The stiff clay can be assumed to be impervious. The void ratio of the silty sand is 0.72 and the specific gravity of the grains is 2.65.

• (a) Estimate the seepage beneath the sheet pile in m3/day per meter.

• (b) What is the pore water pressure at the tip of the sheet pile?

• (c) Is the arrangement safe against piping?

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EXAMPLE

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Solution

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