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٢
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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