Mylan Nejyar

1. A
Datum
7.0 m
5.0 m
•
Example 1):
For the system shown in Fig., find
1. Flow rate per meter length of sheet pile, if k = 1*10-4
m/sec
2. The FS against piping, L=1.5m
3. The effective stress at point A, if void ratio equal
to 0.57 and specific gravity is 2.68.
Solution:
1)
d
f
N
N
Hkq =
/sec/mm10*273.2
11
5
*5*10*1q 344- −
==
2)
ex
cr
i
i
FS = , 07.1
57.01
168.2
e1
1G
i s
cr =
+
−
=
+
−
=
L
Δh
iex = , 455.0
11
5
N
H
h
d
===∆
303.0
1.5
0.455
iex ==
531.3
0.303
1.07
FS ==
3) hp = ht - he
1.5880.455*7.55ht =−= m
he = -7.0 m
hp = 1.588 – (-7.0) = 8.588 m
uA = 8.588 * 9.81 = 84.248 kN/m2
3
w
s
sat kN/m307.20
57.01
81.9*)]57.0*1(68.2[
γ
e1
seG
γ =
+
+
+
+
==
kPa142.1497*20.307σt ==
kPa57.90184.248-142.149σ ==′
Example 2):
The dam and flow net shown below, the dam is 120m long and has two sheet piles.
Datum is at the impervious layer. Required the
1. Quantity of the seepage loss under the dam, k = 20*10-4
cm/sec
2. Factor of safety against boiling.
3. Uplift pressure at point A & F.

2. Solution:
1)
d
f
N
N
Hkq =
/secm10*308.8120*
10.4
3
*12*10*20q 33-6-
==
2)
ex
cr
i
i
FS =
L
Δh
iex = , 154.1
4.10
12
N
H
h
d
===∆
136.0
8.5
1.154
iex ==
353.7
0.136
1.00
FS ==
3) point A
hp = ht - he
37.9611.154*3.542ht =−= m
he = 28.0 m
hp = 37.961 – 28.0 = 9.961 m
uA = 9.961 * 9.81 = 97.717 kN/m2
point F
34.0371.154*6.942ht =−= m
he = 28.0 m
hp = 34.037 – 28.0 = 6.037 m
uF = 6.037 * 9.81 = 59.223 kN/m2

3. Example 3):
For the flow net shown in Figure.
(a) How high would water rise if a piezometer
is placed at (i) A (ii) B (iii) C?
(b) If k = 0.01mm/s, determine the seepage loss
of the dam in (m3
/day/m).
Solution:
(a) point A
m7.5
12
10
*310Δh*lineialequipotentofNo.H =−=− above the ground
surface
point B
m5.833
12
10
*510Δh*lineialequipotentofNo.H =−=−
point C
point A and point C locate at the same equipotential lines so the high of water is
same as in point A and equal to 7.5 m above the ground surface.
(b)
d
f
N
N
Hkq =
/day/mm6.3
12
5
*10*24)*60*(60*10*1q 35-
==
Example 4):
A sheet piling with its corresponding flow net
is shown:
a) Estimate the flow rate in m3
/day.
b) For the element A with L = 1.5 m calculate
the average velocity and the effective stress.
c) Determine the magnitude of the effective stress
at the base and at the right hand side of the sheet pile.
d) Calculate the factor of safety against the piping (quick condition)
k = 0.02 mm/sec and γsat = 20 kN/m3
.
Solution: a)
d
f
N
N
Hkq =
/day/mm356.2
11
5
*3*24)*60*(60*10*2q 35-
==
b) ikv = ,
L
Δh
i = , 273.0
11
3
N
H
h
d
===∆
182.0
1.5
0.273
i ==

4. 1.5m
mm/hr13.104
mm/sec10*3.640.182*0.02v 4-
=
==
m5.318
11
3
*2.51.5)(4.5hp =−+=
uA = 5.318 * 9.81 = 52.17 kN/m2
kPa104.71520*4.59.81*1.5σt =+=
kPa52.54552.17-104.715σ ==′
c)
m4.364
11
3
*66hp =−=
uA = 4.364 * 9.81 = 42.811 kN/m2
kPa0620*3σt ==
kPa.1897142.811-06σ ==′
d)
ex
cr
i
i
FS = , 039.1
81.9
81.920
γ
γ
i
w
cr =
−
=
′
=
L
Δh
iex = , L = 0.6 by scale
455.0
0.6
0.273
iex ==
284.2
0.455
1.039
FS ==
Example 5): For the system shown below compute the
1. Flow rate, (k = 1.1 m/day) and
2. Pore pressure at point B.
Solution: 1)
d
f
N
N
Hkq =
length/day/mm733.0
12
4
*8)-(10*1.1q 3
==
2)
hp = (2+3+3.5) – 10 (2/12) = 6.833 m
uB = 6.833*9.81 = 67.035 kN/m2
Example 6):

5. Given the flow net shown for a sheet Pile cutoff wall. Required the effective stress at
point A and compute the seepage quantity based on k = 4*10-2
cm/sec.
Solution:
Pressure head at A
(13.5+1+2) – 10 (13.5/11) = 4.2m
uA = 4.2*9.81 = 41.2 kN/m2
kPa51.479.81*118.85*2σt =+=
kPa31.641.2-51.47σ ==′
Since 0,σ >′ point A is not "quick" but 6.31 kN/m2
is not large enough to be very
confidence - inspiring.
d
f
N
N
Hkq =
widthwallof/day/mm9.173)24*60*60(*
11
4.1
*13.5*
100
1
*2-10*4q 3
==
Example 7): For the concrete dam shown find the uplift pressure at base of the dam.
Solution:

6. hp = ht - he
For point 1:
0.270.267*144ht =−= m
he = – 1.80 m
hp = 0.27 – (–1.80) = 2.07 m
u1 = 2.07 * 9.81 = 20.307 kN/m2
For point 7.5:
2.00.267*7.54ht =−= m
he = – 2.40 m
hp = 2.0 – (–2.40) = 4.40 m
u7.5 = 4.40 * 9.81 = 43.164 kN/m2
Example 8):
A river bed consists of a layer of sand 8.25m thick overlying impermeable rock; the
depth of water is 2.50 m. A long cofferdam 5.50m wide is formed by driving two lines
of sheet piling to a depth of 6.00m below the level of the river bed and excavation to a
depth of 2.00m below bed level is carried out within the cofferdam. The water level
within the cofferdam is kept at excavation level by pumping. If the flow of water into
the cofferdam is 0.25m3
/h per unit length, what is the coefficient of permeability of the
sand? What is the hydraulic gradient immediately below the excavated surface?
Solution:
The section and flow net appear in the above Figure. In the flow net there are 6.0 flow
channels and 10 equipotential drops.
The total head loss is 4.50 m. The coefficient of permeability is given by
d
f
N
N
Hkq =
cm/sec10*57.2m/h0926.0k
10
6
*4.5*k0.25 3−
==⇒=

7. The distance between the last two equipotentials is measured as 0.9 m. The required
hydraulic gradient is given by
L
Δh
iex = , 45.0
10
5.4
N
H
h
d
===∆
5.0
0.9
0.45
iex ==
Example 9):
The flow net for seepage of water around a single row of sheet piles driven into a
permeable layer is shown in figure below. Calculate the quantity of seepage loss and
factor of safety against downstream heave, given that k = 2.6*10-5
m/sec and γsat = 17.7
kN/m3
. (Note: thickness of permeable layer T = 18 m)
Solution:
d
f
N
N
Hkq =
/sec/mm10*473.1
6
4
*1.5)-(10*10*2.6q
34
5
−
−
=
=
)H(HγC
γD
FS
21wo
−
′
=
D/T = 18/6 = 0.33, from table by interpolation Co = 0.357
59.1
1.5)9.81(10*0.357
9.81)-(17.7*6
FS =
−
=
Example 10):
A concrete dam has a base length of 8m and retains 5m of water. The water level on
the down stream side is at the ground level. Under the dam there exists an 8m thick
layer of anistropic permeable soil with the k values shown in Figure (a). Calculate the
flow rate under the dam in lit/day/m.
Solution:
horizontal scale =
x
z
k
k * vertical scale
= 625.0
10*2.56
10*1
5-
5-
= * vertical scale
Therefore, the length of the transformed base is
8*0.625 = 5m
The dam section must be redrawn according to the new scale, and then the flow nets can
be drawn as shown in Figure (b).

8. 8
4
*5*10*56.2*10*1
N
N
Hkkq 55
d
f
vx
−−
==
3456sec//10*4 35
== −
mm lit/day/m