tutorial on seepage flow nets and full solution chapter 1
TRANSCRIPT
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8/20/2019 Tutorial on Seepage Flow Nets and Full Solution Chapter 1
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Assignment 1 . Flownet Sketching and Interpretation for a Dam
a dam, shown in figure below and in Appendix 1, retains 6 m of water. a
sheet pile wall (cutoff curtain) on the upstream side, which is used to reduce
seepage under the dam, penetrates 4 m into a thick pervious layer of soil.below the that pervious layer, is a a thick deposit of practically impervious
clay. assume the pervious layer is a homogeneous and isotropic.
Your Tasks
(a) draw the flow net under the dam on Appendix I.
(b) determine the flow rate under the dam if the equivalent hydraulic
conductivity is 0.0004 cm/s.
(c) calculate and draw the porewater pressure distribution at the base
of the dam.
(d) determine the uplift force.
(e) determine and draw the porewater pressure distribution on the
upstream and downstream faces of the sheet pile wall.
(f) determine the resultant lateral force on the sheet pile wall due tothe porewater.
(g) determine the maximum hydraulic gradient.
(h) will piping occur if the void ratio of the pervious layer is 0.6?
(i) what is the effect of reducing the depth of penetration of the sheet
pile wall?
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Your Tasks
(a) draw the flow net under the dam on Appendix I.
(b) determine the flow rate under the dam if the equivalent hydraulic
conductivity is 0.0004 cm/s. your solution
(i) the difference between upstream and downstream water level
elevation, H is
(ii) Nd is
(iii) Nf is
(iv) the flow rate
f
d
Nq k H
N
(c) calculate and draw the porewater pressure distribution at the base
of the dam.
your solution
(i) the equal intervals
dam width, B
n
x
(ii) the head loss, h, between each consecutive pair of
equipotential lines
d
Hh
N
(iii) the porewater pressure at each nodal point can be easily
determine by creating a table.
Table 1
(d) determine the uplift force.
Graph 1
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Table 1
under base of dam . measured from point x (m)
x (m) 0 1.75 3.50 5.25 7.00 8.75 10.50 12.25 14 15.75 17.5
Nd (m)
H (m)
h (m)
Nd h
h z (m)
hP (m)= H - NdH - h z
u (kPa) = hp w
x . convenient number of equal intervals; Nd . number of equipotential lines; H . the difference between upstream
and downstream water level elevation; h . the head loss between each consecutive pair of equipotential lines; hz .
the elevation head; hP . the pressure head; u . porewater pressure head.
Graph 1
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Appendix I
6 m
sheet pile
1.25 m
4 m
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Tutorial 1 . Flownet Sketching and Interpretation for a Dam
redo the previous problem on Appendixes II and III.
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Appendix II
6 m
1.25 m
4 m
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Appendix III
6 m
1.25 m
4 m
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51© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
Chapter 8
8.1 Eq. (8.14):
⎟⎟ ⎠
⎞⎜⎜⎝
⎛ +
=
2
2
1
11
112
H
k
H
k H
k hh
( ) ⎟ ⎠
⎞⎜⎝
⎛ +
=
cm15
cm/sec
cm10
cm/sec0.004cm10
cm/sec)cm)(0.004(20cm8
2k
k 2 = 0.009 cm/sec
8.2 The flow net is shown.
k = 4 × 10-4 cm/sec
H = H 1 – H 2
= 6.0 – 1.5 = 4.5 m.
So
/m/daym1077.76 36-×=
×=
⎟ ⎠ ⎞
⎜⎝ ⎛ ×⎟
⎠
⎞⎜⎝
⎛ ×=
−
−
/m/secm109
8
45.4
10
104
36
2
4
q
8.3 The flow net is shown.
N f = 3; N d = 5
⎟⎟ ⎠
⎞⎜⎜⎝
⎛ =
d
f
N
N kH q
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52© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
/m/daym0.5183=×=⎟
⎠ ⎞
⎜⎝ ⎛ −⎟
⎠
⎞⎜⎝
⎛ ×= −
−
/m/secm1065
3)5.03(m/sec
10
104 362
4
q
8.4 Based on the notations in Figure 8.10:
H = (4 – 1.5) m = 2.5 m; S = D = 3.6 m; T ' = D1 = 6 m; S /T
' = 3.6/6 = 0.6
From the figure, 44.0≈kH
q
/m/daym0.38 3=⎟ ⎠
⎞⎜⎝
⎛ ×××
×=
−
m/day24606010
104)5.2)(44.0(
2
4
q
8.5 The flow net is shown.
/m/daym7.2 3=⎟ ⎠ ⎞
⎜⎝ ⎛
⎟ ⎠ ⎞
⎜⎝ ⎛ ×××=⎟⎟
⎠
⎞⎜⎜⎝
⎛ =
12
5)10(m/day246060
10
002.02
d
f
N
N kH q
8.6 Refer to the flow net given in Problem 8.5 and the figure on the next page.
The flow net has 12 potential drops. Also, H = 10 m. So the head loss for each
drop = (10/12) m. Thus,
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Pressure head at D = (10 + 3.34) – (2)(10/12) = 11.67 m
Pressure head at E = (10 + 3.34) – (3)(10/12) = 10.84 m
Pressure head at F = (10 + 1.67) – (3.5)(10/12) = 8.75 m
Pressure head at G = (10 + 1.67) – (8.5)(10/12) = 4.586 m
Pressure head at H = (10 + 3.34) – (9)(10/12) = 5.84 m
Pressure head at I = (10 + 3.34) – (10)(10/12) = 5 m
The pressure heads calculated are shown in the figure. The hydraulic uplift force
per unit length of the structure can now be calculated to be
kN/m1717.5=
++++=
⎥⎥
⎥⎥⎥
⎦
⎤
⎢⎢
⎢⎢⎢
⎣
⎡
⎟ ⎠ ⎞⎜
⎝ ⎛ ++⎟
⎠ ⎞⎜
⎝ ⎛ ++⎟
⎠ ⎞⎜
⎝ ⎛ ++
⎟ ⎠
⎞⎜⎝
⎛ ++⎟
⎠
⎞⎜⎝
⎛ +
=
=
)05.971.816.12236.168.18)(81.9(
)67.1(2
584.5)67.1(2
84.5586.4)32.18(2
586.475.8
)67.1(2
754.884.10)67.1(
2
84.1067.11
)diagram)(1head pressuretheof area(wγ
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54© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
8.7 The flow net is shown. N f = 3; N d = 5.
/m/daym2.13≈×=
⎟ ⎠
⎞⎜⎝
⎛ =⎟
⎠
⎞⎜⎝
⎛ −
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ =
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ =
−
−−
/m/secm10429.2
14
4)5.8)(10(
14
4)5.110(
10
10
35
5
2
3
d
f
N
N kH q
8.8 For this case, T ' = 8 m; S = 4 m; H = H 1 – H 2 = 6 m; B = 8 m; b = B/2 = 4 m.
a. 5.08
4==
′T
S ; x = b – x′ = 4 – 1 = 3 m; 5.0
8
4 ;75.0
4
3==
′−=
T
b
b
x
From Figure 8.11, q/kH = 0.37.
/m/daym1.923≈⎟
⎠ ⎞
⎜⎝ ⎛ ×××= )6(246060
10
001.0)37.0(
2q
b. 5.04
2 m;224 ;5.0 ;5.0 ===−=′−==
′=
′ b
x xb x
T
b
T
S . So q/kH = 0.4.
/m/daym2.07 3≈⎟ ⎠ ⎞
⎜⎝ ⎛ ×××= )6(246060
10
001.0)4.0(
2q
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55© 2014 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.
8.9
α 1 = 35°; α 2 = 40º; H = 7 m; Δ = 7 cot
35 = 10 m. 0.3Δ = 3 m.
m2.24334cot)710(5)40)(cot10(
Δ3.0cot)(cot 11121
=+−++=+−++= α α H H L H d
m94.1
40sin
7
40cos
2.24
40cos
2.24
sincoscos
22
2
2
2
2
2
2
2
=
⎟ ⎠
⎞⎜⎝
⎛ −⎟
⎠
⎞⎜⎝
⎛ −=−−=
α α α
H d d L
/m/daym0.2713≈×=
⎥⎦
⎤⎢⎣
⎡
⎟
⎟
⎠
⎞
⎜
⎜
⎝
⎛ ×==
−
−
/sec/mm10139.3
)40)(sin40(tan)94.1(
10
103sintan
36
2
4
22 α α kLq
8.10 From Problem 8.9, d = 24.2 m; H = 7 m; α 2 = 40º
;46.37
2.24==
H
d m ≈ 0.25 (Figure 8.14)
m2.7240sin
)7)(25.0(sin 2
===α
mH L
/m/daym0.2913≈×=
⎟⎟ ⎠
⎞⎜⎜⎝
⎛ ×==
−
−
/sec/mm1037.3
)40)(sin72.2(10
103sin
36
2
2
4
2
2α kLq