Tutorial on Seepage Flow Nets and Full Solution Chapter 1

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


3/13

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


4/13

Appendix I

6 m

sheet pile

1.25 m

4 m


5/13

Tutorial 1 . Flownet Sketching and Interpretation for a Dam

redo the previous problem on Appendixes II and III.


6/13

Appendix II

6 m

1.25 m

4 m


7/13

Appendix III

6 m

1.25 m

4 m


8/13


9/13

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


N f = 3; N d = 5

⎟⎟ ⎠

⎞⎜⎜⎝

⎛ =

d

f

N

N kH q


10/13


/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


/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,


11/13


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


12/13


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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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

Documents

Tutorial on Seepage Flow Nets and Full Solution Chapter 1