Tutorial on Seepage Flow Nets and Full Solution Chapter 1

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 to the 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

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 N q 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 H h 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

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 hz (m) hP (m) = H - NdH - hz 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.

Appendix I

6 m

sheet pile

1.25 m

Tutorial 1

. Flownet Sketching and Interpretation for a Dam

redo the previous problem on Appendixes II and III.

Appendix II

6 m

sheet pile

1.25 m

Appendix III

6 m

sheet pile

1.25 m

51

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Chapter 8

8.1

Eq. (8.14):

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

+

=

2 2 1 1 1 1 1 2

H

k

H

k

H

k

h

h

(

)

⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ₊ = cm 15 cm/sec cm 10 cm/sec 0.004 cm 10 cm/sec) cm)(0.004 (20 cm 8 2 k

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/day

m

10

77.76

×

-6 3

=

×

=

⎟

⎠

⎞

⎜

⎝

⎛

×

⎟

⎠

⎞

⎜

⎝

⎛ ×

=

− −

/m/sec

m

10

9

8

4

5

.

4

10

10

4

3 6 2 4

q

8.3

The flow net is shown.

N

f

= 3; N

d

= 5

⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = d f N N kH q

52

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/m/day

m

0.518

3

=

×

=

⎟

⎠

⎞

⎜

⎝

⎛

−

⎟

⎠

⎞

⎜

⎝

⎛ ×

=

−

₆

₁₀

−

_m

_/m/sec

5

3

)

5

.

0

3

(

m/sec

10

10

4

6 3 2 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

' = D

1

= 6 m; S/T

' = 3.6/6 = 0.6

From the figure,

≈

0

.

44

kH

q

/m/day

m

0.38

3

=

⎟

⎠

⎞

⎜

⎝

⎛

×

_×

_×

_×

=

−

60

60

24

m/day

10

10

4

)

5

.

2

)(

44

.

0

(

₂ 4

q

8.5

The flow net is shown.

/m/day m 7.2 3 = ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ _{× × ×} = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ = 12 5 ) 10 ( m/day 24 60 60 10 002 . 0 2 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,

53

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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/m 1717.5 = + + + + = ⎥ ⎥ ⎥ ⎥ ⎥ ⎦ ⎤ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣ ⎡ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + + ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ + = = ) 05 . 9 71 . 8 16 . 122 36 . 16 8 . 18 )( 81 . 9 ( ) 67 . 1 ( 2 5 84 . 5 ) 67 . 1 ( 2 84 . 5 586 . 4 ) 32 . 18 ( 2 586 . 4 75 . 8 ) 67 . 1 ( 2 754 . 8 84 . 10 ) 67 . 1 ( 2 84 . 10 67 . 11 ) diagram)(1 head pressure the of area ( w

γ

54

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8.7

The flow net is shown. N

f

= 3; N

d

= 5.

/m/day

m

2.1

3

≈

×

=

⎟

⎠

⎞

⎜

⎝

⎛

=

⎟

⎠

⎞

⎜

⎝

⎛

−

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛

=

− − −

/m/sec

m

10

429

.

2

14

4

)

5

.

8

)(

10

(

14

4

)

5

.

1

10

(

10

10

3 5 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.

0

.

5

8

4 =

=

′

T

S

_{; x = b – x′ = 4 – 1 = 3 m;}

₀

_.

₅

8

4

;

75

.

0

4

3

₌

₌

′

−

=

T

b

b

x

From

Figure

8.11,

q/kH = 0.37.

/m/day m 1.92 3 ≈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ _{× × ×} = 60 60 24 (6) 10 001 . 0 ) 37 . 0 ( ₂ q

b.

0

.

5

4

2

m;

2

2

4

;

5

.

0

;

5

.

0

=

=

−

′

=

−

=

=

=

′

=

′

b

x

x

b

x

T

b

T

S

_{. So q/kH = 0.4.}

/m/day m 2.07 3 ≈ ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ _{× × ×} = 60 60 24 (6) 10 001 . 0 ) 4 . 0 ( ₂ q

55

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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.

m

2

.

24

3

34

cot

)

7

10

(

5

)

40

)(cot

10

(

Δ

3

.

0

cot

)

(

cot

2 1 1 1 1

=

+

−

+

+

=

+

−

+

+

=

H

α

L

H

H

α

d

m

94

.

1

40

sin

7

40

cos

2

.

24

40

cos

2

.

24

sin

cos

cos

2 2 2 2 2 2 2 2 2

=

⎟

⎠

⎞

⎜

⎝

⎛

−

⎟

⎠

⎞

⎜

⎝

⎛

−

=

−

−

=

α

α

α

H

d

d

L

/m/day

m

0.271

3

≈

×

=

⎥

⎦

⎤

⎢

⎣

⎡

⎟⎟

⎠

⎞

⎜⎜

⎝

⎛ ×

=

=

− −

/sec/m

m

10

139

.

3

)

40

)(sin

40

(tan

)

94

.

1

(

10

10

3

sin

tan

3 6 2 4 2 2

α

α

kL

q

8.10 From

Problem

8.9,

d = 24.2 m; H = 7 m;

α

2

= 40º

;

46

.

3

7

2

.

24 =

=

H

d

_{m ≈ 0.25 (Figure 8.14)}

m 2.72 40 sin ) 7 )( 25 . 0 ( sin 2 = = =

α

mH L /m/day m 0.291 3 ≈ × = ⎟⎟ ⎠ ⎞ ⎜⎜ ⎝ ⎛ × = = − − /sec/m m 10 37 . 3 ) 40 )(sin 72 . 2 ( 10 10 3 sin 3 6 2 2 4 2 2

_α

kL q