Geotechnical Engineering

DHARM

SEEPAGE AND FLOW NETS 197

Quantity of seepage per meter

length of wall

UV

W

q = k. H.

n

n

f

d

Impervious

h = 2.5 m

Pervious

IV

III

II I

1

2

3

4

5

6 78 9

10

11

12

13

14

2.5 m

7.5

m

Fig. 6.28 Sheet pile wall (Example 6.6)

= 3 × 10–4 × 2.5 ×

4

14

m^3 /sec/metre run

= 2.143 × 10–4 m^3 /sec/ meter run

= 214.3 ml/sec/metre run.

Exmaple 6.7: An earth dam of homogeneous section with a horizontal filter is shown in Fig. 6.29.
If the coefficient of permeability of the soil is 3 × 10–3 mm/s, find the quantity of seepage per
unit length of the dam.

F

45 m

S

200 m

E

A27 m B

8m

30 m

3:1

32 m

A¢ B¢

3:1

Directrix of

base parabola

90 m

Fig. 6.29 Homogeneous earth dam with horizontal filter at toe (Example 6.7)
AB = 0.3(EB) = 3.0 × 90 = 27 m
With respect of the focus, F (the end of the filter), as origin, the co-ordinates of A, the
starting point of the base parabola, are : x = FA′ = 200 – 90 – 45 + 27 = 92 m

z = A′A = 30 m

The equation to the parabola is

(^) xz^22 + = x + S,
where S is the distance to the directrix from the focus, F.
∴ 9222 + 30 = 92 + S