= = =

Construct the flow net and determine the quantity of seepage through the dam. What is the pore pressure at point P?

Solution The transformed section is obtained by multiplying the horizontal distances by ~kz I kx and by keeping the vertical dimensions unaltered. Fig. Ex. 4.17(a) is a natural section of the dam. The scale factor for transformation in the horizontal direction is

Jf 16x10-8Scale factor = _z =. =0.6kx 4.5 x 10-8 The transformed section of the dam is given in Fig. Ex. 4.17(b). The isotropic equivalent

coefficient of permeability is

ke =~kxkz = (J45 x 1.6) x lO-s =2.7xlO-s mls Confocal parabolas can be constructed with the focus of the parabola at A. The basic parabola

passes through point G such that

GC= 0.3 He = 0.3 x 27 = 8.10 m The coordinates of G are:

x =+40.80 m, Z =+18.0 m

As per Eq. (4.58) (a)

182 -4a2 Substituting for x and z, we get, 40.80 = 0

Simplifying we have, 4a5 + 163.2ao - 324 = 0

Substituting for ao in Eq. (a) above, we can write

h=18.0m

Figure Ex. 4.17

7.6 (b)

By using Eq. (b), the coordinates of a number of points on the basic parabola may be calculated. x(m) z(m)

-1.9 0.0 0.0 3.8

5.0 10.0 20.0 30.0 7.24 9.51 12.9 15.57

The basic parabola is shown in Fig. Ex. 4.17(b). The flownet is completed by making the entry corrections by ensuring that the potential

drops are equal between the successive equipotential lines at the top seepage line level. As per Fig. Ex. 4.17(b), there are 3.8 flow channels and 18 equipotential drops. The seepage

per unit length of dam is

e Nd 18

The quantity of seepage across section Az can also be calculated without the flownet by using Eq. (4.60)

q = keYo = 2k e ao = 2 x 2.7 X 10-8 x 1.9 ::::: 1 X 10-7 m3/sec per meter

Pore pressure at P Let RS be the equipotential line passing through P. The number of equipotential drops up to point P equals 2.4

Total head loss =h =18 m, number of drops =18

18 Head loss per drop = 18 = 1 m . Therefore the head at point P = 18 - 2.4(~h) = 18 - 2.4(1) = 15.6 m Assuming the base of the dam as the datum, the elevation head of point P = 5.50 m. Therefore the pressure head at P = 15.6 - 5.5 = 10.1 m. The pore pressure at P is, therefore, U

w = 10.1 x 9.81 = 99 kN/m2

Example 4. 18 A sheet pile wall was driven across a river to a depth of 6 m below the river bed. It retains a head of water of 12.0 m. The soil below the river bed is silty sand and extends up to a depth of 12.0 m where it meets an impermeable stratum of clay. Flow net analysis gave Nt = 6 and Nd = 12. The hydraulic conductivity of the sub-soil is k = 8 X 10-5 m Imin. The average uplIft pressure head h

a at the bottom

of the pile is 3.5 m. The saturated unit weight of the soil Ysat = 19.5 kN/m3• Determine: (a) The seepage less per meter length of pile per day. (b) The factor of safety against heave on the downstream side of the pile.