Seepage
on the Companion Website. The datum is set at the level of the excavation. The resulting total head distribution is shown in Figure 2.16, and by applying Equation 2.1 the pore water pressure distribution around the tunnel can be plotted.
The flow rate of water into the excavation can be found by considering the flow between the eight adjacent nodes on the discharge boundary. Considering the nodes next to the sheet pile wall, the change in head between the last two nodes is Δh = 0.47. This is repeated along the discharge boundary, and the average Δh between each set of nodes is computed.
Adapting Equation 2.19, and noting that the soil on the discharge boundary has perme- ability k1, the flow rate is then given by:
q k= h
= ´
å
- 1
9 3
3 3 10 D
. m /s
face. Within two hours of this occurring the dam failed catastrophically, causing extensive flood- ing, as shown in Figure 2.18. The direct and indirect costs of this failure were estimated at close to $1 billion. A horizontal under-filter is shown in Figure 2.17. Other possible forms of filter are illustrated in Figures 2.22(a) and (b); in these two cases the discharge surface AD is neither a flow line nor an equipotential since there are components of discharge velocity both normal and tangential to AD.
The boundary conditions of the flow region ABCD in Figure 2.17 can be written as follows:
Equipotential BC: ϕ = −kh Equipotential AD: ϕ = 0
Flow line CD: ψ = q (also, ϕ = −kz) Flow line BA: ψ = 0
The conformal transformation r = w
2Complex variable theory can be used to obtain a solution to the embankment dam problem. Let the complex number w = ϕ+iψ be an analytic function of r = x+iz. Consider the function
r=w2 Thus
( ) ( )
( )
x+ z =
= + -
i +i
i f y
f fy y
2
2 2 2
Equating real and imaginary parts:
x=f2-y2 (2.34)
Figure 2.18 Failure of the Teton Dam, 1976 (photo courtesy of the Bureau of Reclamation).
Seepage
z=2fy (2.35)
Equations 2.34 and 2.35 govern the transformation of points between the r and w planes.
Consider the transformation of the straight lines ψ = n, where n = 0, 1, 2, 3 (Figure 2.19(a)).
From Equation 2.35 f= z
2n
and Equation 2.34 becomes x z
n n
= 22 - 2
4 (2.36)
Equation 2.36 represents a family of confocal parabolas. For positive values of z the parabolas for the specified values of n are plotted in Figure 2.19(b).
Consider also the transformation of the straight lines ϕ = m, where m = 0, 1, 2,…, 6 (Figure 2.19(a)). From Equation 2.35
y= z 2m
and Equation 2.34 becomes
x m z
= 2- m22
4 (2.37)
Equation 2.37 represents a family of confocal parabolas conjugate with the parabolas represented by Equation 2.36. For positive values of z the parabolas for the specified values of m are plotted in Figure 2.19(b). The two families of parabolas satisfy the requirements of a flow net.
Application to embankment dam sections
The flow region in the w-plane satisfying the boundary conditions for the section (Figure 2.17) is shown in Figure 2.20(a). In this case, the transformation function
r Cw= 2
Figure 2.19 Conformal transformation r = w2: (a) w plane, and (b) r plane.
will be used, where C is a constant. Equations 2.34 and 2.35 then become x C
z C
= -
=
(f y ) fy
2 2
2
The equation of the top flow line can be derived by substituting the conditions y
f
=
= - q
kz Thus
z Ckzq
C kq
= -
\ = - 2
1 2 Hence
x kq k z q
x q
k k qz
= − −
= −
1 2 1 2
2 2 2
2
( )
(2.38) Figure 2.20 Transformation for embankment dam section: (a) w plane, and (b) r plane.
Seepage The curve represented by Equation 2.38 is referred to as Kozeny’s basic parabola and is shown in Figure 2.20(b), the origin and focus both being at A.
When z = 0, the value of x is given by
x q
k q kx
0
0
2 2
=
\ =
(2.39)
where 2x0 is the directrix distance of the basic parabola. When x = 0, the value of z is given by z q
k x
0 = =2 0
Substituting Equation 2.39 in Equation 2.38 yields
x x z
= 0- x2
4 0 (2.40)
The basic parabola can be drawn using Equation 2.40, provided the coordinates of one point on the parabola are known initially.
An inconsistency arises due to the fact that the conformal transformation of the straight line ϕ= –kh (representing the upstream equipotential) is a parabola, whereas the upstream equipo- tential in the embankment dam section is the upstream slope. Based on an extensive study of the problem, Casagrande (1940) recommended that the initial point of the basic parabola should be taken at G (Figure 2.21) where GC=0.3HC. The coordinates of point G, substituted in Equation 2.40, enable the value of x0 to be determined; the basic parabola can then be plotted. The top flow line must intersect the upstream slope at right angles; a correction CJ must therefore be made (using personal judgement) to the basic parabola. The flow net can then be completed as shown in Figure 2.21.
If the discharge surface AD is not horizontal, as in the cases shown in Figure 2.22, a further correction KD to the basic parabola is required. The angle β is used to describe the direction of the discharge surface relative to AB. The correction can be made with the aid of values of the ratio MD/MA = Δa/a, given by Casagrande for the range of values of β shown in Table 2.4.
Figure 2.21 Flow net for embankment dam section.
Example 2.4
An homogeneous anisotropic embankment dam section is detailed in Figure 2.23(a), the coefficients of permeability in the x and z directions being 4.5 × 10–8 and 1.6 × 10–8 m/s, respectively. Construct the flow net and determine the quantity of seepage through the dam. What is the pore water pressure at point P?
Figure 2.22 Downstream correction to basic parabola.
Table 2.4 Downstream correction to basic parabola. Reproduced from A. Casagrande (1940) ‘Seepage through dams’, in Contributions to Soil Mechanics
1925–1940, by permission of the Boston Society of Civil Engineers
β 30° 60° 90° 120° 150° 180°
∆a/a (0.36) 0.32 0.26 0.18 0.10 0
Figure 2.23 Example 2.4.
Seepage
Solution
The scale factor for transformation in the x direction is k
k
z x
= 1 6 = 4 5. 0 60
. .
The equivalent isotropic permeability is
¢ =
( )
=
(
´)
´ - = ´ -k k kx z
4 5 1 6. . 108 2 7 10. 8m/s
The section is drawn to the transformed scale in Figure 2.23(b). The focus of the basic parabola is at point A. The basic parabola passes through point G such that
GC=0.3HC=0.3 27.´ 00=8.1 m0 i.e. the coordinates of G are
x= -4 80 0. ,z= +18.00
Substituting these coordinates in Equation 2.40:
-40 80= -18 00
0 4
2
0
. x .
x Hence
x0=1.9 m0
Using Equation 2.40, the coordinates of a number of points on the basic parabola are now calculated:
TABLE E
x (m) 1.90 0 −5.00 −10.00 −20.00 −30.00
z (m) 0 3.80 7.24 9.51 12.90 15.57
The basic parabola is plotted in Figure 2.23(b). The upstream correction is made (JC) and the flow net completed, ensuring that there are equal vertical intervals between the points of intersection of successive equipotentials with the top flow line. In the flow net, there are four flow channels and eighteen equipotential drops. Hence, the quantity of seepage (per unit length) is
q k hN
= ¢ N
= ´ - ´ ´ = ´ -
f d
m /s3
2 7 10 18 4
18 1 1 10
8 7
. .
The quantity of seepage can also be determined from Equation 2.39 (without the necessity of drawing the flow net):
q= k x¢
= ´ ´ - ´ = ´ -
2
2 2 7 10 1 90 1 0 10
0
8 7
. . . m /s3
To determine the pore water pressure at P, Level AD is first selected as datum. An equipo- tential RS is drawn through point P (transformed position). By inspection, the total head at P is 15.60 m. At P the elevation head is 5.50 m, so the pressure head is 10.10 m and the pore water pressure is
up=9 81 10 10 99. ´ . = kPa
Alternatively, the pressure head at P is given directly by the vertical distance of P below the point of intersection (R) of equipotential RS with the top flow line.
Seepage control in embankment dams
The design of an embankment dam section and, where possible, the choice of soils are aimed at reducing or eliminating the detrimental effects of seeping water. Where high hydraulic gradients exist there is a possibility that the seeping water may cause internal erosion within the dam, especially if the soil is poorly compacted (Chapter 12). Erosion can work its way back into the embankment, creating voids in the form of channels or ‘pipes’, thus impairing the stability of the dam. This form of erosion is referred to as piping.
A section with a central core of low permeability, aimed at reducing the volume of seepage, is shown in Figure 2.24(a). Practically all the total head is lost in the core, and if the core is nar- row, high hydraulic gradients will result. There is a particular danger of erosion at the boundary between the core and the adjacent soil (of higher permeability) under a high exit gradient from the core. Protection against this danger can be given by means of a ‘chimney’ drain (Figure 2.24(a)) at the downstream boundary of the core. The drain, designed as a filter (see Section 2.10) to provide a barrier to the migration of soil particles from the core, also serves as an interceptor, keeping the downstream slope in an unsaturated state.
Most embankment dam sections are non-homogeneous owing to zones of different soil types, making the construction of the flow net more difficult. The basic parabola construc- tion for the top flow line applies only to homogeneous sections, but the condition that there must be equal vertical distances between the points of intersection of equipotentials with the top flow line applies equally to a non-homogeneous section. The transfer condition (Equation 2.31) must be satisfied at all zone boundaries. In the case of a section with a central core of low permeability, the application of Equation 2.31 means that the lower the permeability ratio, the lower the position of the top flow line in the downstream zone (in the absence of a chimney drain).
If the foundation soil is more permeable than the dam, the control of underseepage is essential.
Underseepage can be virtually eliminated by means of an ‘impermeable’ cut-off such as a grout curtain (Figure 2.24(b)). Another form of cut-off is the concrete diaphragm wall (see Section 11.7). Any measure designed to lengthen the seepage path, such as an impermeable upstream blanket (Figure 2.24(c)), will result in a partial reduction in underseepage.
An excellent treatment of seepage control is given by Cedergren (1989).
Seepage
Example 2.5
Consider the concrete dam spillway from Example 2.2 (Figure 2.10). Determine the effect of the length of the cut-off wall (Lw) on the reduction in seepage flow beneath the spillway.
Determine also the reduction in seepage flow due to an impermeable upstream blanket of length Lb, and compare the efficacy of the two methods of seepage control.
Figure 2.24 (a) Central core and chimney drain, (b) grout curtain, and (c) impermeable upstream blanket.
Figure 2.25 Example 2.5.
Solution
The seepage flow can be determined using the spreadsheet tool on the Companion Website.
Repeating the calculations for different lengths, Lw, of cut-off wall (Figure 2.25(a)) and separately for different lengths of impermeable blanket, Lb, as shown in Figure 2.25(b), the results shown in Figure 2.25(c) can be determined. By comparing the two methods, it can be seen that for thin layers such as in this problem, cut-off walls are generally more effec- tive at reducing seepage and typically require a lower volume of material (and therefore lower cost) to achieve the same reduction in seepage flow.