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# Craig's Soil Mechanics 7th Edition

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�1 ¼ �k1h1 and �2 ¼ �k2h2 At the common point B, h1 ¼ h2; therefore �1 k1 ¼ �2 k2 Differentiating with respect to s, the direction along the boundary: 1 k1 @�1 @s ¼ 1 k2 @�2 @s i.e. v1s k1 ¼ v2s k2 For continuity of flow across the boundary the normal components of discharge velocity must be equal, i.e. v1n ¼ v2n Figure 2.14 Transfer condition. 54 Seepage Therefore 1 k1 v1s v1n ¼ 1 k2 v2s v2n Hence it follows that tan 1 tan 2 ¼ k1 k2 ð2:26Þ Equation 2.26 specifies the change in direction of the flow line passing through point B. This equation must be satisfied on the boundary by every flow line crossing the boundary. Equation 2.13 can be written as � ¼ �n �s �� i.e. �q ¼ �n �s k�h If �q and �h are each to have the same values on both sides of the boundary then �n �s � � 1 k1 ¼ �n �s � � 2 k2 and it is clear that curvilinear squares are possible only in one soil. If �n �s � � 1 ¼1 then �n �s � � 2 ¼ k1 k2 ð2:27Þ If the permeability ratio is less than 1⁄10 it is unlikely that the part of the flow net in the soil of higher permeability needs to be considered. 2.8 SEEPAGE THROUGH EMBANKMENT DAMS This problem is an example of unconfined seepage, one boundary of the flow region being a phreatic surface on which the pressure is atmospheric. In section the phreatic surface constitutes the top flow line and its position must be estimated before the flow net can be drawn. Seepage through embankment dams 55 Consider the case of a homogeneous isotropic embankment dam on an imper- meable foundation, as shown in Figure 2.15. The impermeable boundary BA is a flow line and CD is the required top flow line. At every point on the upstream slope BC the total head is constant (u/�w and z varying from point to point but their sum remaining constant); therefore, BC is an equipotential. If the downstream water level is taken as datum then the total head on equipotential BC is equal to h, the difference between the upstream and downstream water levels. The discharge sur- face AD, for the case shown in Figure 2.15 only, is the equipotential for zero total head. At every point on the top flow line the pressure is zero (atmospheric), so total head is equal to elevation head and there must be equal vertical intervals �z between the points of intersection between successive equipotentials and the top flow line. A suitable filter must always be constructed at the discharge surface in an embank- ment dam. The function of the filter is to keep the seepage entirely within the dam; water seeping out onto the downstream slope would result in the gradual erosion of the slope. A horizontal underfilter is shown in Figure 2.15. Other possible forms of filter are illustrated in Figure 2.19(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 2.15 can be written as follows: Equipotential BC: � ¼ �kh Equipotential AD: � ¼ 0 Flow line CD: ¼ q (also, � ¼ �kz) Flow line BA: ¼ 0 The conformal transformation r ¼ w2 Complex variable theory can be used to obtain a solution to the embankment dam problem. Let the complex number w ¼ �þ i be an analytic function of r ¼ xþ iz. Consider the function r ¼ w2 Figure 2.15 Homogeneous embankment dam section. 56 Seepage Thus ðxþ izÞ ¼ ð�þ i Þ2 ¼ ð�2 þ 2i� � 2Þ Equating real and imaginary parts: x ¼ �2 � 2 ð2:28Þ z ¼ 2� ð2:29Þ Equations 2.28 and 2.29 govern the transformation of points between the r and w planes. Consider the transformation of the straight lines ¼ n, where n ¼ 0, 1, 2, 3 (Figure 2.16(a)). From Equation 2.29 � ¼ z 2n and Equation 2.28 becomes x ¼ z 2 4n2 � n2 ð2:30Þ Equation 2.30 represents a family of confocal parabolas. For positive values of z the parabolas for the specified values of n are plotted in Figure 2.16(b). Consider also the transformation of the straight lines � ¼ m, where m ¼ 0, 1, 2, . . . , 6 (Figure 2.16(a)). From Equation 2.29 ¼ z 2m and Equation 2.28 becomes x ¼ m2 � z 2 4m2 ð2:31Þ Figure 2.16 Conformal transformation r ¼ w2: (a) w plane and (b) r plane. Seepage through embankment dams 57 Equation 2.31 represents a family of confocal parabolas conjugate with the parabolas represented by Equation 2.30. For positive values of z the parabolas for the specified values of m are plotted in Figure 2.16(b). The two families of parabolas satisfy the requirements of a flow net. Application to embankment dam section The flow region in the w plane satisfying the boundary conditions for the section (Figure 2.15) is shown in Figure 2.17(a). In this case the transformation function r ¼ Cw2 will be used, where C is a constant. Equations 2.28 and 2.29 then become x ¼ Cð�2 � 2Þ z ¼ 2C� The equation of the top flow line can be derived by substituting the conditions ¼ q � ¼ �kz Figure 2.17 Transformation for embankment dam section: (a) w plane and (b) r plane. 58 Seepage Thus z ¼ �2Ckzq ; C ¼ � 1 2kq Hence x ¼ � 1 2kq ðk2z2 � q2Þ x ¼ 1 2 q k � k q z2 � � ð2:32Þ The curve represented by Equation 2.32 is referred to as Kozeny’s basic parabola and is shown in Figure 2.17(b), the origin and focus both being at A. When z ¼ 0 the value of x is given by x0 ¼ q 2k ; q ¼ 2kx0 ð2:33Þ where 2x0 is the directrix distance of the basic parabola. When x ¼ 0 the value of z is given by z0 ¼ q k ¼ 2x0 Substituting Equation 2.33 in Equation 2.32 yields x ¼ x0 � z 2 4x0 ð2:34Þ The basic parabola can be drawn using Equation 2.34, 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 � ¼ �kh (representing the upstream equipotential) is a parabola, whereas the upstream equipotential in the embankment dam section is the upstream slope. Based on an extensive study of the problem, Casagrande [2] recommended that the initial point of the basic parabola should be taken at G (Figure 2.18) where GC ¼ 0:3HC. The coordinates of point G, substituted in Equation 2.34, 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 2.18. If the discharge surface AD is not horizontal, as in the cases shown in Figure 2.19, a further correction KD to the basic parabola is required. The angle � is used to describe the direction of the discharge surface relative to AB. The correction can be made with Seepage through embankment dams 59 the aid of values of the ratio MD/MA ¼ �a/a, given by Casagrande for the range of values of � (Table 2.4). 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. Erosion can work its way back into the embankment, creating voids in the form of channels or ‘pipes’, and thus impairing the stability of the dam. This form of erosion is referred to as piping. Figure 2.18 Flow net for embankment dam section. Figure 2.19 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