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# Craig's Soil Mechanics 7th Edition

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Mechanics 1925–1940, by permission of the Boston Society of Civil Engineers � 30� 60� 90� 120� 150� 180� �a/a (0.36) 0.32 0.26 0.18 0.10 0 60 Seepage A section with a central core of low permeability, aimed at reducing the volume of seepage, is shown in Figure 2.20(a). Practically all the total head is lost in the core and if the core is narrow, high hydraulic gradients will result. There is a particular danger of erosion at the boundary between the core and the adjacent soil (of higher per- meability) under a high exit gradient from the core. Protection against this danger can be given by means of a ‘chimney’ drain (Figure 2.20(a)) at the downstream boundary of the core. The drain, designed as a filter to provide a barrier to 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 construction 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 Figure 2.20 (a) Central core and chimney drain, (b) grout curtain and (c) impermeable upstream blanket. Seepage through embankment dams 61 of equipotentials with the top flow line applies equally to a non-homogeneous section. The transfer condition (Equation 2.26) must be satisfied at all zone boundaries. In the case of a section with a central core of low permeability, the application of Equation 2.26 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 2.20(b)). Another form of cut-off is the concrete diaphragm wall (Section 6.9). Any measure designed to lengthen the seepage path, such as an impermeable upstream blanket (Figure 2.20(c)), will result in a partial reduction in underseepage. An excellent treatment of seepage control is given by Cedergren [3]. Filter design Filters used to control seepage must satisfy certain fundamental requirements. The pores must be small enough to prevent particles from being carried in from the adjacent soil. The permeability must be high enough to ensure the free drainage of water entering the filter. The capacity of a filter should be such that it does not become fully saturated. In the case of an embankment dam, a filter placed down- stream from the core should be capable of controlling and sealing any leak which develops through the core as a result of internal erosion. The filter must also remain stable under the abnormally high hydraulic gradient which is liable to develop adjacent to such a leak. Based on extensive laboratory tests by Sherard et al. [7, 8] and on design experience, it has been shown that filter performance can be related to the size D15 obtained from the particle size distribution curve of the filter material. Average pore size, which is largely governed by the smaller particles in the filter, is well represented byD15. A filter of uniform grading will trap all particles larger than around 0.11D15; particles smaller than this size will be carried through the filter in suspension in the seeping water. The characteristics of the adjacent soil, in respect of its retention by the filter, can be represented by the size D85 for that soil. The following criterion has been recom- mended for satisfactory filter performance: ðD15Þf ðD85Þs < 5 ð2:35Þ where (D15)f and (D85)s refer to the filter and the adjacent (upstream) soil, respectively. However, in the case of filters for fine soils the following limit is recommended for the filter material: D15 � 0:5mm Care must be taken to avoid segregation of the component particles of the filter during construction. 62 Seepage To ensure that the permeability of the filter is high enough to allow free drainage, it is recommended that ðD15Þf ðD15Þs > 5 ð2:36Þ Graded filters comprising two (or more) layers with different gradings can also be used, the finer layer being on the upstream side. The above criterion (Equation 2.35) would also be applied to the component layers of the filter. Example 2.4 A homogeneous anisotropic embankment dam section is detailed in Figure 2.21(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? The scale factor for transformation in the x direction is ﬃﬃﬃﬃﬃ kz kx s ¼ ﬃﬃﬃﬃﬃﬃﬃ 1:6 4:5 r ¼ 0:60 The equivalent isotropic permeability is k0 ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðkxkzÞ p ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ð4:5� 1:6Þ p � 10�8 ¼ 2:7� 10�8 m=s Figure 2.21 Example 2.4. Seepage through embankment dams 63 The section is drawn to the transformed scale as in Figure 2.21(b). The focus of the basic parabola is at point A. The basic parabola passes through point G such that GC ¼ 0:3HC ¼ 0:3� 27:00 ¼ 8:10m i.e. the coordinates of G are x ¼ �40:80; z ¼ þ18:00 Substituting these coordinates in Equation 2.34: �40:80 ¼ x0� 18:00 2 4x0 Hence x0 ¼ 1:90m Using Equation 2.34 the coordinates of a number of points on the basic parabola are now calculated: x 1.90 0 �5.00 �10.00 �20.00 �30.00 z 0 3.80 7.24 9.51 12.90 15.57 The basic parabola is plotted in Figure 2.21(b). The upstream correction is made 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 3.8 flow channels and 18 equipotential drops. Hence, the quantity of seepage (per unit length) is q ¼ k0h Nf Nd ¼ 2:7� 10�8 � 18� 3:8 18 ¼ 1:0� 10�7 m3=s The quantity of seepage can also be determined from Equation 2.33 (without the necessity of drawing the flow net): q ¼ 2k0x0 ¼ 2� 2:7� 10�8 � 1:90 ¼ 1:0� 10�7 m3=s Level AD is selected as datum. An equipotential RS is drawn through point P (transformed position). By inspection the total head at P is 15.60m. At P the elevation head is 5.50m, so the pressure head is 10.10m and the pore water pressure is up ¼ 9:8� 10:10 ¼ 99 kN=m2 64 Seepage 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. Example 2.5 Draw the flow net for the non-homogeneous embankment dam section detailed in Figure 2.22 and determine the quantity of seepage through the dam. Zones 1 and 2 are isotropic, having coefficients of permeability 1:0� 10�7 and 4:0� 10�7 m/s, respectively. The ratio k2/k1 ¼ 4. The basic parabola is not applicable in this case. Three funda- mental conditions must be satisfied in the flow net: 1 There must be equal vertical intervals between points of intersection of equipo- tentials with the top flow line. 2 If the part of the flow net in zone 1 consists of curvilinear squares then the part in zone 2 must consist of curvilinear rectangles having a length/breadth ratio of 4. 3 For each flow line the transfer condition (Equation 2.26) must be satisfied at the inter-zone boundary. The flow net is shown in Figure 2.22. In the flow net there are 3.6 flow channels and 8 equipotential drops. The quantity of seepage per unit length is given by q ¼ k1h Nf Nd ¼ 1:0� 10�7 � 16� 3:6 8 ¼ 7:2� 10�7 m3=s (If curvilinear squares are used in zone 2, curvilinear rectangles having a length/ breadth ratio of 0.25 must be used in zone 1 and k2 must be used in the seepage