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# Craig's Soil Mechanics 7th Edition

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is calculated at the points of intersection of the equipoten- tials with the base of the dam. The total head at each point is obtained from the flow net and the elevation head from the section. The calculations are shown in Table 2.3 and the pressure diagram is plotted in Figure 2.9. Example 2.3 A river bed consists of a layer of sand 8.25m thick overlying impermeable rock; the depth of water is 2.50m. A long cofferdam 5.50m wide is formed by driving two lines of sheet piling to a depth of 6.00m below the level of the river bed and excavation to a depth of 2.00m below bed level is carried out within the cofferdam. The water level within the cofferdam is kept at excavation level by pumping. If the flow of water into Table 2.2 Level z (m) hb (m) ub/�w (m) hf (m) uf /�w (m) ub � uf (kN/m2) 1 0 2.30 2.30 0 0 22.6 2 �2.70 2.10 4.80 0 2.70 20.6 3 �4.00 2.00 6.00 0 4.00 19.6 4 �5.50 1.83 7.33 0.21 5.71 15.9 5 �7.10 1.68 8.78 0.50 7.60 11.6 6 �8.30 1.51 9.81 0.84 9.14 6.6 7 �8.70 1.25 9.95 1.04 9.74 2.1 Flow nets 47 the cofferdam is 0.25m3/h per unit length, what is the coefficient of permeability of the sand? What is the hydraulic gradient immediately below the excavated surface? The section and flow net appear in Figure 2.10. In the flow net there are 6.0 flow channels and 10 equipotential drops. The total head loss is 4.50m. The coefficient of permeability is given by k ¼ q h Nf=Ndð Þ Table 2.3 Point h (m) z (m) h� z (m) u ¼ �w(h� z) (kN/m2) 1 0.27 �1:80 2.07 20.3 2 0.53 �1:80 2.33 22.9 3 0.80 �1:80 2.60 25.5 4 1.07 �2:10 3.17 31.1 5 1.33 �2:40 3.73 36.6 6 1.60 �2:40 4.00 39.2 7 1.87 �2:40 4.27 41.9 7 1 2 2.00 �2:40 4.40 43.1 Figure 2.9 Example 2.2. 48 Seepage ¼ 0:25 4:50� 6=10� 602 ¼ 2:6� 10 �5 m=s The distance (�s) between the last two equipotentials is measured as 0.9m. The required hydraulic gradient is given by i ¼ �h �s ¼ 4:50 10� 0:9 ¼ 0:50 2.5 ANISOTROPIC SOIL CONDITIONS It will now be assumed that the soil, although homogeneous, is anisotropic with respect to permeability. Most natural soil deposits are anisotropic, with the coefficient of permeability having a maximum value in the direction of stratification and a minimum value in the direction normal to that of stratification; these directions are denoted by x and z, respectively, i.e. kx ¼ kmax and kz ¼ kmin Figure 2.10 Example 2.3. Anisotropic soil conditions 49 In this case the generalized form of Darcy’s law is vx ¼ kxix ¼ �kx @h @x ð2:18aÞ vz ¼ kziz ¼ �kz @h @z ð2:18bÞ Also, in any direction s, inclined at angle to the x direction, the coefficient of permeability is defined by the equation vs ¼ �ks @h @s Now @h @s ¼ @h @x @x @s þ @h @z @z @s i.e. vs ks ¼ vx kx cos þ vz kz sin The components of discharge velocity are also related as follows: vx ¼ vs cos vz ¼ vs sin Hence 1 ks ¼ cos 2 kx þ sin 2 kz or s2 ks ¼ x 2 kx þ z 2 kz ð2:19Þ The directional variation of permeability is thus described by Equation 2.19 which represents the ellipse shown in Figure 2.11. Given the generalized form of Darcy’s law (Equation 2.18) the equation of con- tinuity (2.6) can be written: kx @2h @x2 þ kz @ 2h @z2 ¼ 0 ð2:20Þ 50 Seepage or @2h ðkz=kxÞ@x2 þ @2h @z2 ¼ 0 Substituting xt ¼ x ﬃﬃﬃﬃﬃ kz kx s ð2:21Þ the equation of continuity becomes @2h @x2t þ @ 2h @z2 ¼ 0 ð2:22Þ which is the equation of continuity for an isotropic soil in an xt�z plane. Thus, Equation 2.21 defines a scale factor which can be applied in the x direction to transform a given anisotropic flow region into a fictitious isotropic flow region in which the Laplace equation is valid. Once the flow net (representing the solution of the Laplace equation) has been drawn for the transformed section the flow net for the natural section can be obtained by applying the inverse of the scaling factor. Essential data, however, can normally be obtained from the transformed section. The necessary transformation could also be made in the z direction. The value of coefficient of permeability applying to the transformed section, referred to as the equivalent isotropic coefficient, is k0 ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðkxkzÞ p ð2:23Þ A formal proof of Equation 2.23 has been given by Vreedenburgh [9]. The validity of Equation 2.23 can be demonstrated by considering an elemental flow net field through Figure 2.11 Permeability ellipse. Anisotropic soil conditions 51 which flow is in the x direction. The flow net field is drawn to the transformed and natural scales in Figure 2.12, the transformation being in the x direction. The discharge velocity vx can be expressed in terms of either k0 (transformed section) or kx (natural section), i.e. vx ¼ �k0 @h @xt ¼ �kx @h @x where @h @xt ¼ @hﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃðkz=kxÞp @x Thus k0 ¼ kx ﬃﬃﬃﬃﬃ kz kx s ¼ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ðkxkzÞ p 2.6 NON-HOMOGENEOUS SOIL CONDITIONS Two isotropic soil layers of thicknesses H1 and H2 are shown in Figure 2.13, the respective coefficients of permeability being k1 and k2; the boundary between the layers is horizontal. (If the layers are anisotropic, k1 and k2 represent the equivalent isotropic coefficients for the layers.) The two layers can be considered as a single homogeneous anisotropic layer of thickness (H1 þH2) in which the coefficients in the directions parallel and normal to that of stratification are kx and kz, respectively. For one-dimensional seepage in the horizontal direction, the equipotentials in each layer are vertical. If h1 and h2 represent total head at any point in the respective layers, then for a common point on the boundary h1 ¼ h2. Therefore, any vertical line through the two layers represents a common equipotential. Thus, the hydraulic gradients in the two layers, and in the equivalent single layer, are equal; the equal hydraulic gradients are denoted by ix. Figure 2.12 Elemental flow net field. 52 Seepage The total horizontal flow per unit time is given by qx ¼ ðH1 þH2Þkxix ¼ ðH1k1 þH2k2Þix ; kx ¼ H1k1 þH2k2 H1 þH2 ð2:24Þ For one-dimensional seepage in the vertical direction, the discharge velocities in each layer, and in the equivalent single layer, must be equal if the requirement of continuity is to be satisfied. Thus vz ¼ kziz ¼ k1i1 ¼ k2i2 where iz is the average hydraulic gradient over the depth (H1 þH2). Therefore i1 ¼ kz k1 iz and i2 ¼ kz k2 iz Now the loss in total head over the depth (H1 þH2) is equal to the sum of the losses in total head in the individual layers, i.e. izðH1 þH2Þ ¼ i1H1 þ i2H2 ¼ kziz H1 k1 þH2 k2 � � ; kz ¼ H1 þH2 H1 k1 � � þ H2 k2 � � ð2:25Þ Similar expressions for kx and kz apply in the case of any number of soil layers. It can be shown that kx must always be greater than kz, i.e. seepage can occur more readily in the direction parallel to stratification than in the direction perpendicular to stratification. 2.7 TRANSFER CONDITION Consideration is now given to the condition which must be satisfied when seepage takes place diagonally across the boundary between two isotropic soils 1 and 2 having Figure 2.13 Non-homogeneous soil conditions. Transfer condition 53 coefficients of permeability k1 and k2, respectively. The direction of seepage approach- ing a point B on the boundary ABC is at angle 1 to the normal at B, as shown in Figure 2.14; the discharge velocity approaching B is v1. The components of v1 along the boundary and normal to the boundary are v1s and v1n respectively. The direction of seepage leaving point B is at angle 2 to the normal, as shown; the discharge velocity leaving B is v2. The components of v2 are v2s and v2n. For soils 1 and 2 respectively