Page 1 NPTEL- Advanced Geotechnical Engineering Dept. of Civil Engg. Indian Institute of Technology, Kanpur 1 Module 2 Lecture 9 Permeability and Seepage -5 Topics 1.2.7 Numerical Analysis of Seepage 1.2.8 Seepage Force per Unit Volume of Soil Mass 1.2.9 Safety of Hydraulic Structures against Piping 1.2.10 Calculation of Seepage through an Earth Dam Resting on an Impervious Base ? Dupuit’s solution ? Schaffernak’s solution ? Casagrande’s solution ? Pavlovsky’s solution 1.2.7 Numerical Analysis of Seepage In this section, we develop some approximate finite-difference equation s for solving seepage problems. We start from Lalplace’s equation, which was derived in section 2.2.1: for two- dimensional seepage (1.86) Figure 2.41 shows a part of a region in which flow is taking place. For flow in the horizontal direction, using Taylor’s series we can write Page 2 NPTEL- Advanced Geotechnical Engineering Dept. of Civil Engg. Indian Institute of Technology, Kanpur 1 Module 2 Lecture 9 Permeability and Seepage -5 Topics 1.2.7 Numerical Analysis of Seepage 1.2.8 Seepage Force per Unit Volume of Soil Mass 1.2.9 Safety of Hydraulic Structures against Piping 1.2.10 Calculation of Seepage through an Earth Dam Resting on an Impervious Base ? Dupuit’s solution ? Schaffernak’s solution ? Casagrande’s solution ? Pavlovsky’s solution 1.2.7 Numerical Analysis of Seepage In this section, we develop some approximate finite-difference equation s for solving seepage problems. We start from Lalplace’s equation, which was derived in section 2.2.1: for two- dimensional seepage (1.86) Figure 2.41 shows a part of a region in which flow is taking place. For flow in the horizontal direction, using Taylor’s series we can write NPTEL- Advanced Geotechnical Engineering Dept. of Civil Engg. Indian Institute of Technology, Kanpur 2 (2.141) And (2.142) Adding equation (2.141) and (2.142) we obtain (2.143) Assuming to be small, can neglect the third and subsequent terms on the right-hand side of equation (2.143). thus (2.144) Similarly, for flow in the z direction we can obtain (2.145) Substitution of equation (2.144) and (2.145) into equation (1.86) gives (2.146) If . Equation (2.146) simplifies to Figure 2.41 Hydraulic heads for flow in a region Page 3 NPTEL- Advanced Geotechnical Engineering Dept. of Civil Engg. Indian Institute of Technology, Kanpur 1 Module 2 Lecture 9 Permeability and Seepage -5 Topics 1.2.7 Numerical Analysis of Seepage 1.2.8 Seepage Force per Unit Volume of Soil Mass 1.2.9 Safety of Hydraulic Structures against Piping 1.2.10 Calculation of Seepage through an Earth Dam Resting on an Impervious Base ? Dupuit’s solution ? Schaffernak’s solution ? Casagrande’s solution ? Pavlovsky’s solution 1.2.7 Numerical Analysis of Seepage In this section, we develop some approximate finite-difference equation s for solving seepage problems. We start from Lalplace’s equation, which was derived in section 2.2.1: for two- dimensional seepage (1.86) Figure 2.41 shows a part of a region in which flow is taking place. For flow in the horizontal direction, using Taylor’s series we can write NPTEL- Advanced Geotechnical Engineering Dept. of Civil Engg. Indian Institute of Technology, Kanpur 2 (2.141) And (2.142) Adding equation (2.141) and (2.142) we obtain (2.143) Assuming to be small, can neglect the third and subsequent terms on the right-hand side of equation (2.143). thus (2.144) Similarly, for flow in the z direction we can obtain (2.145) Substitution of equation (2.144) and (2.145) into equation (1.86) gives (2.146) If . Equation (2.146) simplifies to Figure 2.41 Hydraulic heads for flow in a region NPTEL- Advanced Geotechnical Engineering Dept. of Civil Engg. Indian Institute of Technology, Kanpur 3 or (2.147) Equation (2.147) can also be derived considering Darcy’s law, . For the rate of flow from point 1 to point 0 through the channel shown hatched in Figure 2.42a, we have (2.148) Similarly, (2.149) (2.150) (2.151) Since the total rate of flow into point 0 is equal to the total rate of flow out of point . Hence + (2.152) Taking and substituting equation (2.148) to (2.151) into equation (2.152), we get If the point 0 is located on the boundary of a pervious and an impervious layer as shown in Figure 2.42b. equation (2.147) must be modified as follows: (2.153) (2.154) (2.155) For continuity of flow, (2.156) We , combining equations (2.153) to (2.156) gives Or (2.157) When point 0 is located at the bottom of a pilling (Figure 2.42c), the equation for the hydraulic head for flow continuity can be given by Page 4 NPTEL- Advanced Geotechnical Engineering Dept. of Civil Engg. Indian Institute of Technology, Kanpur 1 Module 2 Lecture 9 Permeability and Seepage -5 Topics 1.2.7 Numerical Analysis of Seepage 1.2.8 Seepage Force per Unit Volume of Soil Mass 1.2.9 Safety of Hydraulic Structures against Piping 1.2.10 Calculation of Seepage through an Earth Dam Resting on an Impervious Base ? Dupuit’s solution ? Schaffernak’s solution ? Casagrande’s solution ? Pavlovsky’s solution 1.2.7 Numerical Analysis of Seepage In this section, we develop some approximate finite-difference equation s for solving seepage problems. We start from Lalplace’s equation, which was derived in section 2.2.1: for two- dimensional seepage (1.86) Figure 2.41 shows a part of a region in which flow is taking place. For flow in the horizontal direction, using Taylor’s series we can write NPTEL- Advanced Geotechnical Engineering Dept. of Civil Engg. Indian Institute of Technology, Kanpur 2 (2.141) And (2.142) Adding equation (2.141) and (2.142) we obtain (2.143) Assuming to be small, can neglect the third and subsequent terms on the right-hand side of equation (2.143). thus (2.144) Similarly, for flow in the z direction we can obtain (2.145) Substitution of equation (2.144) and (2.145) into equation (1.86) gives (2.146) If . Equation (2.146) simplifies to Figure 2.41 Hydraulic heads for flow in a region NPTEL- Advanced Geotechnical Engineering Dept. of Civil Engg. Indian Institute of Technology, Kanpur 3 or (2.147) Equation (2.147) can also be derived considering Darcy’s law, . For the rate of flow from point 1 to point 0 through the channel shown hatched in Figure 2.42a, we have (2.148) Similarly, (2.149) (2.150) (2.151) Since the total rate of flow into point 0 is equal to the total rate of flow out of point . Hence + (2.152) Taking and substituting equation (2.148) to (2.151) into equation (2.152), we get If the point 0 is located on the boundary of a pervious and an impervious layer as shown in Figure 2.42b. equation (2.147) must be modified as follows: (2.153) (2.154) (2.155) For continuity of flow, (2.156) We , combining equations (2.153) to (2.156) gives Or (2.157) When point 0 is located at the bottom of a pilling (Figure 2.42c), the equation for the hydraulic head for flow continuity can be given by NPTEL- Advanced Geotechnical Engineering Dept. of Civil Engg. Indian Institute of Technology, Kanpur 4 (2.158) Note that 2’ and 2” are two points at the same elevation on the opposite sides of the sheet pile with hydraulic heads of , respectively. For this condition we can obtain (for ), through a similar procedure to that above, (2.159) Seepage in layered soils. equation (2.147), which we derived above, is valid for seepage in homogeneous soils. However, for the case of flow across the boundary of one homogeneous soil layer to another, equation (2.147) must be modified. Referring to Figure 2.42d, since the flow region is located half in soil 1 with a coefficient of permeability and half in soil 2 with a coefficient of permeability , we can say that (2.160) Now, if we replace soil 2 by soil 1, it will have a hydraulic head of in place of . For the velocity to remain the same, (2.161) Figure 2.42 Page 5 NPTEL- Advanced Geotechnical Engineering Dept. of Civil Engg. Indian Institute of Technology, Kanpur 1 Module 2 Lecture 9 Permeability and Seepage -5 Topics 1.2.7 Numerical Analysis of Seepage 1.2.8 Seepage Force per Unit Volume of Soil Mass 1.2.9 Safety of Hydraulic Structures against Piping 1.2.10 Calculation of Seepage through an Earth Dam Resting on an Impervious Base ? Dupuit’s solution ? Schaffernak’s solution ? Casagrande’s solution ? Pavlovsky’s solution 1.2.7 Numerical Analysis of Seepage In this section, we develop some approximate finite-difference equation s for solving seepage problems. We start from Lalplace’s equation, which was derived in section 2.2.1: for two- dimensional seepage (1.86) Figure 2.41 shows a part of a region in which flow is taking place. For flow in the horizontal direction, using Taylor’s series we can write NPTEL- Advanced Geotechnical Engineering Dept. of Civil Engg. Indian Institute of Technology, Kanpur 2 (2.141) And (2.142) Adding equation (2.141) and (2.142) we obtain (2.143) Assuming to be small, can neglect the third and subsequent terms on the right-hand side of equation (2.143). thus (2.144) Similarly, for flow in the z direction we can obtain (2.145) Substitution of equation (2.144) and (2.145) into equation (1.86) gives (2.146) If . Equation (2.146) simplifies to Figure 2.41 Hydraulic heads for flow in a region NPTEL- Advanced Geotechnical Engineering Dept. of Civil Engg. Indian Institute of Technology, Kanpur 3 or (2.147) Equation (2.147) can also be derived considering Darcy’s law, . For the rate of flow from point 1 to point 0 through the channel shown hatched in Figure 2.42a, we have (2.148) Similarly, (2.149) (2.150) (2.151) Since the total rate of flow into point 0 is equal to the total rate of flow out of point . Hence + (2.152) Taking and substituting equation (2.148) to (2.151) into equation (2.152), we get If the point 0 is located on the boundary of a pervious and an impervious layer as shown in Figure 2.42b. equation (2.147) must be modified as follows: (2.153) (2.154) (2.155) For continuity of flow, (2.156) We , combining equations (2.153) to (2.156) gives Or (2.157) When point 0 is located at the bottom of a pilling (Figure 2.42c), the equation for the hydraulic head for flow continuity can be given by NPTEL- Advanced Geotechnical Engineering Dept. of Civil Engg. Indian Institute of Technology, Kanpur 4 (2.158) Note that 2’ and 2” are two points at the same elevation on the opposite sides of the sheet pile with hydraulic heads of , respectively. For this condition we can obtain (for ), through a similar procedure to that above, (2.159) Seepage in layered soils. equation (2.147), which we derived above, is valid for seepage in homogeneous soils. However, for the case of flow across the boundary of one homogeneous soil layer to another, equation (2.147) must be modified. Referring to Figure 2.42d, since the flow region is located half in soil 1 with a coefficient of permeability and half in soil 2 with a coefficient of permeability , we can say that (2.160) Now, if we replace soil 2 by soil 1, it will have a hydraulic head of in place of . For the velocity to remain the same, (2.161) Figure 2.42 NPTEL- Advanced Geotechnical Engineering Dept. of Civil Engg. Indian Institute of Technology, Kanpur 5 Or (2.162) Thus, based on equation (1.86), we can write (2.163) Taking and substituting equation (2.162) into equation (2.163), (2.163a) Or (2.164) The application of the equations developed in this section can best be demonstrated by the use of a numerical example. Consider the problem of determining the hydraulic heads at various points below the dam shown in Figure 2.30. let . Since the flow net below the dam will be symmetrical, we will consider only the left-half. The steps for determining the values of h at various points in the permeable soil layers are as follows: 1. Roughly sketch out a flow net. 2. Based on the rough flow net (step 1), assign some values for the hydraulic heads at various grid points. These are shown in Figure 2.43a. Note that the values of h assigned here are in percent. 3. Consider the heads for row 1 (i.e., i =1). The are 100 in Figure 2.43a; these are correct values based on the boundary conditions. The are estimated values. The flow condition for these grid points is similar to that shown in Figure 2.42b; and according to equation (2.157), (2.165) Since the hydraulic heads in Figure 2.43 are assumed values, equation (2.165) will not be satisfied. For example, for the grid point . If these values are substituted in equation (2.165) we get [68 + 2(78) +100]-4(84) = -12, instead of zero. If we set -12 equal to R (where R stands for residual) and add , equation (2.165) will be satisfied. So the new, corrected value of is equal to 84 + (-3) = 81, are shown in Figure 2.43b. This is called relaxation process. Similarly, the corrected head for the grid point can be found as follows: .Read More

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