Lecture 9 - Numerical Analysis of Seepage Notes | EduRev

: Lecture 9 - Numerical Analysis of Seepage Notes | EduRev

 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: 
                          
    
      
                  . 
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