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EXAMPLE 3.3 (Craig)

The flow net for seepage under a sheet pile wall is shown in Figure 3.8 a), the saturated unit weight of the soil being 20 kN/m 3 . Determine the values of effective stress at A and B. First, direction of seepage is left to right. The total head driving seepage is 8 m.

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EXAMPLE 3.3 (Craig)

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  1. The flow net for seepage under a sheet pile wall is shown in Figure 3.8 a), the saturated unit weight of the soil being 20 kN/m3. Determine the values of effective stress at A and B. First, direction of seepage is left to right. The total head driving seepage is 8 m. Then, number the equipotentials: EXAMPLE 3.3 (Craig) 12 11 0 + + 10 1 2 9 3 8 4 7 5 6 sat = 20 kN/m3

  2. The volume of column AD is 11m x 1m x 1m = 11 m3 Therefore,column AD weighs 20kN/m3 x 11 m3 = 220kN  The 4m of water above point D exerts another boundary water force equal to 9.8 x 4 = 39.2 kN The resultant body force (effective stress, σA’) acting at A is: 220kN  + 39.2kN  - 122.5kN = 136.7kN  First for Point A consider the column of soil AD, 11 m high and 1m long and 1m wide Therefore, uA = w(hA-zA) = 9.8(5.5+7) = 122.5kPa The boundary water force on the bottom 1m2 = 122.5kN The pore water pressure acting on the column at A is uA the elevation head at A, the total head at A: EXAMPLE 3.3 (Craig) 1 m2 39.2 kN D 12 11 0 11.00 m + + 10 1 8.2 σA’ = 136.7 kN 220 kN A 2 9 3 8 122.5kN 4 7 5 6 sat = 20 kN/m3

  3. The seepage force JA = iwVAD, VAD = 11 m3 ∆h from D to A is 8.0 – 5.5 = 2.5 m, over a distance of 11 m: JA = 0.227x9.8x11=24.5kN ’ = 20-9.8 = 10.2 kN/m3 and the volume of column AD is 11 m3 The effective weight of column AD is 10.2 x 11 = 112.2 kN  We could have solved forσA’ using the Effective Weight of column AD and the seepage force, JA The resultant body force (effective stress, σA’) acting at A is: 112.2kN  + 24.5kN  = 136.7kN  EXAMPLE 3.3 (Craig) 1 m2 D 12 11 0 11.00 m + + 10 1 8.2 112.2 kN A 2 9 3 8 4 7 5 6 24.5 kN σA’ = 136.7 kN sat = 20 kN/m3

  4. Now for Point B consider the column of soil BC, 6 m high and 1m long and 1m wide The volume of column BC is 6 m x 1 m x 1 m = 6 m3 Therefore,column BC weighs 20 kN/m3 x 6 m3 = 120kN  The 1m of water above point C exerts another boundary water force equal to 9.8 x 1 = 9.8 kN Therefore, uB = w(hB-zB) = 9.8(1.6+7) = 84.28 kPa The boundary water force on the bottom 1m2 = 84.3 kN The pore water pressure acting on the column at B is uB the elevation head at B, the total head at B: The resultant body force (effective stress, σB’) acting at B is: 120 kN  + 9.8 kN  - 84.3 kN = 45.5 kN  EXAMPLE 3.3 (Craig) 1 m2 12 9.8 kN C 11 0 6.00 m + + 10 1 2.4 120 kN σB’ = 45.5 kN B 2 9 3 84.3 kN 8 4 7 5 6 sat = 20 kN/m3

  5. Now findingσB’ using the Effective Weight of column BC and the seepage force, JB The resultant body force (effective stress, σB’) acting at B is: 61.2 kN  - 15.7 kN  = 45.5 kN  The seepage force JB = iwVBC, VBC = 6 m3 ∆h from B to C is 1.6 – 0.0 = 1.6 m, over a distance of 6 m: JA = 0.266x9.8x6=15.7 kN ’ = 20-9.8 = 10.2 kN/m3 and the volume of column BC is 6 m3 The effective weight of column BC is 10.2 x 6 = 61.2 kN  EXAMPLE 3.3 (Craig) 12 1 m2 C 11 0 6.00 m + + 10 1 2.4 61.2 kN B 2 9 3 8 4 7 5 6 15.7 kN σA’ = 45.5 kN sat = 20 kN/m3

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