EXAMPLE 5.1 SOLUTION 5.1 STEP1: Draw the dam to scale. …Next Slide…. 1 STEP1: Draw the dam to scale. 2 . Divide the base into a convenient number of equal intervals. use a spreadsheet. Select the downstream end EF as datum. that is: 30. Use a table for convenience or. H 10 m. better yet. N d 14.6 x 3.057 cm 3 /s Nd 14 STEP5: Determine the pore water pressure under the base of the dam. N f 4 Nf 4 q k H 2 10 4 (10 100) 0.STEP2: Identify the permeable and impermeable boundaries. H 10 h 0. Let us use 10 intervals.06 m 10 Determine the pore water pressure at each nodal point. Draw the flow net. AB and EF are permeable: equipotential lines.714 m N d 14 3 . STEP4: Calculate the flow. BCIJDE and GH are impermeable: flow STEP3: lines. 47 3.69 6.60 5.80 6.40 -2.00 8.7 49.6 Nd 5.42 24.12 5.35 7.93 -2.2 69.18 12.40 -2.00 4.80 9.40 -2.71 6.Parameters x (m) Under base of dam 0 3.1 73.5 59. Using Simpson’s rule (Eq.69 5.5.06 6.28 5.9 49.50 Nd h (m) 4.97 7.3 18.30 11.7 59.3 34.05 4.90 7.28 6.9 73.40 -2.1 69.10 12.40 8.71 7.1 2(78.3 80.40 -2.9 55.9 34.40 -2.06 82.26 7.93 8.4) 4(80.40 -2.7 43.47 7.24 15.40 -2.36 21.40 hp(m)= H N d h hz 8.14 4.12 6.7 65.40 -2.4 kN/m 3 4 .2 65.54 30.12 9.48 u (kPa) = hp w 82.17): Pw 3.4 43.1 hz(m) STEP6: Calculate the uplift force.20 6.40 8.40 -2.43 4.9 78.93 5.5 55.48 27.40 10.9) 1946. STEP7: Determine the pore water pressure distribution on the sheet pile wall. Six intervals were chosen because it is convenient for scaling using the scale that was used to draw the flow net.76 15.00 1.85 13.83 7.40 -2.52 17.33 3.70 1. The greater the interval the greater the accuracy.00 0.00 Nd 0. Only one interval is used for the back face of the wall because there are no equipotential lines that meet there.17 m and the back face into one interval.00 5.83 8. The pore water pressure distribution at the back of the wall is a trapezoid and the area is readily calculated.73 -5.71 2.71 16.1 155.26 15.3 144.9 169. Use a table or spreadsheet to compute the pore water pressure distribution and the hydrostatic forces.30 1.57 -4.40 u (kPa) = hp w 116.40 -9.36 1.14 1.3 Pw (kN/m) Front Back Difference 1011.1 82.90 2.90 -7.90 12.40 -3.67 5.7 830.00 5.23 -9. Divide the front face of the wall into six intervals of 7/6 = 1. Parameters z (m) Front of wall Back of wall 0 1.0 135.40 3.60 Nd h (m) 0.71 0.00 7.0 161.50 4. Use Simpson’s rule to calculate the hydrostatic force on the front face of the wall.50 0.60 1.07 -8.6 126.81 14.14 3.57 4.17 2.00 hz(m) -2.8 5 .6 154.9 180.93 1.40 hp(m)= H N d h hz 11. 714 0. the value of h increases and imax is likely to increase. piping will not occur.8 Since imax<icr.94 1 e 1 0.6 Front Back STEP8: Determine the maximum hydraulic gradient. If the depth is reduced. imax h 0. The smallest value of L occurs at the exit.36 Lmin 2 STEP9: Determine if piping would occur.7 1 0. icr STEP10: State the effect of reducing the depth of penetration of the sheet pile wall. By measurement Lmin=2 m. . G s 1 2. 4 (0. 0. 1 9615.3 1 1 0.EXAMPLE 4. (4.4 (0.001) 5.3 5000 0.001 Step 2: Solve the equation.8 kN/m 6 .0005) 0. Step 3: Calculate the lateral force per unit length 3 x 6 Px 5.3 ( 1 0 . 3 )( 1 2 0 .3 0. The soil element is likely to be under the plane strain condition. (ε2 = 0).1 SOLUTION 4. use Eq.3 1 0.21).7 0.3 (0.7 (0.001) 0.3(dx 1) 0 5.1 Step 1: Determine the appropriate stress condition and write the appropriate equation.3 x 0 31.5 kPa 3 9615.3 kPa the negative sign means reduction.0005 0.0005) 0. 3 ) 3 0. 24) 0.9 10 3 z 0 5 0 9. Step 2: Decide on a stress condition. The element is directly under the center of the tank.2 SOLUTION 4. Use Eq.6 20 1. 7 20 10 3 1 1 50 0. Step 3: Choose the appropriate equation and solve. 3 50 0 .EXAMPLE 4.5 10 3 m 9.5 mm .6 50 1 1 1 20 3 0 . 3 0 .2 Step 1: Draw a diagram of the problem. so the the axisymmetric condition prevails. (4. 7 20 5 10 20 10 3 1 Step 4: Calculate the vertical displacement 1 z 5 z z dz 1.9 10 3 3 20 10 1 5 3 0 .