Solution Manual Vol 7

# The parabola passes through point g such that gc 03

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Unformatted text preview: oordinates of a number of points on the basic parabola are calculated, i.e. x 2:0 x z 2.0 0 z2 8:0 0 4.00 5:0 7.48 10:0 9.80 20:0 13.27 30:0 16.00 The basic parabola is plotted in Figure Q2.7. The upstream correction is drawn using personal judgement. No downstream correction is required in this case since 180 . If required, the top flow line can be plotted back onto the natural section, the x coordinates above being divided by the scale transformation factor. The quantity of seepage can be calculated using Equation 2.33, i.e. q 2k0 x0 2 4:5 106 2:0 1:8 105 m3 =s per m 2.8 The flow net is drawn in Figure Q2.8, from which Nf 3.3 and Nd 7. The overall loss in total head is 2.8 m. Then, 12 Seepage Figure Q2.8 q kh Nf 3:3 4:5 105 2:8 7 Nd 5 3 5:9 10 m =s per m 2.9 The two isotropic soil layers, each 5 m thick, can be considered as a single homogeneous anisotropic layer of thickness 10 m in which the coefficients of permeability in the horizontal and vertical directions, respectively, are given by Equations 2.24 and 2.25, i.e. kx kz H1 k1 H2 k2 106 f5 2:0 5 16g 9:0 106 m=s H1 H2 10 H1 H2 10 H1 H2 5 5 6 k1 k2 2 10 16 106 3:6 106 m=s Then the scale transformation factor is given by pffiffiffiffiffi pffiffiffiffiffiffiffi kz 3:6 xt x pffiffiffiffiffi x pffiffiffiffiffiffiffi 0:63x 9:0 kx Thus in the transformed section the dimension 10.00 m becomes 6.30 m; vertical dimensions are unchanged. The transformed section is shown in Figure Q2.9 and the flow net is drawn as for a single isotropic layer. From the flow net, Nf 5.6 and Nd 11. The overall loss in total head is 3.50 m. The equivalent isotropic permeability is Seepage 13 Figure Q2.9 k0 qffiffiffiffiffiffiffiffiffiffiffiffiffi pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi kx kz 9:0 3:6 106 5:7 106 m=s Then the quantity of seepage is given by q k0 h Nf 5:6 5:7 106 3:50 11 Nd 5 3 1:0 10 m =s per m Chapter 3 Effective stress 3.1 Buoyant unit weight: 0 sat w 20 9:8 10:2 kN=m3 Effective vertical stress: 0v 5 10:2 51 kN=m2 Total vertical stress: v 2 9:8 5 20 119:6 kN=m2 Pore water pressure: u 7 9:8 68:6 kN=m2 Effective vertical stress: 0v v u 119:6 68:6 51 kN=m2 or 3.2 Buoyant unit weight: 0 sat w 20 9:8 10:2 kN=m3 Effective vertical stress: 0v 5 10:2 51 kN=m2 or Effective stress 15 Figure Q3.1/3.2 Total vertical stress: v 200 9:8 5 20 2060 kN=m2 Pore water pressure: u 205 9:8 2009 kN=m2 Effective vertical stress: 0v v u 2060 2009 51 kN=m2 3.3 At top of the clay: v 2 16:5 2 19 71:0 kN=m2 u 2 9:8 19:6 kN=m2 0v v u 71:0 19:6 51:4 kN=m2 Alternatively, 0 sand 19 9:8 9:2 kN=m3 0v 2 16:5 2 9:2 51:4 kN=m2 At bottom of the clay: v 2 16:5 2 19 4 20 151:0 kN=m2 u 12 9:8 117:6 kN=m2 0v v u 151:0 117:6 33:4 kN=m2 NB The alternative method of calculation is not applicable because of the artesian condition. 16 Effective stress Figure Q3.3 Figure Q3.4 3.4 0 20 9:8 10:2 kN=m3 At 8 m depth: 0v 2:5 16 1:0 20 4:5 10:2 105:9 kN=m2 3.5 0 sand 19 9:8 9:2 kN=m3 0 clay 20 9:8 10:2 kN=m3 Effective stress 17 Figure Q3.5 (a) Immediately after WT rise: At 8 m depth, pore water pressure is governed by the new WT level because the permeability of the sand is high. ; 0v 3 16 5 9:2 94:0 kN=m2 At 12 m depth, pore water pressure is governed by the old WT level...
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## This note was uploaded on 09/21/2009 for the course CVEN 3718 taught by Professor Dobroslavznidarcic during the Spring '08 term at Colorado.

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