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# Craig's Soil Mechanics 7th Edition

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the flow region, use of the finite element method is usually necessary. The fundamental condition to be satisfied in a flow net is that every intersection between a flow line and an equipotential must be at right angles. In addition, it is convenient to construct the flow net such that � has the same value between any two adjacent flow lines and �� has the same value between any two adjacent equipoten- tials. It is also convenient to make �s ¼ �n in Equation 2.13, i.e. the flow lines and equipotentials form ‘curvilinear squares’ throughout the flow net. Then for any curvilinear square � ¼ �� Now, � ¼ �q and �� ¼ k�h, therefore: �q ¼ k�h ð2:14Þ For the entire flow net, h is the difference in total head between the first and last equipotentials, Nd the number of equipotential drops, each representing the same total head loss �h, and Nf the number of flow channels, each carrying the same flow �q. Then, �h ¼ h Nd ð2:15Þ and q ¼ Nf�q Hence, from Equation 2.14 q ¼ kh Nf Nd ð2:16Þ 42 Seepage Equation 2.16 gives the total volume of water flowing per unit time (per unit dimen- sion in the y direction) and is a function of the ratio Nf/Nd. Between two adjacent equipotentials the hydraulic gradient is given by i ¼ �h �s ð2:17Þ Example of a flow net As an illustration the flow net for the problem detailed in Figure 2.7(a) will be considered. The figure shows a line of sheet piling driven 6.00m into a stratum of soil 8.60m thick, underlain by an impermeable stratum. On one side of the piling the depth of water is 4.50m; on the other side the depth of water (reduced by pumping) is 0.50m. The first step is to consider the boundary conditions of the flow region. At every point on the boundary AB the total head is constant, so AB is an equipotential; similarly CD is an equipotential. The datum to which total head is referred may be any level but in seepage problems it is convenient to select the downstream water level as datum. Then, the total head on equipotential CD is zero (pressure head 0.50m; elevation head �0:50m) and the total head on equipotential AB is 4.00m (pressure head 4.50m; elevation head �0:50m). From point B, water must flow down the upstream face BE of the piling, round the tip E and up the down-stream face EC. Water from point F must flow along the impermeable surface FG. Thus BEC and FG are flow lines. The shapes of other flow lines must be between the extremes of BEC and FG. The first trial sketching of the flow net (Figure 2.7(b)) can now be attempted using a procedure suggested by Casagrande [2]. The estimated line of flow (HJ) from a point on AB near the piling is lightly sketched. This line must start at right angles to equipotential AB and follow a smooth curve round the bottom of the piling. Trial equipotential lines are then drawn between the flow lines BEC and HJ, intersecting both flow lines at right angles and forming curvilinear squares. If necessary the position of HJ should be altered slightly so that a whole number of squares is obtained between BH and CJ. The procedure is continued by sketching the estimated line of flow (KL) from a second point on AB and extending the equipotentials already drawn. The flow line KL and the equipotential extensions are adjusted so that all intersections are at right angles and all areas are square. The procedure is repeated until the boundary FG is reached. At the first attempt it is almost certain that the last flow line drawn will be inconsistent with the boundary FG as, for example, in Figure 2.7(b). By studying the nature of this inconsistency the position of the first flow line (HJ) can be adjusted in a way that will tend to correct the inconsistency. The entire flow net is then adjusted and the inconsistency should now be small. After a third trial the last flow line should be consistent with the boundary FG, as shown in Figure 2.7(c). In general, the areas between the last flow line and the lower boundary will not be square but the length/breadth ratio of each area should be constant within this flow channel. In constructing a flow net it is a mistake to draw too many flow lines; typically, four to five flow channels are sufficient. Once experience has been gained and flow nets for various seepage situations have been studied, the detailed procedure described above Flow nets 43 can be short-circuited. Three or four flow lines, varying in shape between the extremes of the two boundaries, can be sketched before or together with the equipotentials. Subsequent adjustments are made until a satisfactory flow net is achieved. In the flow net in Figure 2.7(c) the number of flow channels is 4.3 and the number of equipotential drops is 12; thus the ratioNf/Nd is 0.36. The equipotentials are numbered from zero at the downstream boundary; this number is denoted by nd. The loss in total head between any two adjacent equipotentials is �h ¼ h Nd ¼ 4:00 12 ¼ 0:33m Figure 2.7 Flow net construction: (a) section, (b) first trial and (c) final flow net. 44 Seepage The total head at every point on an equipotential numbered nd is nd�h. The total volume of water flowing under the piling per unit time per unit length of piling is given by q ¼ kh Nf Nd ¼ k� 4:00� 0:36 ¼ 1:44km3=s A piezometer tube is shown at a point P on the equipotential denoted by nd ¼ 10. The total head at P is hp ¼ nd Nd h ¼ 10 12 � 4:00 ¼ 3:33m i.e. the water level in the tube is 3.33m above the datum. The point P is at a distance zp below the datum, i.e. the elevation head is �zp. The pore water pressure at P can then be calculated from Bernoulli’s theorem: up ¼ �wfhp � ð�zpÞg ¼ �wðhp þ zpÞ Figure 2.7 (continued) Flow nets 45 The hydraulic gradient across any square in the flow net involves measuring the average dimension of the square (Equation 2.17). The highest hydraulic gradient (and hence the highest seepage velocity) occurs across the smallest square and vice versa. Example 2.1 The section through a sheet pile wall along a tidal estuary is given in Figure 2.8. At low tide the depth of water in front of the wall is 4.00m; the water table behind the wall lags 2.50m behind tidal level. Plot the net distribution of water pressure on the piling. The flow net is shown in the figure. The water level in front of the piling is selected as datum. The total head at water table level (the upstream equipotential) is 2.50m (pressure head zero; elevation head þ2:50m). The total head on the soil surface in front of the piling (the downstream equipotential) is zero (pressure head 4.00m; elevation head �4:00m). There are 12 equipotential drops in the flow net. The water pressures are calculated on both sides of the piling at selected levels numbered 1–7. For example, at level 4 the total head on the back of the piling is hb ¼ 8:8 12 � 2:50 ¼ 1:83m and the total head on the front is hf ¼ 1 12 � 2:50 ¼ 0:21m The elevation head at level 4 is �5:5m. Figure 2.8 Example 2.1. 46 Seepage Therefore, the net pressure on the back of the piling is ub � uf ¼ 9:8ð1:83þ 5:5Þ � 9:8ð0:21þ 5:5Þ ¼ 9:8ð7:33� 5:71Þ ¼ 15:9 kN=m2 The calculations for the selected points are shown in Table 2.2 and the net pressure diagram is plotted in Figure 2.8. Example 2.2 The section through a dam is shown in Figure 2.9. Determine the quantity of seepage under the dam and plot the distribution of uplift pressure on the base of the dam. The coefficient of permeability of the foundation soil is 2:5� 10�5 m/s. The flow net is shown in the figure. The downstream water level is selected as datum. Between the upstream and downstream equipotentials the total head loss is 4.00m. In the flow net there are 4.7 flow channels and 15 equipotential drops. The seepage is given by q ¼ kh Nf Nd ¼ 2:5� 10�5 � 4:00� 4:7 15 ¼ 3:1� 10�5 m3=s ðper mÞ The pore water pressure