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# Craig's Soil Mechanics 7th Edition

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¼ �wb2 sin ���h�wb ¼ vector de Boundary water force on BC (static case) ¼ �wb2 sin � ¼ vector dc Resultant boundary water force (seepage case) ¼ vector be Resultant boundary water force (static case) ¼ �wb2 ¼ vector bc Seepage force ¼ �h�wb ¼ vector ce Resultant body force (seepage case) ¼ vector ae Resultant body force (static case) ¼ vector ac ¼ �0b2 The resultant body force can be obtained by one or other of the following force combinations: 1 Total (saturated) weightþ resultant boundary water force, i.e. vector abþ vector be. 2 Effective (buoyant) weight þ seepage force, i.e. vector ac þ vector ce. Only the resultant body force contributes to effective stress. A component of seepage force acting vertically upwards will therefore reduce a vertical effective stress compon- ent from the static value. A component of seepage force acting vertically downwards will increase a vertical effective stress component from the static value. A problem may be solved using either force combination 1 or force combination 2, but it may be that one combination is more suitable than the other for a particular problem. Combination 1 involves consideration of the equilibrium of the whole soil mass (solids þ water), while combination 2 involves consideration of the equilibrium of the soil skeleton only. The quick condition Consider the special case of seepage vertically upwards. The vector ce in Figure 3.6(c) would then be vertically upwards and if the hydraulic gradient were high enough the resultant body force would be zero. The value of hydraulic gradient corresponding to 82 Effective stress zero resultant body force is called the critical hydraulic gradient (ic). For an element of soil of volume V subject to upward seepage under the critical hydraulic gradient, the seepage force is therefore equal to the effective weight of the element, i.e. ic�wV ¼ �0V Therefore ic ¼ � 0 �w ¼ Gs � 1 1þ e ð3:9Þ The ratio �0/�w, and hence the critical hydraulic gradient, is approximately 1.0 for most soils. When the hydraulic gradient is ic, the effective normal stress on any plane will be zero, gravitational forces having been cancelled out by upward seepage forces. In the case of sands the contact forces between particles will be zero and the soil will have no strength. The soil is then said to be in a quick condition (quick meaning ‘alive’) and if the critical gradient is exceeded the surface will appear to be ‘boiling’ as the particles are moved around in the upward flow of water. It should be realized that ‘quicksand’ is not a special type of soil but simply sand through which there is an upward flow of water under a hydraulic gradient equal to or exceeding ic. In the case of clays, the quick condition may not necessarily result when the hydraulic gradient reaches the critical value given by Equation 3.9. Conditions adjacent to sheet piling Highupwardhydraulic gradientsmaybe experienced in the soil adjacent to thedownstream face of a sheet pile wall. Figure 3.7 shows part of the flow net for seepage under a sheet pile wall, the embedded length on the downstream side being d. A mass of soil adjacent to the pilingmay become unstable and be unable to support thewall.Model tests have shown that failure is likely to occur within a soil mass of approximate dimensions d � d/2 in section (ABCD in Figure 3.7). Failure first shows in the form of a rise or heave at the surface, associated with an expansion of the soil which results in an increase in permeability. This in turn leads to increased flow, surface ‘boiling’ in the case of sands and complete failure. The variation of total head on the lower boundary CD of the soil mass can be obtained from the flow net equipotentials, but for purposes of analysis it is sufficient to determine the average total head hm by inspection. The total head on the upper boundary AB is zero. The average hydraulic gradient is given by im ¼ hm d Since failure due to heaving may be expected when the hydraulic gradient becomes ic, the factor of safety (F ) against heaving may be expressed as F ¼ ic im ð3:10Þ Influence of seepage on effective stress 83 In the case of sands, a factor of safety can also be obtained with respect to ‘boiling’ at the surface. The exit hydraulic gradient (ie) can be determined by measuring the dimension �s of the flow net field AEFG adjacent to the piling: ie ¼ �h �s where �h is the drop in total head between equipotentials GF and AE. Then, the factor of safety is F ¼ ic ie ð3:11Þ There is unlikely to be any appreciable difference between the values of F given by Equations 3.10 and 3.11. The sheet pile wall problem shown in Figure 3.7 can also be used to illustrate the two methods of combining gravitational and water forces. Figure 3.7 Upward seepage adjacent to sheet piling. 84 Effective stress 1 Total weight of mass ABCD ¼ 1 2 �satd 2 Average total head on CD ¼ hm Elevation head on CD ¼ �d Average pore water pressure on CD ¼ (hm þ d )�w Boundary water force on CD ¼ d 2 (hm þ d )�w Resultant body force of ABCD ¼ 1 2 �satd 2 � d 2 ðhm þ d Þ�w ¼ 1 2 ð�0 þ �wÞd2 � 1 2 ðhmd þ d2Þ�w ¼ 1 2 �0d2 � 1 2 hm�wd 2 Effective weight of mass ABCD ¼ 1 2 �0d2 Average hydraulic gradient through ABCD ¼ hm d Seepage force on ABCD ¼ hm d �w d2 2 ¼ 1 2 hm�wd Resultant body force of ABCD¼1⁄2� 0d2� 1⁄2hm�wd as in method 1 above. The resultant body force will be zero, leading to heaving, when 1 2 hm�wd ¼ 1 2 �0d2 The factor of safety can then be expressed as F ¼ 1 2 �0d2 1 2 hm�wd ¼ � 0d hm�w ¼ ic im If the factor of safety against heaving is considered inadequate, the embedded length d may be increased or a surcharge load in the form of a filter may be placed on the surface AB, the filter being designed to prevent entry of soil particles. If the effective weight of the filter per unit area is w0 then the factor of safety becomes F ¼ � 0d þ w0 hm�w Example 3.3 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 vertical stress at A and B. Influence of seepage on effective stress 85 1 First consider the combination of total weight and resultant boundary water force. Consider the column of saturated soil of unit area between A and the soil surface at D. The total weight of the column is 11�sat (220 kN). Due to the change in level of the equipotentials across the column, the boundary water forces on the sides of the column will not be equal although in this case the difference will be small. There is thus a net horizontal boundary water force on the column. However, as the effective vertical stress is to be calculated, only the vertical component of the resultant body force is required and the net horizontal boundary water force need not be considered. The boundary water force on the top surface of the column is only due to the depth of water above D and is 4�w (39 kN). The boundary water force on the bottom surface of the column must be determined from the flow net, as follows: Number of equipotential drops between the downstream soil surface and A ¼ 8:2. There are 12 equipotential drops between the upstream and downstream soil surfaces, representing a loss in total head of 8m. Total head at A, hA ¼ 8:2 12 � 8 ¼ 5:5m. Elevation head at A, zA ¼ �7:0m. Pore water pressure at A, uA ¼ �w(hA � zA) ¼ 9:8(5:5þ 7:0) ¼ 122 kN=m2 i.e. boundary water force on bottom surface ¼ 122 kN. Net vertical boundary water force ¼ 122� 39 ¼ 83 kN. Total weight of the column ¼ 220 kN. Vertical component of resultant body force ¼ 220� 83 ¼ 137 kN i.e. effective vertical stress at A ¼ 137 kN/m2. (a) (b) Figure 3.8 Examples 3.3 and 3.4. 86 Effective stress It should be