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# Craig's Soil Mechanics 7th Edition

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value of this factor. Example 6.4 Details of a cantilever retaining wall are shown in Figure 6.19, the water table being below the base of the wall. The unit weight of the backfill is 17 kN/m3 and a surcharge pressure of 10 kN/m2 acts on the surface. Characteristic values of the shear strength parameters for the backfill are c0 ¼ 0 and �0 ¼ 36�. The angle of friction between the base and the foundation soil is 27� (i.e. 0:75�0). Is the design of the wall satisfactory according to (a) the traditional approach and (b) the limit state (EC7) approach? The position of the base reaction is determined by calculating the moments of all forces about the toe of the wall, the unit weight of concrete being taken as 23.5 kN/m3. The active thrust is calculated on the vertical plane through the heel of the wall, thus � ¼ 0 and the Rankine value of Ka is appropriate. Gravity walls 189 (a) For �0 ¼ 36� and � ¼ 0,Ka ¼ 0:26. The calculations are set out in Table 6.3. Lever arm of base resultant is given by �M V ¼ 243:5 212:3 ¼ 1:15m i.e. the resultant acts within the middle third of the base. The factor of safety against overturning is: 397:3/153:8 ¼ 2:58 Eccentricity of base reaction, e ¼ 1:50� 1:15 ¼ 0:35m The maximum and minimum base pressures are given by Equation 6.27, i.e. p ¼ 212:3 3:0 1� 6� 0:35 3:0 � � Figure 6.19 Example 6.4. Table 6.3 (per m) Force (kN) Arm (m) Moment (kNm) (1) 0:26� 10� 5:40 ¼ 14:0 2.70 37.9 (2) 1 2 � 0:26� 17� 5:402 ¼ 64.4 1.80 115.9 H ¼ 78:4 MH ¼ 153:8 (Stem) 5:00� 0:30� 23:5 ¼ 35:3 1.10 38.8 (Base) 0:40� 3:00� 23:5 ¼ 28:2 1.50 42.3 (Soil) 5:00� 1:75� 17 ¼ 148.8 2.125 316.2 V ¼ 212:3 MV ¼ 397:3 153.8 �M ¼ 243:5 190 Lateral earth pressure Thus, pmax ¼ 120 kN/m2 (at the toe of the wall) and pmin ¼ 21 kN/m2 (at the heel). The factor of safety against sliding is given by F ¼ V tan � H ¼ 212:3 tan 27 � 78:4 ¼ 1:38 The design of the wall is satisfactory. (b) The design value of �0 ¼ tan�1 (tan 36�/1:25) ¼ 30�, for which Ka ¼ 0:33. Case C is relevant, therefore a partial factor of 1.30 is applied to force (1), the surcharge pressure being a variable unfavourable action. The partial factor for all other forces is 1.00. Then H ¼ ð0:33� 10� 5:40� 1:30Þ þ 1 2 � 0:33� 17� 5:402 � � ¼ 23:2þ 81:8 ¼ 105:0 kN MH ¼ ð23:2� 2:70Þ þ ð81:8� 1:80Þ ¼ 209:9 kNm V ¼ 212:3 kN ðas beforeÞ MV ¼ 397:3 kNm ðas beforeÞ �M ¼ 187:4 kNm Lever arm of base resultant ¼ 187:4/212:3 ¼ 0:88m Eccentricity of base reaction, e ¼ 1:50� 0:88 ¼ 0:62m Then from Equation 6.27, pmax ¼ 159 kN=m2 and pmin ¼ �17 kN=m2 The resultant acts outside the middle third of the base, giving a negative value of pmin. The design value of � is (0:75� 30�) ¼ 22:5� The sliding limit state is not satisfied, the resisting force V tan � (88:0 kN) being less than the disturbing force H (105.0 kN). The width of the base would have to be increased because of the negative base pressure at the heel of the wall and the sliding resistance limit state not being satisfied. The overturning limit state is satisfied, the resisting momentMV being greater than the disturbing moment MH. It should be noted that passive resistance in front of the wall has been neglected to allow for unplanned excavation. It is likely that the design would be satisfactory if passive resistance could be relied on throughout the life of the wall. Case B is likely to govern the structural design of the wall, with partial factors of 1.50 and 1.35 being applied to forces (1) and (2), respectively, and shear strength being unfactored. Example 6.5 Details of a gravity-retaining wall are shown in Figure 6.20, the unit weight of the wall material being 23.5 kN/m3. The unit weight of the backfill is 18 kN/m3 and design values of the shear strength parameters are c0 ¼ 0 and �0 ¼ 33�. The value of � between Gravity walls 191 wall and backfill and between wall and foundation soil is 26�. The pressure on the foundation soil should not exceed 250 kN/m2. Is the design of the wall satisfactory? As the back of the wall and the soil surface are both inclined, the value of Ka will be calculated from Equation 6.17. The values of the angles in this equation are ¼ 100�,� ¼ 20�,� ¼ 33� and � ¼ 26�. Thus, Ka ¼ sin 67 �=sin 100�ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ sin 125� p þ ﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃsin 58� sin 13�= sin 80�p !2 ¼ 0:48 Then, from Equation 6.16, Pa ¼ 1 2 � 0:48� 18� 62 ¼ 155:5 kN=m acting at 1⁄3 height and at 26 � above the normal, or 36� above the horizontal. Moments are considered about the toe of the wall, the calculations being set out in Table 6.4. Lever arm of base resultant is given by �M V ¼ 287:1 312:5 ¼ 0:92m The overturning limit state is satisfied since the restoring moment (MV) is greater than the disturbing moment (MH). Eccentricity of base reaction, e ¼ 1:375� 0:92 ¼ 0:455m The maximum and minimum base pressures are given by Equation 6.27, i.e. p ¼ 312:5 2:75 1� 6� 0:455 2:75 � � Figure 6.20 Example 6.5. 192 Lateral earth pressure Thus pmax ¼ 226 kN/m2 and pmin ¼ 1 kN/m2. The maximum base pressure is less than the allowable bearing capacity of the foundation soil, therefore the bearing resistance limit state is satisfied. With respect to base sliding, the restoring force, V tan � ¼ 312:5 tan 26� ¼ 152:4 kN. The disturbing force, H is 125.8 kN. Therefore the sliding limit state is satisfied. The design of the wall is satisfactory, the relevant limit states being satisfied in terms of design values of the shear strength parameters. Example 6.6 Details of a retaining structure, with a vertical drain adjacent to the back surface, are shown in Figure 6.21(a), the saturated unit weight of the backfill being 20kN/m3. The design parameters for the backfill are c0 ¼ 0, �0 ¼ 38� and � ¼ 15�. Assuming a failure plane at 55� to the horizontal, determine the total horizontal thrust on the wall when the backfill becomes fully saturated due to continuous rainfall, with steady seepage towards the drain. Determine also the thrust on the wall (a) if the vertical drain were replaced by an inclined drain below the failure plane and (b) if there were no drainage system behind the wall. The flow net for seepage towards the vertical drain is shown in Figure 6.21(a). Since the permeability of the drain must be considerably greater than that of the backfill, the drain remains unsaturated and the pore pressure at every point within the drain is zero (atmospheric). Thus, at every point on the boundary between the drain and the backfill, total head is equal to elevation head. The equipotentials, therefore, must intersect this boundary at points spaced at equal vertical intervals �h: the boundary itself is neither a flow line nor an equipotential. The combination of total weight and boundary water force is used. The values of pore water pressure at the points of intersection of the equipotentials with the failure plane are evaluated and plotted normal to the plane. The boundary water force (U ), acting normal to the plane, is equal to the area of the pressure diagram, thus U ¼ 55 kN=m Table 6.4 (per m) Force (kN) Arm (m) Moment (kNm) Pa cos 36 � ¼ 125.8 2.00 251.6 H ¼ 125.8 MH ¼ 251:6 Pa sin 36 � ¼ 91.4 2.40 219.4 Wall 1 2 � 1:05� 6� 23:5 ¼ 74.0 2.05 151.7 0:70� 6� 23:5 ¼ 98.7 1.35 133.2 1 2 � 0:50� 5:25� 23:5 ¼ 30.8 0.83 25.6 1:00� 0:75� 23:5 ¼ 17.6 0.50 8.8 V ¼ 312.5 MV ¼ 538:7 251.6 �M ¼ 287:1 Gravity walls 193 The water forces on the other two boundaries of the soil wedge are zero. The total weight (W ) of the soil wedge is now calculated, i.e. W ¼ 252 kN=m The forces acting on the wedge are shown in Figure 6.21(b). Since the directions of the four forces are known, together with the values ofW and U, the force polygon can be drawn, from which Pa ¼