Solution Manual Vol 7

# 28 si 078 041 68 consolidation settlement layer 1 2 3

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Unformatted text preview: herefore overall stability is assured. Stability of slopes 75 7.4 m 14 1 2 13 1 2 2.8 m Q C 2.1 m D Soil (1) Soil (2) 73 3.9 m B W A Figure Q9.1 9.2 u 0 Depth factor, D 11 1:22 9 Using Equation 9.2 with F 1:0: Ns cu 30 0:175 F H 1:0 19 9 Hence from Figure 9.3: 50 For F 1:2: 30 0:146 1:2 19 9 ; 27 Ns 9.3 Refer to Figure Q9.3: Slice No. 1 2 3 4 5 6 7 8 h cos (m) 0.5 1.2 2.4 3.4 4.3 4.9 5.3 5.7 h sin (m) 0.1 0.2 0.5 0.9 1.4 1.8 u/ w (m) 0.7 1.7 3.0 3.9 4.7 5.1 5.7 5.8 u (kN/m2) 6.9 16.7 29.4 38.2 46.1 50.0 55.9 56.8 l (m) 1.2 2.0 2.0 2.0 2.1 2.1 2.1 2.1 ul (kN/m) 8 33 59 76 97 105 117 119 76 Stability of slopes Slice No. 9 10 11 12 13 14 15 h cos (m) 5.9 5.9 5.6 5.2 4.6 3.4 01.6 59.9 h sin (m) 2.4 2.9 3.3 3.5 3.7 3.2 0 .90 1 25.6 u/ w (m) 5.9 6.0 5.7 5.2 4.5 3.4 1.8 u (kN/m2) 57.8 58.8 55.9 51.0 44.1 33.3 17.6 l (m) 2.2 2.2 2.3 2.4 2.5 2.7 02.9 32.8 ul (kN/m) 127 129 129 122 110 90 0051 1372 Figure Q9.3 Stability of slopes 77 X X W cos b X X h cos 21 2 59:9 2516 kN=m W sin b h sin 21 2 25:6 1075 kN=m X W cos ul 2516 1372 1144 kN=m Arc length; La 1 57 32:6 32:7 m 180 2 The factor of safety is given by F c0 La tan 0 W cos ul W sin 8 32:7 tan 32 1144 1075 0:91 According to the limit state method: 0d tan1 c0 tan 32 26:5 1:25 8 5 kN=m2 1:60 Design resisting moment 5 32:7 tan 26:5 1144 734 kN=m Design disturbing moment 1075 kN=m The design resisting moment is less than the design disturbing moment, therefore a slip will occur. 9.4 X 1 sec 0 0 F fc b W ub tan g W sin 1 tan tan 0 =F c0 8 kN=m2 0 32 c0 b 8 2 16 kN=m W bh 21 2 h 42h kN=m Try F 1:00 tan 0 0:625 F 78 Stability of slopes Values of u are as obtained in Figure Q9.3. Slice No. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 h (m) 0.5 1.3 2.4 3.4 4.3 5.0 5.5 6.0 6.3 6.5 6.5 6.3 5.9 4.6 2.5 W bh (kN/m) 21 55 101 143 181 210 231 252 265 273 273 265 248 193 105 W sin (kN/m) 2 3 0 10 24 40 58 80 99 120 136 148 154 132 0078 1074 ub (kN/m) 8 33 59 76 92 100 112 114 116 118 112 102 88 67 35 c0 b (W ub) tan 0 (kN/m) 24 30 42 58 72 85 90 102 109 113 117 118 116 95 59 sec 1 (tan tan 0 )/F Product (kN/m) 26 31 42 56 67 77 80 89 94 97 99 100 99 82 0052 1091 6 31 2 0 4 71 2 11 1 142 181 2 22 26 30 34 381 2 43 48 1.078 1.042 1.000 0.960 0.931 0.907 0.889 0.874 0.861 0.854 0.850 0.847 0.853 0.862 0.882 F Thus 1091 1:02 (assumed value 1.00) 1074 F 1:01 9.5 X 1 sec 0 F fW1 ru tan g W sin 1 tan tan 0 =F 0 33 ru 0:20 W bh 20 5 h 100h kN=m 1 ru tan 0 0:80 tan 33 0:520 Try F 1:10 tan 0 tan 33 0:590 F 1:10 Stability of slopes 79 Referring to Figure Q9.5: Slice No. 1 2 3 4 5 6 7 8 h (m) 1.5 3.1 4.5 5.3 6.0 5.0 3.4 1.4 W bh (kN/m) 75 310 450 530 600 500 340 28 W sin (kN/m) 5 48 120 190 282 287 232 21 1185 W(1 ru ) tan 0 (kN/m) 20 161 234 276 312 260 177 3 sec 1 ( tan tan 0 )/F 0.963 0.926 0.892 0.873 0.862 0.864 0.882 0.908 Product (kN/m) 19 149 209 241 269 225 156 3 1271 4 9 151 2 21 28 35 43 49 Figure Q9.5 80 Stability of slopes F 1271 1:07 1185 The trial value was 1.10, therefore take F to be 1.08. 9.6 (a) Water table at surface; the factor of safety is given by Equation 9.12 F pti:e: 1:5 0 tan 0 sat tan 9:2 tan 36 19 tan ; tan 0:234 13 Water table well below surface; the factor of safety is given by Equation 9.11 tan 0 tan tan 36 tan 13 3:1 tan 36 (b) 0d tan1 30 1:25 F Depth of potential failure surface z Design resisting moment per unit area; Rd u tan 0 0 z cos2 tan 0d 9:2 z cos2 13 tan 30 5:04z kN Design disturbing moment per unit area; Sd sat sin cos 19 z sin 13 cos 13 4:16z kN Rd > Sd , therefore the limit state for overall stability is satisfied....
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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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