Solution Manual Vol 7

Chapter 7 consolidation theory 71 total change in

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Unformatted text preview: layers: d1 9:5 mm and d2 2500 mm 2 d2 2 d1 ; for U 0:50, t2 t1 20 25002 2:63 years 60 24 365 9:52 for U < 0:60, Tv U 2 (Equation 7.24(a)) 4 0:302 ; t0:30 t0:50 0:502 2:63 0:36 0:95 years 54 Consolidation theory 7.4 The layer is open, ;d 8 4m 2 cv t 2:4 3 Tv 2 0:450 d 42 ui 84 kN=m2 The excess pore water pressure is given by Equation 7.21: ue m1 X m0 2ui Mz sin expM 2 Tv d M In this case, z d: ; sin where M M 2 3 2 Mz sin M d 3 5 ; ; ;... 2 2 2 sin M 1 1 M2Tv 1.110 9.993 exp (M2 Tv ) 0.329 4:57 105 ; ue 2 84 2 1 0:329 35:2 kN=m2 other terms negligible 7.5 The layer is open, ;d 6 3m 2 cv t 1:0 3 Tv 2 0:333 d 32 The layer thickness will be divided into six equal parts, i.e. m 6. Consolidation theory 55 For an open layer: n m2 0:333 62 3:00 ; n 4 Tv 4 The value of n will be taken as 12 (i.e. t 3/12 1/4 year), making 0:25. The computation is set out below, all pressures having been multiplied by 10: ui; j1 ui; j 0:25ui1; j ui1; j 2ui; j i j 0 0 1 2 3 4 5 6 0 500 400 300 200 100 0 1 0 350 400 300 200 100 0 2 0 275 362 300 200 100 0 3 0 228 325 292 200 100 0 4 0 195 292 277 198 100 0 5 0 171 264 261 193 99.5 0 6 0 151 240 245 189 98 0 7 0 136 219 230 180 96 0 8 0 123 201 215 171 93 0 9 0 112 185 201 163 89 0 10 0 102 171 187 154 85 0 11 0 94 158 175 145 81 0 12 0 87 146 163 137 77 0 The initial and 3-year isochrones are plotted in Figure Q7.5. Area under initial isochrone 180 units Area under 3-year isochrone 63 units The average degree of consolidation is given by Equation 7.25. Thus U 1 63 0:65 180 Figure Q7.5 56 Consolidation theory 7.6 At the top of the clay layer the decrease in pore water pressure is 4 w . At the bottom of the clay layer the pore water pressure remains constant. Hence at the centre of the clay layer, 0 2 w 2 9:8 19:6 kN=m2 The final consolidation settlement (one-dimensional method) is sc mv 0 H 0:83 19:6 8 130 mm 1 40 Corrected time, t 2 1:615 years 2 52 ; Tv cv t 4:4 1:615 0:444 d2 42 From Figure 7.18 (curve 1), U 0:73. Settlement after 2 years Usc 0:73 130 95 mm 7.7 The clay layer is thin relative to the dimensions of the raft, and therefore the one-dimensional method is appropriate. The clay layer can be considered as a whole (see Figure Q7.7) Figure Q7.7 Consolidation theory 57 Point 1 2 3 4 m 30 1:5 20 60 3:0 20 60 3:0 20 30 1:5 20 n 20 1:0 20 20 1:0 20 40 2:0 20 40 2:0 20 Ir 0:194 (4) 0:204 (2) 0:238 (1) 0:224 (2) (kN/m2 ) 113 59 35 65 s (mm) c 124 65 38 72 Note * sc mv 0 H 0:22 0 5 1:10 (mm) (0 ). 7.8 Due to the thickness of the clay layer relative to the size of the foundation, there will be significant lateral strain in the clay and the SkemptonBjerrum method is appropriate. The clay is divided into six sublayers (Figure Q7.8) for the calculation of consolidation settlement. (a) Immediate settlement: H 30 0:86 B 35 D 2 0:06 B 35 Figure Q7.8 58 Consolidation theory From Figure 5.15 (circle), 1 0:32 and 0 1:0: si 0 1 qB 105 35 30 mm 1:0 0:32 Eu 40 (b) Consolidation settlement: Layer 1 2 3 4 5 6 z (m) 2.5 7.5 12.5 17.5 22.5 27.5 D/z 14 4.67 2.80 2.00 1.55 1....
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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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