Solution Manual Vol 7

# Tan1 nq 12 n 10 design bearing resistance rd 42 112

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Unformatted text preview: .10, is 90 kN/m2. This value is increased by 50%. Thus qa 90 1:5 135 kN=m2 (c) Burland and Burbidge. Using Figure 8.12, for B 3:5 m, z1 2:5 m. Use N values between depths of 1.2 and 3.7 m, the average value being 10. From Equation 8.18: Ic 1:71 0:068 101:4 From Equation 8.19(a), with s 25 mm: q 3:50:7 25 150 kN=m2 0:068 8.10 Peak value of strain influence factor occurs at a depth of 2.7 m and is given by Izp 0:5 0:1 130 16 2:7 0:5 0:67 Refer to Figure Q8.10: E 2:5qc Layer 1 2 3 4 5 z (m) 1.2 0.8 1.6 1.6 0.8 qc (MN/m2) 2.6 5.0 4.0 7.2 12.4 E 2:5qc (MN/m2) 6.5 12.5 10.0 18.0 31.0 Iz 0.33 0.65 0.48 0.24 0.07 Iz z (mm3/MN) E 0.061 0.042 0.077 0.021 0.002 0.203 C1 1 0:5 00 0:5 1:2 16 0:93 1 130 qn C2 1 say X Iz z 0:93 1 130 0:203 25 mm ; s C1 C2 qn E 68 Bearing capacity 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 Iz 14 q (MN/m2) c 0 2 4 6 8 10 12 1 Iz 2 (1) 3 (2) qc z (m) 4 (3) 5 (4) 6 (5) 7 8 Figure Q8.10 8.11 At pile base level: cu 220 kN=m2 ; qb cu Nc 220 9 1980 kN=m2 Disregard skin friction over a length of 2B above the under-ream. Between 4 and 17.9 m: 00 10:95 0 10:95 11:2 122:6 kN=m2 ; qs 00 0:7 122:6 86 kN=m2 Then Q f A b qb A s qs 32 1980 1:05 13:9 86 4 13 996 3941 17 937 kN Bearing capacity 69 Allowable load: a Qf 17 937 8968 kN 2 2 A b qb 13 996 3941 8606 kN A s qs 3 3 b i.e. allowable load 8600 kN. Adding 1/3( DAb W), the allowable load becomes 9200 kN. According to the limit state method: Characteristic undrained strength at base level, cuk Characteristic base resistance, qbk 9cuk 9 Characteristic shaft resistance, qsk 220 kN=m2 1:50 220 1320 kN=m2 1:50 00 86 57 kN=m2 1:50 1:50 Characteristic base and shaft resistances: Rbk 32 1320 9330 kN 4 86 2629 kN Rsk 1:05 13:9 1:50 For a bored pile the partial factors are b 1:60 and s 1:30 Design bearing resistance, Rcd 9330 2629 1:60 1:30 5831 2022 7850 kN Adding DAb W) the design bearing resistance becomes 9650 kN. 8.12 a qb 9cu 9 145 1305 kN=m2 qs cu 0:40 105 42 kN=m2 For a single pile: Q f A b qb A s qs 0:62 1305 0:6 15 42 4 369 1187 1556 kN 70 Bearing capacity Assuming single pile failure and a group efficiency of 1, the ultimate load on the pile group is (1556 36) 56 016 kN. The width of the group is 12.6 m, and hence the ultimate load, assuming block failure and taking the full undrained strength on the perimeter, is given by 12:62 1305 4 12:6 15 105 207 180 79 380 286 560 kN (Even if the remoulded strength were used, there would be no likelihood of block failure.) Thus the load factor is (56 016/21 000) 2:7. (b) Design load, Fcd 15 (6 1:30) 22:8 MN qbk 9cuk 9 qsk cuk Rbk 220 1320 kN=m2 1:50 105 28 kN=m2 0:40 1:50 0:602 1320 373 kN 4 Rsk 0:60 15 28 791 kN Rcd 373 791 233 608 841 kN 1:60 1:30 Design bearing resistance of pile group, Rcd 841 36 1:0 30 276 kN 30:27 MN Rcd > Fcd therefore the bearing resistance limit state is satisfied. (c) Settlement is estimated using the equivalent raft concept. The equivalent raft is located 10 m ( 2/3 15 m) below the top of the piles and is 17.6 m wide (see Figure Q8.12). Assume that the load on the...
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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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