Solution Manual Vol 7

# 2 08 16 16 08 qc mnm2 26 50 40 72 124 e 25qc mnm2

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Unformatted text preview: equivalent raft is spread at 2:1 to the underlying clay. Thus the pressure on the equivalent raft is q 21 000 68 kN=m2 17:62 Immediate settlement: H 15 0:85 B 17:6 D 13 0:74 B 17:6 L 1 B Hence from Figure 5.15: 0 0:78 and 1 0:41 Bearing capacity 71 Figure Q8.12 Thus, using Equation 5.28: si 0:78 0:41 68 Consolidation settlement: Layer 1 2 3 z (m) 2.5 7.5 12.5 Area (m2) 20.12 25.12 30.12 (kN/m2) 52.0 33.3 23.2 mv H (mm) 20.8 13.3 9.3 43.4 (sod ) 17:6 6 mm 65 Equivalent diameter 19:86 m; thus H/B 15/19:86 0:76. Now A 0:28, hence from Figure 7.12, 0:56. Then, from Equation 7.10: sc 0:56 43:4 24 mm The total settlement is (6 24) 30 mm. 8.13 At base level, N 26. Then using Equation 8.30: qb 40N Db 2 8320 kN=m2 40 26 0:25 B Check: &lt;400N; i:e: 400 26 10 400 kN=m2 72 Bearing capacity Over the length embedded in sand: 18 24 N 21 i:e: 2 Using Equation 8.31: qs 2N 2 21 42 kN=m2 For a single pile: Q f A b qb A s qs 0:252 8320 4 0:25 2 42 520 84 604 kN For the pile group, assuming a group efficiency of 1.2: X Qf 1:2 9 604 6523 kN Then the load factor is F 6523 2:1 2000 1000 (b) Design load, Fcd 2000 (1000 1:30) 3300 kN 8320 5547 kN/m2 Characteristic base resistance per unit area, qbk 1:50 42 28 kN/m2 Characteristic shaft resistance per unit area, qsk 1:50 Characteristic base and shaft resistances for a single pile: Rbk 0:252 5547 347 kN Rsk 4 0:25 2 28 56 kN For a driven pile the partial factors are b s 1:30 Design bearing resistance, Rcd 347 56 310 kN 1:30 1:30 For the pile group, Rcd 1:2 9 310 3348 kN Rcd &gt; Fcd (3348 &gt; 3300), therefore the bearing resistance limit state is satisfied. (c) Referring to Figure Q8.13, the equivalent raft is 2.42 m square. For the serviceability limit state, the design load, Fcd 2000 1000 3000 kN The pressure on the equivalent raft, q 3000=2:422 512 kN/m2 From Figure 8.12, for B 2:42 m, the value of zI is 1.8 m. Therefore N values between depths of 1.33 and 3.13 m should be used. Thus Bearing capacity 73 Figure Q8.13 N 24 26 34 28 3 1:71 Ic 1:4 0:016 Equation 8:18 28 s 512 2:420:7 0:016 15 mm Equation 8:19a The settlement is less than 20 mm, therefore the serviceability limit state is satisfied. 8.14 Using Equation 8.41: Tf DLcu D2 d 2 cu Nc 4 0:2 5 0:6 110 0:22 0:12 110 9 4 207 23 230 kN Chapter 9 Stability of slopes 9.1 Referring to Figure Q9.1: W 41:7 19 792 kN=m Q 20 2:8 56 kN=m 73 9:0 11:5 m Arc length, AB 180 Arc length, BC 28 9:0 4:4 m 180 The factor of safety is given by F rcu La 9:030 4:4 45 11:5 1:67 Wd1 Qd2 792 3:9 56 7:4 2cu 2 20 2:1 m 19 Depth of tension crack; z0 Arc length, BD F 1 13 9:0 2:1 m 180 2 9:030 2:1 45 11:5 1:49 792 3:9 56 7:4 The surcharge is a variable action, therefore a partial factor of 1.30 is applied. In the limit state method, the design values of undrained strength (cud) are (30/1.40) and (45/ 1.40) kN/m2. X 9:0 30 4:4 45 11:5 4175 kN=m Design resisting moment r cud La 1:4 Design disturbing moment Wd1 Qd2 792 3:9 56 1:30 7:4 3628 kN=m The design resisting moment is greater than the design disturbing moment, t...
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