Solution Manual Vol 7

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Unformatted text preview: tion level): 0 20 9:8 10:2 kN=m3 Bearing capacity 63 qnf 17 1 63 0:4 10:2 2:5 95 1071 969 2040 kN=m2 F 2040 5:3 383 (c) Water table at ground level with upward hydraulic gradient 0.2: 0 j 10:2 0:2 9:8 8:2 kN=m3 qnf 8:2 1 63 0:4 8:2 2:5 95 517 779 1296 kN=m2 F 1296 3:3 392 8.5 The following partial factors are used: dead load, 1.0; imposed load, 1.3; shear strength ( tan 0 ), 1.25. Design load, Vd 4000 1:3 1500 5950 kN 0 1 tan 39 Design value of tan 33 1:25 For 0 33 , Nq 26 and N 29: Design bearing resistance; Rd 32 10:2 1:5 26 0:4 10:2 3:0 29 32 398 355 6777 kN Rd > Vd , therefore the bearing resistance limit state is satisfied. 8.6 (a) Undrained shear, for u 0: Nc 5:14; Nq 1; N 0 qnf 1:2cu Nc 1:2 100 5:14 617 kN=m2 qn qnf 617 206 kN=m2 3 F q qn D 206 21 227 kN=m2 64 Bearing capacity Drained shear, for 0 32 : Nq 23; N 25 0 21 9:8 11:2 kN=m3 qnf 0 DNq 1 0:4 0 BN 11:2 1 22 0:4 11:2 4 25 246 448 694 kN=m2 q 694 21 231 21 252 kN=m2 3 Design load 42 227 3632 kN (b) Design undrained strength, cud 100 71 kN/m2 1:4 Design bearing resistance; Rd 1:2cud Ne area 1:2 71 5:14 42 For drained shear, 0d tan1 ; Nq 12; N 10 Design bearing resistance; Rd 42 11:2 1 12 0:4 11:2 4 10 42 134 179 5008 kN (c) Consolidation settlement: the clay will be divided into three sublayers (Figure Q8.6): Layer 1 2 3 z (m) 2 6 10 m, n 1.00 0.33 0.20 Ir 0.175 0.044 0.017 0 (kN/m2) 0.700qn 0.176qn 0.068qn sod (mm) 0.182qn 0.046qn 0.018qn 0.246qn 7007 kN tan 32 26 1:25 Diameter of equivalent circle, B 4:5 m H 12 2:7 B 4:5 ; 0:60 and A 0:42 from Figure 7:12 mm sc 0:60 0:246qn 0:147qn Bearing capacity 65 Figure Q8.6 For sc 30 mm: qn 30 204 kN=m2 0:147 q 204 21 225 kN=m2 Design load 42 225 3600 kN The design load is 3600 kN, settlement being the limiting criterion. 8.7 D 8 2:0 B 4 From Figure 8.5, for a strip, Nc 7:1. For a depth/breadth ratio of 2, Equation 8.12 should be used. F cu Nc 40 7:1 1:8 20 8 D 8.8 Design load for ultimate limit state, Vd 2500 (1250 1:30) 4125 kN 0 1 tan 38 Design value of tan 32 1:25 66 Bearing capacity For 0 32 , Nq 23 and N 25 Design bearing resistance; Rd 2:502 17 1:0 23 0:4 10:2 2:50 25 2:502 391 255 4037 kN The design bearing resistance is (slightly) less than the design load, therefore the bearing resistance limit state is not satisfied. To satisfy the limit state, the dimension of the foundation should be increased to 2.53 m. Design load for serviceability limit state 2500 1250 3750 kN 3750 17 583 kN=m2 2:502 1:26 0:21 From Figure 5:10, m n 6 ; Ir 0:019 For B 2:50 m, qn Stress increment, 4 0:019 583 44 kN=m2 Consolidation settlement, sc mv H 0:15 44 2 13 mm The settlement would be less than 13 mm if an appropriate value of settlement coefficient (Figure 7.12) was applied. The settlement is less than 20 mm, therefore the serviceability limit state is satisfied. 8.9 Depth (m) 0.70 1.35 2.20 2.95 3.65 4.40 5.15 6.00 N 6 9 10 8 12 13 17 23 0v (kN/m2)* 23 37 50 58 65 CN 1.90 1.55 1.37 1.28 1.23 N1 17 15 11 15 16 Note * Using 17 kN/m3 and 0 10 kN/m3 . (a) Terzaghi and Peck. Use N1 values between depths of 1.2 and 4.7 m, the average value being 15. For B 3:5 m and N 15, the provisional value of bearing capacity, using Figure 8.10, is 150 kN/m2. The water table correction factor (Equation 8.16) is 3:0 Cw 0:5 0:5 0:82 4:7 Bearing capacity 67 Thus qa 150 0:82 120 kN=m2 (b) Meyerhof. Use uncorrected N values between depths of 1.2 and 4.7 m, the average value being 10. For B 3:5 m and N 10, the provisional value of bearing capacity, using Figure 8...
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