Solution Manual Vol 7

5 0 6 0 151 240 245 189 98 0 7 0 136 219 230 180 96 0

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 27 I c 0.997 0.930 0.804 0.647 0.505 0.396 (kN/m2 ) 107 98 84 68 53 42 sy (mm) od 73.5 68.6 58.8 47.6 37.1 29.4 315.0 Notes * From Figure 5.9. y sod mv 0 H 0:14 0 5 0:700 (0 ). Now H 30 0:86 and A 0:65 B 35 ; from Figure 7.12; 0:79 ; sc sod 0:79 315 250 mm Total settlement: s si sc 30 250 280 mm 7.9 Without sand drains: Uv 0:25 ; Tv 0:049 from Figure 7.18 Tv d 2 0:049 82 cv cv With sand drains: ;t R 0:564S 0:564 3 1:69 m R 1:69 11:3 n r 0:15 ch t ch 0:049 82 Tr and ch cv 4R2 4 1:692 cv 0:275 ; Ur 0:73 (from Figure 7.30) Consolidation theory 59 Using Equation 7.40: 1 U 1 Uv 1 Ur 1 0:251 0:73 0:20 ; U 0:80 7.10 Without sand drains: Uv 0:90 ; Tv 0:848 ;t Tv d 2 0:848 102 8:8 years cv 9:6 With sand drains: R 0:564S 0:564 4 2:26 m R 2:26 n 15 r 0:15 Tr ch d 2 same t Tv cv 4R2 ; Tr 14:0 102 7:14 9:6 4 2:262 Tv 1 Using Equation 7.40: 1 U 1 Uv 1 Ur ; 1 0:90 1 Uv 1 Ur ; 1 Uv 1 Ur 0:10 2 An iterative solution is required using (1) and (2), an initial value of Uv being estimated. Uv 0.40 0.30 0.29 0.295 Tv 0.1256 0.0707 0.0660 0.0683 Tr 7:14Tv 0.897 0.505 0.471 0.488 Ur 0.97 0.87 0.85 0.86 (1 Uv )(1 Ur ) 0:60 0:03 0:018 0:70 0:13 0:091 0:71 0:15 0:107 0:705 0:14 0:099 Thus Uv 0:295 and Ur 0:86 0:0683 0:7 years ; t 8:8 0:848 Chapter 8 Bearing capacity 8.1 (a) The ultimate bearing capacity is given by Equation 8.3 1 qf cNc DNq BN 2 For u 0: Nc 5:14; Nq 1; N 0 ; qf 105 5:14 21 1 1 540 21 kN=m2 The net ultimate bearing capacity is qnf qf D 540 kN=m2 The net foundation pressure is qn q D 425 21 1 192 kN=m2 2 The factor of safety (Equation 8.6) is F qnf 540 2:8 192 qn (b) For 0 28 : Nc 26; Nq 15; N 13 21 9:8 11:2 kN=m 0 3 from Figure 8.4 1 11:2 2 13 2 ; qf 10 26 11:2 1 15 260 168 146 574 kN=m2 Bearing capacity 61 qnf 574 11:2 563 kN=m2 563 2:9 F 192 (qn 192 kN/m2 assumes that backfilled soil on the footing slab is included in the load of 425 kN/m.) 8.2 For 0 38 : Nq 49; N 67 1 from Equation 8.3 ; qnf DNq 1 BN 2 1 18 1:5 67 18 0:75 48 2 648 905 1553 kN=m2 500 18 0:75 320 kN=m2 qn 1:5 qnf 1553 4:8 ;F 320 qn tan 38 0d tan1 32 , therefore Nq 23 and N 25. 1:25 1 18 1:5 25 Design bearing resistance; Rd 1:5 18 0:75 23 2 1:5310 337 970 kN=m Design load (action); Vd 500 kN=m The design bearing resistance is greater than the design load, therefore the bearing resistance limit state is satisfied. 8.3 D 3:50 1:55 B 2:25 From Figure 8.5, for a square foundation: Nc 8:1 62 Bearing capacity For a rectangular foundation (L 4:50 m; B 2:25 m): Nc B 0:84 0:16 8:1 7:45 L Using Equation 8.10: qnf qf D cu Nc 135 7:45 1006 kN=m2 For F 3: qn 1006 335 kN=m2 3 ; q qn D 335 20 3:50 405 kN=m2 ; Design load 405 4:50 2:25 4100 kN Design undrained strength; cud 135 96 kN=m2 1:4 7241 kN Design load; Vd 4100 kN Rd > Vd , therefore the bearing resistance limit state is satisfied. Design bearing resistance; Rd cud Nc area 96 7:45 4:50 2:25 8.4 For 0 40 : Nq 64; N 95 qnf DNq 1 0:4 BN (a) Water table 5 m below ground level: qnf 17 1 63 0:4 17 2:5 95 1071 1615 2686 kN=m2 qn 400 17 383 kN=m2 F 2686 7:0 383 (b) Water table 1 m below ground level (i.e. at founda...
View Full Document

Ask a homework question - tutors are online