Solution Manual Vol 7

# 5 hydrostatic pressure on the two sides of the wall

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Unformatted text preview: 30 sin 85 The total active thrust (acting at 25 above the normal) is given by Equation 6.16 Pa 1 0:50 19 7:502 267 kN=m 2 40 Lateral earth pressure Horizontal component: Ph 267 cos 40 205 kN=m Vertical component: Pv 267 sin 40 172 kN=m Consider moments about the toe of the wall (Figure Q6.6) (per m): Force (kN) (1) (2) (3) (4) (5) 1 1:75 6:50 23:5 2 0:50 6:50 23:5 1 0:70 6:50 23:5 2 1:00 4:00 23:5 1 0:80 0:50 23:5 2 Pa sin 40 Pa cos 40 133:7 76:4 53:5 94:0 4:7 172.0 V 525 H 205 Arm (m) 2.58 1.75 1.27 2.00 0.27 3.33 2.50 Moment (kN m) 345 134 68 188 1 573 MV 1307 MH 512 M 795 Lever arm of base resultant: M 795 1:51 m V 525 Eccentricity of base resultant: e 2:00 1:51 0:49 m Figure Q6.6 Lateral earth pressure 41 Base pressures (Equation 6.27): p 525 6 0:49 1 4 4 228 kN=m2 and 35 kN=m2 The overturning limit state is satisfied, the restoring moment (1307 kN m) being greater than the overturning moment (512 kN m). The bearing resistance limit state is satisfied, the ultimate bearing capacity of the foundation soil (250 kN/m2) being greater than the maximum base pressure (228 kN/m2). The sliding limit state is satisfied, the restoring force (525 tan 25 245 kN) being greater than the disturbing force (205 kN). 6.7 For 0 35 , Ka 0:27; for 0 27 , Ka 0:375 and Kp 2:67; for soil, 0 11:2 kN/m3 ; for backfill, 0 10:2 kN/m3 . The pressure distribution is shown in Figure Q6.7. Hydrostatic pressure is balanced. Consider moments about the anchor point (A), per m: Force (kN) (1) (2) (3) (4) (5) (6) (7) (8) 1 0:27 17 52 57:4 2 0:27 17 5 3 68:9 1 2 0:27 10:2 3 12:4 2 0:375 f(17 5) (10:2 3)g d 43:4d 1 0:375 11:2 d2 2:1d2 2 pffiffiffiffiffiffiffiffiffiffiffi 2 10 0:375 d 12:2d 1 2:67 2 11:2 d 7:5d2 2 2 pffiffiffiffiffiffiffiffiffi 2:67 d 16:3d 2 10 2 Tie rod force per m T Arm (m) 1.83 5.00 5.50 d/2 6:50 2d/3 6:50 d/2 6:50 2d/3 6:50 d/2 6:50 0 Moment (kNm) 105.0 344.5 68.2 21:7d2 282:1d 1:4d3 13:7d2 6:1d2 79:3d 5:0d3 48:8d2 8:2d2 106:0d 0 X M 3:6d 3 27:7d 2 96:8d 517:7 0 ; d 3 7:7d 2 26:9d 143:8 0 ; d 4:67 m Depth of penetration 1:2d 5:60 m 42 Lateral earth pressure Figure Q6.7 Algebraic sum of forces for d 4:67 m is X F 57:4 68:9 12:4 202:7 45:8 57:0 163:5 76:1 T 0 ; T 90:5 kN=m Force in each tie rod 2:5T 226 kN 6.8 (a) For 0 36 , Ka 0:26 and Kp 3:85; 0 21 9:8 11:2 kN=m3 The pressure distribution is shown in Figure Q6.8. In this case the net water pressure at C is given by uC 15:0 1:5 9:8 13:4 kN=m2 16:5 The average seepage pressure is j Hence, 0 j 11:2 0:9 12:1 kN=m3 0 j 11:2 0:9 10:3 kN=m3 1:5 9:8 0:9 kN=m3 16:5 Lateral earth pressure 43 Consider moments about the anchor point A (per m): Force (kN) (1) 10 0:26 15:0 1 0:26 18 4:52 2 (3) 0:26 18 4:5 10:5 1 (4) 0:26 12:1 10:52 2 1 (5) 13:4 1:5 2 (6) 13:4 3:0 1 (7) 13:4 6:0 2 (2) (8) Ppm 39:0 47:4 221:1 173:4 10:1 40:2 40:2 571 Arm (m) 6.0 1.5 8.25 10.0 4.0 6.0 9.5 11.5 Moment (kN m) 234.0 71.1 1824.0 1734.0 40.4 241.2 381.9 4527 11.5PPm X M0 4527 394 kN=m 11:5 ; Ppm Available passive resistance: 1 Pp 3:85 10:3 62 714 kN=m 2 Factor of safety: Fp Pp 714 1:8 Ppm 394 Force in each tie 2T 2(571 394) 354 kN Figure Q6.8 44 Lateral ear...
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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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