Solution Manual Vol 7

# 83 500 550 d2 650 2d3 650 d2 650 2d3 650 d2 650

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Unformatted text preview: th pressure (b) 0 tan1 ( tan 36 /1:25) 30 ; therefore, Ka 0:33 and Kp 3:0. The surcharge is a variable action; therefore, a partial factor of 1.30 applies to force (1). In this calculation the depth d in Figure Q6.8 is unknown. Consider moments (per m) about the tie point A: Force (kN) (1) 0:33 10 (d 9:0) 1:30 4:3d 38:6 1 0:33 18 4:52 60:1 2 (3) 0:33 18 4:5 (d 4:5) 26:7d 120:3 1 (4) 0:33 12:1 (d 4:5)2 2:00d2 18:0d 40:4 2 1 10:1 (5) 13:4 1:5 2 (6) 13:4 3:0 40:2 1 6:7d (7) 13:4 d 2 1 (8) 3:0 10:3 d2 15:45d2 2 (2) Arm (m) d/2 3:0 1.5 d/2 5:25 2d/3 6:0 4.0 6.0 d/3 7:5 2d/3 7:5 Moment (kN m) (1) (2) (3) (4) (5) (6) (7) (8) 2:15d2 32:2d 115:8 90:2 13:35d2 200:3d 631:6 1:33d3 24:0d2 134:9d 242:4 40:4 241:2 2:20d2 50:2d 10:3d3 115:9d2 X M 8:97d 3 74:2d 2 417:6d 1361:6 0 ; d 3 8:27d 2 46:6d 151:8 0 By trial, d 5:44 m The minimum depth of embedment required is 5.44 m. 6.9 For 0 30 and 15 , Ka 0:30 and Kp 4:8; 0 20 9:8 10:2 kN=m3 The pressure distribution is shown in Figure Q6.9. Assuming uniform loss in total head along the wall, the net water pressure at C is Lateral earth pressure 45 Figure Q6.9 uC 14:7 2:6 9:8 21:6 kN=m2 17:3 and the average seepage pressure around the wall is j 2:6 9:8 1:5 kN=m3 17:3 Consider moments about the prop (A) (per m): Force (kN) (1) (2) (3) (4) (5) (6) (7) (8) 1 0:3 17 2:72 2 0:3 17 2:7 5:3 1 0:3 (10:2 1:5) 5:32 2 0:3 f(17 2:7) (11:7 5:3)g 6:0 1 21:6 2:6 2 21:6 2:7 1 21:6 6:0 2 1 f4:8 (10:2 1:5) 0:3 (10:2 1:5)g 6:02 2 18.6 73.0 49.3 194:2 28.1 58.3 64.8 688:5 Arm (m) 0.20 3.35 4.23 9.00 2.43 4.65 8.00 10.00 Moment (kN m) 3:7 244.5 208.5 1747.8 68.4 271.2 518.4 3055 6885 Factor of safety: Fr 6885 2:25 3055 46 Lateral earth pressure 6.10 For 0 40 , Ka 0:22. The pressure distribution is shown in Figure Q6.10. p 0:65Ka H 0:65 0:22 19 9 24:5 kN=m2 Strut load 24:5 1:5 3 110 kN (a load factor of at least 2.0 would be applied to this value). Using the recommendations of Twine and Roscoe p 0:2 H 0:2 19 9 34:2 kN=m2 Strut load 34:2 1:5 3 154 kN (this value would be multiplied by a partial factor of 1.35). Figure Q6.10 6.11 18 kN=m3 ; H 3:50 m; 0 34 mH 1:85 m nH 3:35 m; Consider a trial value of F 2:0. Refer to Figure 6.35. tan 34 0m tan1 18:6 2:0 Then, 45 0m 54:3 2 1 W 18 3:502 cot 54:3 79:2 kN=m 2 Lateral earth pressure 47 1 s 3:352 5:61 s kN=m 2 1 U 9:8 1:852 cosec 54:3 20:6 kN=m 2 P Equations 6.30 and 6.31 then become 5:61 s N 20:6 tan 18:6 cos 54:3 N sin 54:3 0 79:2 N 20:6 tan 18:6 sin 54:3 N cos 54:3 0 i.e. 5:61 s 0:616N 4:05 0 79:2 0:857N 5:63 0 84:8 98:9 kN=m ;N 0:857 Then, 5:61 s 60:9 4:05 0 64:9 11:6 kN=m3 ; s 5:61 The calculations for trial values of F of 2.0, 1.5 and 1.0 are summarized below: F 2.0 1.5 1.0 0m 18.6 24.2 34 54.3 57.1 62 W (kN/m) 79.2 71.3 58.6 U (kN/m) 20.6 19.9 19.1 N (kN/m) 98.9 85.6 65.7 s (kN/m3) 11.6 9.9 7.7 s is plotted against F in Figure Q6.11. From Figure Q6.11, for s 10:6 kN/m3 , F 1:7 Figure Q6.11 48 Lateral earth pressure 6.12 For 0 36 , Ka 0:26 and K0 1 sin 36 0:41 45 0 63 2 For the retained material between 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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