Solution Manual Vol 7

Calculation of total thrust and its point of

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Unformatted text preview: ffiffiffiffiffiffiffiffiffiffiffiffiffiffi Kac 2 0:35 1:5 1:45 Kp 3:7 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi Kpc 2 3:7 1:5 4:7 At the lower end of the piling: pa Ka q Ka 0 z Kac c0 0:35 18 3 0:35 10:2 4 1:45 10 18:9 14:3 14:5 18:7 kN=m2 pp Kp 0 z Kpc c0 3:7 10:2 4 4:7 10 151 47 198 kN=m2 Equation 6:19 Equation 6:24 Lateral earth pressure 37 6.4 (a) For 0 38 , Ka 0:24 0 20 9:8 10:2 kN=m3 The pressure distribution is shown in Figure Q6.4. Consider moments (per m length of wall) about the toe. Force (kN) (1) 0:24 10 6:6 15.9 1 2 31.0 (2) 0:24 17 3:9 2 (3) 0:24 17 3:9 2:7 43.0 1 2 (4) 0:24 10:2 2:7 8.9 2 1 35.7 (5) 9:8 2:72 2 H 134.5 (6) (7) (8) (9) (10) 6:2 0:4 23:5 4:0 0:4 23:5 3:9 2:6 17 2:3 2:6 20 100 58.3 37.6 172:4 119:6 100.0 V 487.9 Arm (m) 3.3 4.00 1.35 0.90 0.90 1.20 2.00 2.70 2.70 1.20 Moment (kN m) 52.5 124.0 58.0 8.0 32.1 MH 274.6 70.0 75.2 465.5 322.9 120.0 MV 1053.6 X M MV MH 779:0 kN m Lever arm of base resultant: M 779 1:60 V 488 Eccentricity of base resultant: e 2:00 1:60 0:40 m (10) 10 kN/m2 0.4 m (6) 2.6 m (8) 3.9 m (1) (2) WT (9) 2.7 m 0.4 m (7) 4.0 m (4) (3) (5) Hydrostatic Figure Q6.4 38 Lateral earth pressure Base pressures (Equation 6.27): V 6e 1 p B B 488 1 0:60 4 195 kN=m2 and 49 kN=m2 Factor of safety against sliding (Equation 6.28): F V tan 488 tan 25 1:7 H 134:5 (b) Using a partial factor of 1.25 the design value of 0 is tan1 ( tan 38 /1:25) 32 . Therefore, Ka 0:31 and the forces and moments are: H 163:3 kN V 487:9 kN MH 345:3 kN m MV 1053:6 kN m The overturning limit state is satisfied, the restoring moment (MV ) being greater than the overturning moment (MH ). The sliding limit state is satisfied, the resisting force (V tan 227:5 kN) being greater than the disturbing force (H ). 6.5 For 0 36 , Ka 0:26 and Kp 3:85. Kp 3:85 2 F 0 20 9:8 10:2 kN=m3 The pressure distribution is shown in Figure Q6.5; hydrostatic pressure on the two sides of the wall balances. Consider moments about X (per m), assuming d &gt; 0: Force (kN) 1 0:26 17 4:52 2 (2) 0:26 17 4:5 d 1 (3) 0:26 10:2 d2 2 1 3:85 17 1:52 (4) 2 2 3:85 (5) 17 1:5 d 2 1 3:85 (6) 10:2 d2 2 2 (1) 44:8 19:9d 1:33d2 36:8 49:1d 9:82d2 Arm (m) d 1:5 d/2 d/3 d 0:5 d/2 d/3 Moment (kN m) 44:8d 67:2 9.95d2 0.44d3 36:8d 18:4 24:55d2 3:27d3 Lateral earth pressure 39 Figure Q6.5 X M 2:83d 3 14:6d 2 8:0d 48:8 0 ; d 3 5:16d 2 2:83d 17:24 0 ; d 1:79 m Depth of penetration 1:21:79 1:50 3:95 m X F 0; hence R 71:5 kN substituting d 1:79 m Over additional 20% embedded depth: pp pa 3:85 17 4:5 0:26 17 1:5 3:85 0:2610:2 2:12 365:5 kN=m2 Net passive resistance 365:5 0:66 241 kN &gt;R 6.6 The design value of 0 36 , i.e. the partial factor has been applied. The active pressure coefficient is given by Equation 6.17, in which 105 , 20 , 36 and 25 2 32 6 7 sin 69=sin 105 6 7 pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi7 0:50 Ka 6 sin 16 )5 4pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ( sin 61 pffiffiffiffiffiffiffiffiffiffiffiffiffiffi sin 1...
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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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