Example 8-1.doc

Example 8-1 - Example 8.1 The cross section of a cantilever retaining wall is shown in Figure 8.14 r the factors of safety with respect to

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Unformatted text preview: Example 8.1 The cross section of a cantilever retaining wall is shown in Figure 8.14. r the factors of safety with respect to overturning, sliding, and bearing capacity}: Solution From the figure, ['1’ = H1 +H2+H3=2.6tan10°+6+0.7 = 0.458 + 6 + 0.7 = 7,158m The Rankine active force per unit length of wall = P, = £71H'2Ka. For 4:; = and a = 10°, K, is equal to 0.350. (See 'Ihble 7.1.) Thus, P. = %(18) (7.158)2(0.35) = 161.4 kN/m P. = 1",, sin10° = 161.4(sin10°) = 28.03 kN/m 31‘, : . ‘1. and Pb = P0008100 = 161.4(cos10°) = 158.95 kN/m Factor of Safety against Overturning . g The following table can now be prepared for determining the resistingnio Weight/unit Moment arm Section Area length from point e Moment , " ‘=i_ no.‘ (m‘) (RN/m) (ml . lkN-m‘lm) ' ~._ 1 6 x 05 = 3 70.74 1.15 81.35 ' f; 2 %(0.2)6 = 0.6 14.15 0.833 ' . 11.79 . 3 4 x 0.7 = 2.8 66.02 2.0 132.04 4 6 x 2.6 = 15.6 280.80 27 758.16 I 5 §(2.6) (0.458) = 0595 10.71 3.13 33.52 P. = 28.03 4.0 112.12 1 2V = 470.45 21128.98 = SM ‘For section numbers, refer to Figure 8.14 7mm = 23.58 kN/m3 The overturning nioment M0 = pkg) = 158950—1553) = 37925 kN-m/ - . -_.._——._——— —.-——_______..__..__.: . -I: i H3=0.7m C _______ _- k-o.7m->l+0.7m->k—2.6m—>J rz=19kNlm3 ¢é=20° cé = 40kN/m2 Fianna 14 Calculation of stability of a retaining wall "and_ _ EMR _ 1128.98 _ Fsmmm — Mo — 379.25 — 2.98 > 2,01( Factor ofASafety against Sliding From (8.11), * ' FS . = (2V)tan(k,¢§) + Bkzcg + Pp ‘ (W) P“ cosa Letki 4-: k2 = §.A1so, ' P, = éprzDz + zen/va K, = tan2(45 + = m2(45 + 10) = 2.04 and D= 1.5m So Pp = §(2.04) (19) (1.5)2 + 2(40) (V2.04) (15) = 43.61 + 171.39 = 215 kN/m 20) + (4)(§)(40) + 215 158.95 Hence, X (470.45) tan ( 2 “(my = 111.5 + 106.67 + 215 -— 158.95 — 2.73 > 1.5, 0K Note: For some designs, the depth D in a passive pressure calculation may: taken to be equal to the thickness of the base slab. ‘ H Factor of Safety against Bearing Capacity Failure Combining Eqs. (8.13), (8.14), and (8.15) yields B EMR - 2M0 4 1128.98 - 379.25 “13- 2v =2" 470.45 =0406m<~§=§=0666m Again, from Eqs. (8.17) and (8.18), 2v 6e 470.45 6 x 0.406 I 4131:; = 79—(1 : E) = 4 (1 1 ———4—-——) = 189.2 kN/mz (toe) = 45.99kN/m2 (had) _'_ The ultimate bearing capacity of the soil can be determiner} from Eq. (8.19 For 41; = 20° (see Table 3.4), N. = 14.83, N, = 6.4, and N7 = 539. A180, _‘ q = 720 = (19) (1.5) = 28.5 kN/mz ' ' B' = 13‘2‘—’=4-2(0.406) =3.188m - 1 +1.1(g)=1 1 1.4.1:.) - 1.111 3.188 D 1.5 = I _ ' r 2 __ z ________. = E”, 1 + 2tan¢2(l sunk) (3,) 1 + 0.315(3188) 1.148 _ Fyd = 1 0 2 'Fci=Fqi=(1—9¢(I)o) and EV 470.45 So 18.67 2 Fa—Fqi-(l- 90 ) —0.628 and __ «It 2 _ 18.67 2 F""(1 abs)"(1 20 ) 0 Hence, qu = (40) (14.83) (1.188) (0.628) + (28.5) (6.4) (1.148) (0.628) + §(19) (5.93) (3.188) (1) (O) = 442.57 + 131.50 + 0 = 574.07 kN/m2 and u 574.07 FS(beafingmpacity) = q— : -——- = 3.03 > 3, 0K qm 189.2 ...
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This test prep was uploaded on 04/03/2008 for the course CIVE 410 taught by Professor Chang during the Spring '08 term at Drexel.

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Example 8-1 - Example 8.1 The cross section of a cantilever retaining wall is shown in Figure 8.14 r the factors of safety with respect to

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