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Another Prob. 8 Solution - Long Term For silty clay 1111...

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Unformatted text preview: Long Term For silty clay: 1111:: lOOpcf Wsat: lOSpcf 1W: 62.4pcf ’53“: 500psf (1):: 15 Assume saturated silty clay for both the long term and shor term analysis. B :2 5ft L := B = SAft D :2 15ft. Df :2 2.5ft PS I: 3 w w Fort: =15 degree: NO: 10.98 Nq:: 3.94 N1: 1.129 Shape Factor (for 4: >= 10): 2 2103::1+0.20-E-t 45+i .l 21.34 L 2 180 B {b 11' 2 xqs :: 1 + 0.10~[E]-|:tau|:[45 + EJ-Efl : 1.17 Mg :: )th : 1.17 Depth Factor (for ct: >= 10): Df xcd::1+ 0.20. — ~t 45+2 -l : 1.13 B 2 180 A 1+010 Df t 45+¢ " 1065 )1 1065 :: _ . — . a — .— : _ 2: : _ Cid B 2 180 Md qd Inclination Factor (for or = O): )‘ci:: 1.00 )‘qii: 1.00 )‘ofi :: 1.00 ”1'2: 'Tsat , 1w : 45.6-pcf q :: T-Df : 114-psf 1 (111:: GNU-xcsotcdotci + q-quqsotqdotqi + 3-1 -B-N,Y-)\,.YS-)\,Yd->\Wi : 9.033ka 2 -B A:: L : 19.635-ft2 ”W 4 Q11 :: qu-A : 177.37-kip Qnet :: qu-A , q-A: 175.132-kip Qnet Qnet_allowable :: E : 58-377-kip Short Term For silty clay: "inf: 10(1ch Al'sat := lfltglcf 1w:= 62.4pcf (311:: 95(3351' (13:11:: 0 Assume saturated silty clay for both the long term and shor term analysis. B := 5ft ‘3“: B: S-fl DW:= 15ft Df := 2.5fl FS := 3 For = 0 de ree: -= -= -= t 9 NC . 5.14 Nq . 1.00 N.f . 0 Shape Factor (for q: = 0): kg := 1+ 0.208] = 1.2 kqs := 1 M5 := 1 Depth Factor (for q: = 0}: ID . f M13: 1+ 0.20-[—] = 1.1 )‘qd :2 1 Md :2 1 Inclination Factor (for o = 0}: he“: 1.00I Aqi :2 1.00 A1“: 1.00 q := r‘lsat'Df = 270-psf qu == C‘u-Nc')\t:s"\cc1')*ci+ q'Nq'kqs'qu'kqi = 63mm 2 ”IT-B := T = 19.635112 Q1] := qu-A= 131.86-kip Qnet := qu-A— q-A= 126558-ij ant ant_allowablc :2 E = 42-1351“? Short term condition is more critical and should be used for design because Qnet of the short term is less. ...
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