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181_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011

# 181_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011 -...

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o to maximize the developed shear take the derivative of Cd: VaG: = 0 o lind critical lailure plane angle 8, = ~ (i + ~,) 1 )ti Si{:' )s{-:') o , .. on the critical plane: c - - (-) d - 2 sini COS\$d SL7 l-cos'(~ ) cos ¢l d o so: o hall angle relationship: a ~1+cosa . cos-= where a=I-<i'd 2 2 ,----,----~ _ )ti('-COS(i-~,)) u~ 4c, ( sinicos~, ')1 c,-- oro-- 4 sinicos,;" 'I 1-cos(i-~,) o to solve for the critical height set Cd = c and \$11 = ¢l o to solve 10' the maximum safe height set c, ~(F~) aDd ~,= tan-fa;~ o Substitute y for y if slope submerged, use othe' methods 10' seepage o If calculating FS for a cut then assume FS, calculate ~, = tan-'( t~;~) ,use equation to find Cd, and check FS = C/Cd. Ite,ate to find FS Ihat works. Example C.ullman Slope Calculation ? ...
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Unformatted text preview: ? ... =io'" o A vertical cut is made in sandy clay with Y= 17.3 kN/m', c = 28.7 kPa and ~ = 24°. Determine the sale depth 01 cut using a FS = 2. o Cd =(2~.7)=14.4kPa and <Pd =tan-1 ('an 2 24°)=12.6 a maximum depth of cut: H = 4c d ( sin i c~s <Pd ) = 4(14.4kPa) ( sin 90° cos 12.6° ) = 4.16m 'Y 1-COS(I- <Pd) 17.3kN/m' 1-cos(90°-12.6°) o if slope sUbmerged, maximum depth of cut: H = 4c d ( sin i COS<Pd ) = 4(14.4kPa) (Sin900 cos12.6° ) = 9.59m y' 1-cos(i - <Pd) (17.3 - 9.8)kN / m' 1- cos(90°-12.6°) (but must keep cut full of water!)...
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