181_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
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.
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ? ... =io'&quot; 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 &lt;Pd =tan-1 ('an 2 24)=12.6 a maximum depth of cut: H = 4c d ( sin i c~s &lt;Pd ) = 4(14.4kPa) ( sin 90 cos 12.6 ) = 4.16m 'Y 1-COS(I- &lt;Pd) 17.3kN/m' 1-cos(90-12.6) o if slope sUbmerged, maximum depth of cut: H = 4c d ( sin i COS&lt;Pd ) = 4(14.4kPa) (Sin900 cos12.6 ) = 9.59m y' 1-cos(i - &lt;Pd) (17.3 - 9.8)kN / m' 1- cos(90-12.6) (but must keep cut full of water!)...
View Full Document

Ask a homework question - tutors are online