{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

200_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011

200_CEG 4012 Notes Fall 2011 CEG 4012 Notes Fall 2011 - 1.0...

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

SL26 Cousins Example Find the factor of safety for a slope where the. following conditions exist: Slope inclination, i = 30·, height = 40 ft (12 m) Cohesion, C' = 200 psf (9.6 kN/m 2 ) Angle of internal friction, cl> = 20· Unit weight, 'Y = 110 pcf (17.3 kN/m 3 ) Pore pressure ratio,'u = 0.2 (assume that value was estimated from a pore pressure ratio diagram ) Find A = 'Y H tan cl> = (110 pcf)(4O ft)(tan 20 0 ) = 8 c+ c' 200 psf Checking possibility of base failure Checking possibility of toe failure For 'u = 0, use Fig. e, For,u = 0, use Fig. b, get D = I, N s = 26 get D = 1.03, N s = 27 For 'u = 0.25, use Fig. f, For 'u = 0.25, use Fig. c, get D = I, N. = 22 get D = 1.03, N s = 22 ~ Failure arc passes very close to the toe (based on least N,), use D =
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1.0 ( + ). If r u = 0.2, N s for stability computation = 23 (by interpolation of above N s values). FS = N,c' = (23)(200 psf) = 1 05 "fH (110 pcf)(4O ft) . Coordinates of center for failure circle if a toe failure is assumed as a close approximation: For r u = 0, (X/H) tan j = 0.10 (Y/H) tan j = 1.2 For,u = 0.25, (XIH) tan i = 0.15 (YIH) tan i = 1.1 (from Fig. h) (from Fig. i) Therefore, for 'u = 0.2, use (Y/H) tan i = 1.18, or Y = (1.18)(40 ft)(.an 1 30") = 81.8 ft (X/H) tan i = 0.14, or X = (0.14)(40 ft)(tan\oo) = 9.7 ft...
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online