Rotational failure analysis

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Rotational failure analysis

  1. 1. By Charchit Jain
  2. 2. Condition for Rotational failure  Failure is along the line of least resistance through the slope , observation of various failure suggests that the failure surface is generally takes the form of a circle and most of the theories are based on this observation.  The individual particles in the rockmass are very small as compared with the size of the slope.  These particles are not interlocked.
  3. 3. Derivation of Rotational failure charts  The following assumptions are made in deriving the stability charts : 1. Material forming the slope is assumed to be homogeneous. 2. τe = c + σ tan φ 3. Failure surface passes through the toe of the slope. 4. Location of the tension crack and the failure surface are such that the FOS of the slope is minimum for the slope geometry. 5. A range of ground water condition, from dry to fully saturated condition are considered.
  4. 4.  Factor of safety of the slope is defined as = and rearranging this equation, we obtain τe = (c + σ tan φ)/FS shear stress available to resist sliding depends upon the distribution of normal stress (σ ) along this surface . )surface(sliponmequilibriuforrequiredstressshear )tan(cslidingresisttoavailablestrengthshear e  FS
  5. 5.  Failure charts are produced depending upon the ground water condition.  Location of tension crack and the failure surface is determined.
  6. 6. Example  A 50 ft. high slope with the face angle of 40° is to be excavated In the overburden soil with density ɗ = 100 lb/ft3 ,angle of friction is 30°, cohesive strength of c = 800 lb/ft2 , find the factor of safety of slope assuming it to be a fully drained slope.  value of (c/ɗ.H.tanØ) is 0.28 , so the corresponding tanØ/F value for slope of 40° is 0.28 . Hence , the Factor of safety is 2.061 .
  7. 7. Location of failure surface and tension crack  Charts are used for the construction of drawings of potential slides.  Charts are also used for the determination of location of tension crack.  Ground water condition is considered while selecting the chart.
  8. 8. Location of critical failure surface and tension crack for drained slope
  9. 9. Location of critical failure surface and tension crack in the presence of ground water
  10. 10. Example  A slope having face angle of 30° in a drained soil with angle of friction is 20°. Locate the centre of critical failure surface and tension crack .  X = 0.2H Y = 1.85H b = 0.1H
  11. 11. Short term stability of soils with Ø = 0  Failure will then occur along the circular arc of radius R, and soil is assumed to be homogeneous. F = Resisting moment / disturbing moment Resisting moment = R2ØCu Disturbing moment = Wx
  12. 12. Method of slicing  For soil having Ø ≠ 0 , a more elaborate analysis is required.  Dividing the failure mass into number of slices, preferably more than 5.
  13. 13. Force acting on ith slice
  14. 14. Overturning moment = R W i=1 n i i sin  Restoring moment = R T i=1 n i F sisting Moment Overturning Moment c l N W i i i i i n i i i n           Re [ tan ] sin    1 1 Case of drained slope is considered so Ui = 0
  15. 15. Swedish Method of Slices This method is based on the assumption that the resultant of the inter-slice forces acts perpendicular to the normal force N. Ei Ei 1 Uii Uii  1 Xi Xi  1 Ti Ni Ui Wi Ti Ni Ui Wi R N = N + U = Wi i i i i cos 
  16. 16. F = [ c l + (W - U ) W i=1 n i i i i i i i=1 n i i     cos tan ] sin    F = [ c l + W t W i=1 n ui i i i ui i=1 n i i    cos an ] sin    For effective stress analysis For total stress (undrained) analysis
  17. 17. Example
  18. 18. Calculations for slice 6 1. Measure l from the figure and width of the slice 6 26 ft l = 28.5 ft Embankment foundation
  19. 19. Example 4. Determine the self-weight of the slice 6 26 ft embankment foundation 53.6ft 5 ft W = A g = 257 pound/ft 3 10.4 ft
  20. 20.  Total resisting force = 640 kips  Total driving force = 524 kips Factor of safety = 640 / 524 = 1.22
  21. 21. Bishop’s simplified method Assumption: The vertical inter-slice forces are equal and opposite. The resultant thus acts perpendicular to W Ei Ei 1 Uii Uii  1 Xi Xi  1 Ti Ni Ui Wi Ti Ni Ui Wi R W = T + N + u xi i i i i i isin cos  
  22. 22. W = T + N + u xi i i i i i isin cos   T = c l F + N F i i i i i   tan          N = W - u x - (1/ F) c x 1 + F i i i i i i i i i i   tan cos tan tan     now hence
  23. 23. Substitution of the expression for N’ into the equation for the factor of safety F sisting Moment Overturning Moment c l N W i i i i i n i i i n           Re [ tan ] sin    1 1 leads to F = ( c x + ( W - u x ) ) 1 M ( ) W i=1 n i i i i i i i i=1 n i i            tan sin    where M ( ) = [ 1 + F ]i i i i     cos tan tan 
  24. 24. Janbu simplified method of slice  It is based on the assumption that the failure surface can be of general shape.  Analysis of this method is similar to that of bishop’s method.
  25. 25. The main methods for Rotational failure analysis

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