ch16 Soil Mechanics [PDF]

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Exercise 16.1 A clay slope fails as shown in Fig. P16.1. Derive an equation for the undrained shear strength of the clay.



Solution 16.1 B



C



A



Sine rule : AB  b



b sin( s   )







AB sin 



Ho sin  s



H o sin( s   ) sin  sin  s



1 1  H o2 sin( s   ) W   bH o  2 2 sin  sin  s 1  H o2 sin( s   ) 1  H o2 sin( s   ) sin     W sin   2 sin  sin  s 2 sin  s L=



Ho sin 



Shear resistance along L is:



suHo sin 



Ho 1  H o2 sin( s   )  su  sin  2 sin  s su 



1  H o sin( s   )sin  2 sin  s



Exercise 16.2



Derive an equation for the factor of safety of the slope in Fig. P16.2 using the mechanism shown.



Solution 16.2



F F



z



 0 : W1  R1 sin   1 sin   N1 cos1  0



(1)



x



 0 : 1 cos1  N1 sin 1  R1 cos   0



(2)



Now 1  N 1



tan   FS



(3)



Substituting (3) into (1) and (2) gives  tan    W1  R1 sin   N1  sin 1  cos1    0  FS 



(1a)



  tan   N1  cos1  sin 1   R1 cos   0   FS



(2a)



From (1a)



N1 



R1 sin   W1   tan 1 sin 1  cos 1     FS



(1b)



From (2a)



N1 



R 1 cos  tan    cos1  sin 1    FS 



(2b)



tan   tan   sin 1  cos 1 and B = cos 1  sin 1 FS FS and solving for R1 from (1b) and (2b), we get R1 sin   W1 R1 cos   A B



Putting A =



R1 sin  R1 cos  W1   A B A  sin  cos   W1  R1    B  A  A W1 A and, R1   sin  cos      B   A



________________________________________________________________________



F F



z



 0 : W3  R3 sin   3 sin  3  N 3 cos 3  0



(4)



x



 0 : R3 cos  3 cos 3  N 3 sin  3  0



(5)



Now 3  N 3



tan   FS



Substituting (6) into (4) and (5) gives  tan    sin  3  cos 3   0 W3  R3 sin   N 3   FS 



(6)



(4a)



R3 sin   W3 R sin   W3  3 tan   C sin  3  cos 3 FS tan   where C = sin  3  cos  3 FS   tan   cos 3  sin  3   0 R3 cos   N 3    FS N3 



(4b)



(5a)



 R3 cos  (5b) D tan   where D = cos  3  sin  3 FS Equating (4b) and (5b) gives R3 sin   W3  R3 cos   C D W3 C  R3  sin  cos   C D ________________________________________________________________________ N3 



F F



z



 0 : W2  N 2  R3 sin   R1 sin   0



 0 : R1 cos   2  R3 cos   0 tan   Now 2  N 2 FS Substituting (9) into (7) and (8) and solving for N 2 gives N 2  W2  R3 sin   R1 sin  x



N2 



FS R3 cos   R1 cos   tan  



(7) (8) (9) (7a) (8a)



Equating (7a) and (8a) gives FS R3 cos   R1 cos    W2  R3 sin   R1 sin  tan   tan   FS = W2  R3 sin   R1 sin   R3 cos   R1 cos 







The solution of FS is solved by iteration, assuming values of  and  . The maximum value of either  and  is   . tan   Usually tan   tan   FS NOTE: W1 , W2 , and W3 , are dependent on  s



Exercise 16.3 Figure P16.3 shows the profile of a beach on a lake. It is proposed to draw down the lake by 2 m. Determine the slope angle of the beach below the high water level after the drawdown. You may assume an infinite slope failure mechanism. The critical state friction angle of the sand is 30°.



Solution 16.3    17.8  9.8  8 kN m 3



It can be assumed that the angle of repose is 28o – the slope of the upper portion of the beach. The slope angle of the lower portion within the proposed drawdown depth is 33o. The slope is stable at this higher angle because of the lateral stress of the water. Therefore, for the worst-case scenario you can take the friction angle as 28o. The maximum stable slope from the high water level to the drawdown level is given by equation (16.8).  8  αs =tan -1  tan   tan -1  0.45  tan   17.8 



For  '  28 ,



 s  tan 1  0.45  tan 28  13.5



Exercise 16.4 A cut for a highway is shown in Figure P16.4. Determine the factor of safety of the slope using an ESA and a TSA. Assume a center of rotation, O, such that the slip surface passes through the toe of the slope.



Solution 16.4 Bishop's simplified method x Assume that no tension cracks Homogenous soil su 40 kPa 29 deg. ' 3 9.8 kN/m w



b



R = 14.8m



11 9



z 1



3



sat z'cr zs R TCM/R FS



18.5



kN/m



9.20



m



0 14.8 171.6 3.1



m m kN assumed



Slice



b



z



W=bz



zw



1 2 3 4 5 6



m 3 3 3 3 3 1.5



m 2.7 7.2 10.2 12.2 13.8 14.4



kN 149.9 399.6 566.1 677.1 765.9 399.6



m 2.8 5.8 7.7 8.5 8.7 8.5



10



8 2



3 zw



4



5



6 7







5m



ru







mj



Wsin



0.55 0.43 0.40 0.37 0.33 0.31



deg -55 -38 -20 -12 0 11



2.34 1.48 1.14 1.06 1.00 0.98



-122.7 -246.0 -193.6 -140.8 0.0 76.2



ESA W(1 ru)tan' mj



TSA su b/cos



87.7 187.3 214.4 251.7 282.8 149.9



209.2 152.3 127.7 122.7 120.0 61.1



7 8 9 10 11



1.5 3 3.5 2.5 1.2



14.2 13.4 11.3 7.7 3.3



394.1 743.7 731.7 356.1 73.3



8.1 7.5 5.3 1.7 0.0



0.30 0.30 0.25 0.12 0.00



13 22.5 40 52 76.5



0.99 1.01 1.13 1.32 2.45



88.6 284.6 470.3 280.6 71.2



150.2 292.2 345.9 230.4 99.7



61.6 129.9 182.8 162.4 205.6



Sum



568.5



2292.1



1535.3



FS



3.10



2.07



Exercise 16.5 Determine the factor of safety of the slope shown in Figure P16.5 using an ESA and a TSA. The point of rotation is indicated by O and the line representing the top of the stiff soil is a tangent to the failure plane.



Solution 16.5 No tension crack Soil 1 Soil 2 su 70 25 22 28 ' 9.8 9.8 w sat zcr z'cr zs R TCM/R FS



b



kPa deg. 3 kN/m



18.5



18



kN/m m m m m kN



1.16



2.78 6.10 3 13.2 13.9 assumed



R = 13.2m



3



11 z 4



1 2 3 



zw 5



6



9



10 soil 2



8



7 5m soil 1 ESA



Slice



1 2 3 4 5 6 7



b



z



W=bz



zw



m



m



kN



m



3 0.5 2 2 2 2 2



1.2 2 3.5 5.2 6.7 8 8.9



64.8 18.0 126.0 187.2 241.2 288.0 320.4



1.2 2 3.5 5.2 6.5 7 6.4



ru







mj



Wsin



W(1 - ru)tan' mj



TSA su b/cos



1.65 1.38 1.22 1.08 1.00 0.95 0.92



-34.3 -7.3 -36.8 -26.1 0.0 42.6 91.0



25.9 6.0 37.1 48.9 60.5 75.9 95.2



88.4 13.7 52.3 50.5 50.0 50.6 52.1



deg 0.54 0.54 0.54 0.54 0.53 0.48 0.39



-32 -24 -17 -8 0 8.5 16.5



8 9 10 11



2 2 2 1.8



8.6 7.2 5.2 2.5



309.6 259.2 187.2 81.0



6 5 3.0 0.2



0.38 0.38 0.31 0.04



26 36.5 51.5 66.5



0.91 0.93 1.02 1.22



135.7 154.2 146.5 74.3



92.8 79.6 69.6 50.3



55.6 62.2 80.3 112.9



Sum



539.7



641.8



668.6



FS



1.16



1.21



Exercise 16.6 A compacted earth fill is constructed on a soft, saturated clay (Figure P16.6). The fill was compacted to an average dry unit weight of 19 kN/m3 and water content of 15%. The shearing strength of the fill was determined by CU tests on samples compacted to representative field conditions. The shear strength parameters are su = 45 kPa, p = 34°, and cs = 28°. The variation of undrained shear strength of the soft clay with depth as determined by simple shear tests is shown in Figure P16.6, and the friction angle at the critical state is cs = 30°. The average water content of the soft clay is 40%. Compute the factor of safety using Bishop’s simplified method. Assume that a tension crack will develop in the fill.



Solution 16.6 Soil 1 d = 19 kN/m3 (



Soil 2



)



(



Under GWE soil is saturated. Also assumed that Gs = 2.7



(



)



(



)



)



Tension cracks filled with water Two soil layers Soil 1 Soil 2 10 35 30 28 9.8 9.8



su ' w sat zcr z'cr zs R TCM/R FS Slice



1 2 3 4 5 6 7 8 9



17.8



2.45



21.9 3.20 5.79



kN/m



4 14.4 21.3 assumed



m m kN



1 2 3 4 5 6 7 8 9



3



m m



b



z1



z2



W=bz







m



m



m



kN



deg



3 3 3 3 3 3 2 1.8 2



1.3 3 4 4.1 3.8 2.6 1 0 0



0.5 1.4 2.5 3.5 4.5 5.5 6 5.2 3



102.3 252.2 377.9 448.9 498.6 500.2 298.4 205.0 131.4



-36 -22 -9 1 14.5 27 38 46 61



No tension crack TCM/R 21.3 FS 2.89 assumed Slice



kPa deg. 3 kN/m



mj



Wsin



ESA



TSA



Wtan' mj



su



su b/cos



kPa 1.49 1.19 1.05 1.00 0.97 1.00 1.09 1.18 1.48



-60.1 -94.5 -59.1 7.8 124.8 227.1 183.7 147.5 114.9



88.0 173.5 229.4 258.1 280.3 289.4 187.0 0.0 0.0



Sum



592.1



1505.8



377.0



FS



2.45



0.61



15 18 18 18 18 18 15 0 0



55.6 58.2 54.7 54.0 55.8 60.6 38.1 0.0 0.0



kN



b



z1



z2



W=bz







m



m



m



kN



deg



3 3 3 3 3 3 2 1.8 2



1.3 3 4 4.1 3.8 2.6 1 0 0



0.5 1.4 2.5 3.5 4.5 5.5 6 5.2 3



102.3 252.2 377.9 448.9 498.6 500.2 298.4 205.0 131.4



-36 -22 -9 1 14.5 27 38 46 61



mj



Wsin



ESA



TSA



Wtan' mj



su



su b/cos



kPa 1.45 1.17 1.05 1.00 0.98 1.02 1.11 1.21 1.55



-60.1 -94.5 -59.1 7.8 124.8 227.1 183.7 147.5 114.9



85.4 170.8 228.1 258.3 282.7 294.1 191.1 143.1 117.5



Sum



592.1



1771.1



612.1



FS



2.89



1.00



15 18 18 18 18 18 15 35 35



55.6 58.2 54.7 54.0 55.8 60.6 38.1 90.7 144.4



Exercise 16.7 A cross section of a canal is shown in Figure P16.7. Determine the factor of safety for (a) the existing condition and (b) a rapid drawdown of the water level in the canal. Use Bishop’s method. The center of rotation of the sliding mass is at coordinates x = 113 m and y = 133 m. The rock surface is tangent to the slip plane. The properties of the soil are as follows:



Solution 16.7



Solution 16.7 (a) Factor of Safety for existing condition



R = 34m



Assume failure surface is tangent to rock surface



Tension cracks filled with water Two soil layers Soil 1 Soil 2 su 21 kPa 34 30 28 deg. f' gw 9.8 9.8 kN/m3 gsat 18 17.8 kN/m3 zcr 3.82 m z'cr 8.50 m zs 18.6 m R 34 m TCM/R 44.5 kN Moment from water in canal 2 Fx = 0.5 x 9.8 x 4.8 = Fz = 0.5 x 9.8 x 4.8 x 11.2 = TMR/R FS 2.04 Slice 1 2 3 4 5 6 7 8 9 10



b m 5.2 4 4 4 4 4 4 3 3 2



Fz 8 3.7m



7



11.2m 5



4.8m Fx 1.6m



1



23.2o



112.896 kN 263.424 kN



2



3



m 1.8 4.4 6.2 5.5 4.6 3.1 0.9 0 0 0



m 0 0 0 2 3.7 5.4 7.1 7 4.6 1.5



W=gbz kN 168.5 316.8 446.4 538.4 594.6 607.7 570.3 373.8 245.6 53.4



Soil 1



Rock Moment arm (m) Moment (kN.m) 31.2 3522.4 9.3 2449.8 Total moment (TMR) 5972.2



q deg -14.5 -6 1 9 16 22.5 31.5 35 45 52



mj



Wsinq



1.11 -42.2 1.03 -33.1 1.00 7.8 0.97 84.2 0.97 163.9 0.98 232.5 1.01 298.0 1.02 214.4 1.10 173.7 1.19 42.1 Sum 1141.3 FS



TSA ESA Wtanf' mj su b/cosq 99.2 174.1 236.3 278.4 306.1 315.7 306.7 202.5 144.0 0.0 2063.0 2.04



Soil 2 4m



4



175.7 kN



z2



10



6



assumed z1



9



112.8 84.5 84.0 85.0 87.4 90.9 98.5 124.5 144.2 0.0 911.9 0.90



1.0



Solution 16.7 (a) Factor of Safety for existing condition



R = 34m



Assume failure surface is tangent to rock surface



No tension crack Two soil layers Soil 1 Soil 2 su 21 34 30 28 f' gw 9.8 9.8 gsat 18 zcr z'cr zs R TCM/R



kPa deg. kN/m3 kN/m3 m m m m kN



17.8 3.82 8.50 18.6 34 44.5



Moment from water in canal 2 Fx = 0.5 x 9.8 x 4.8 = Fz = 0.5 x 9.8 x 4.8 x 11.2 = TMR/R FS 2.08 Slice 1 2 3 4 5 6 7 8 9 10



b m 5.2 4 4 4 4 4 4 3 3 2



Fz 8 3.7m



7



11.2m 5



4.8m Fx 1.6m



1



23.2o



112.896 kN 263.424 kN



2



3



m 1.8 4.4 6.2 5.5 4.6 3.1 0.9 0 0 0



m 0 0 0 2 3.7 5.4 7.1 7 4.6 1.5



W=gbz kN 168.5 316.8 446.4 538.4 594.6 607.7 570.3 373.8 245.6 53.4



Soil 1



Rock Moment arm (m) Moment (kN.m) 31.2 3522.4 9.3 2449.8 Total moment (TMR) 5972.2



q deg -14.5 -6 1 9 16 22.5 31.5 35 45 52



mj



Wsinq



1.11 -42.2 1.03 -33.1 1.00 7.8 0.97 84.2 0.97 163.9 0.98 232.5 1.01 298.0 1.02 214.4 1.11 173.7 1.20 42.1 Sum 1141.3 FS



TSA ESA Wtanf' mj su b/cosq 99.1 174.1 236.3 278.6 306.4 316.2 307.5 203.1 144.6 34.0 2099.9 2.08



Soil 2 4m



4



175.7 kN



z2



10



6



assumed z1



9



112.8 84.5 84.0 85.0 87.4 90.9 98.5 124.5 144.2 110.5 1022.4 1.01 1.0



Solution 16.7 (b) Factor of Safety for existing condition



R = 34m



Assume failure surface is tangent to rock surface



Tension cracks filled with water Two soil layers Soil 1 Soil 2 su 21 kPa 34 30 28 deg. f' 3 gw 9.8 kN/m 9.8 3 gsat 18 kN/m 17.8 zcr 3.82 m z'cr 8.50 m zs 18.6 m R 34 m TCM/R 44.5 kN FS Slice 1 2 3 4 5 6 7 8 9 10



1.75 b m 5.2 4 4 4 4 4 4 3 3 2



8



9



Soil 2



7



5 1



2



3



10



6



4m



4



Soil 1



Rock



assumed z1 m 1.8 4.4 6.2 5.5 4.6 3.1 0.9 0 0 0



z2 m 0 0 0 2 3.7 5.4 7.1 7 4.6 1.5



W=gbz kN 168.5 316.8 446.4 538.4 594.6 607.7 570.3 373.8 245.6 53.4



q deg -14.5 -6 1 9 16 22.5 31.5 35 45 52



mj



Wsinq



1.12 -42.2 1.04 -33.1 0.99 7.8 0.97 84.2 0.96 163.9 0.96 232.5 0.99 298.0 0.99 214.4 1.06 173.7 1.14 42.1 Sum 1141.3 FS



TSA ESA Wtanf' mj su b/cosq 100.4 175.0 236.1 276.5 302.6 310.6 299.8 197.1 138.9 32.4 2069.5 1.75



112.8 84.5 84.0 85.0 87.4 90.9 98.5 124.5 144.2 110.5 1022.4 0.86 1.0



Solution 16.7 (b) Factor of Safety for existing condition



Tension cracks filled with water Two soil layers Soil 1 Soil 2 su 21 kPa 34 30 28 deg. f' 3 gw 9.8 kN/m 9.8 3 gsat 18 kN/m 17.8 zcr 3.82 m z'cr 8.50 m zs 18.6 m R 34 m TCM/R 44.5 kN FS Slice 1 2 3 4 5 6 7 8 9 10



1.72 b m 5.2 4 4 4 4 4 4 3 3 2



4m



R = 34m



Assume failure surface is tangent to rock surface



8



1



2



3



10 Soil 2



7 5



9



6



4



Soil 1



Rock



assumed z1



z2



m 1.8 4.4 6.2 5.5 4.6 3.1 0.9 0 0 0



m 0 0 0 2 3.7 5.4 7.1 7 4.6 1.5



W=gbz kN 168.5 316.8 446.4 538.4 594.6 607.7 570.3 373.8 245.6 53.4



q deg -14.5 -6 1 9 16 22.5 31.5 35 45 52



mj



Wsinq



1.12 1.04 0.99 0.97 0.96 0.96 0.99 0.99 1.06 1.14 Sum



-42.2 -33.1 7.8 84.2 163.9 232.5 298.0 214.4 173.7 42.1 1141.3 FS



TSA ESA Wtanf' mj su b/cosq 100.6 175.1 236.1 276.3 302.1 310.0 299.0 196.5 138.3 0.0 2034.0 1.72



112.8 84.5 84.0 85.0 87.4 90.9 98.5 124.5 144.2 0.0 911.9 0.77



1.0



Exercise 16.8 Use Janbu’s method to determine the factor of safety of the slope shown in Figure P16.8. Assume the soil above the groundwater level is saturated.



Solution 16.8 ' w



Soil 1 20 9.8



Soil 2 25



sat d l d/l FS



18.5 7.6 27.0 0.28 0.59



18



kN/m m



fo assumed



1.05



Slice



b



z1



z2



W=bz



zw



m



m



m



kN



m



3.82 2.8 14 5



3 6 3 0



0 2 4.6 2.5



212.0 411.6 1936.2 225.0



0 4.8 6.2 2.5



1 2 3 4



9.8



deg. 3 kN/m 3



ru







mj



Wtan



W(1 - ru)tan' cos mj



0.83 0.93 0.95 3.69



332.8 587.8 193.3 -225.0



44.1 54.2 372.2 97.4



Sum



888.9



567.9



FS



0.67



deg 0.00 0.32 0.44 0.54



57.5 55 5.7 -45



Exercise 16.9 Use Taylor’s method to determine the factor of safety of the slope shown in Figure P16.9.



Solution 16.9 From figure 16.2 D/H = 4/6 = 0.67 From Taylor’s chart: No = 6.2 FS = 6.2 x 20/ (18 x 6) = 1.15



Exercise 16.10 Use Taylor’s method to determine the slope in Figure P16.10 for FS = 1.25.



Solution 16.10 From figure 16.10. D/H = 6/8 = 0.75 No = 1.25 x 18 x 8/32 = 5.63 From Taylor’s chart: s = 42o



Exercise 16.11 Determine the factor of safety of a 2:1 slope with ru = 0.25, ’cs = 27° using the Bishop– Morgenstern method. Solution 16.11 Given:



Slope: 2:1 ru = 0.25, '  27



From chart (Fig 16.13) values of m and n for the Bishop-Morgenstern method: n = 1.25



FS  m  n.ru FS = 1 – 1.25 x 0.25 = 0.69



m = 1,



Exercise 16.12 The soil at a site is shown in Figure P16.12. A slope will be cut to facilitate the construction of a roadway. One possible failure surface is shown in Figure P16.12. Determine the factor of safety. The shear strength parameters were obtained from direct simple shear tests.



Solution 16.12 ' su



Soil 1 22 48



Soil 2 25 22



deg. kPa



w



9.8



9.8



kN/m



3



sat d l d/l FS



18 17.6 83.2 0.21 1.44



17



kN/m m



3



fo assumed



1.045



Slice



b



z1



z2



W=bz







mj



Wtan



Wtan' cos mj



1 2 3 4 5 6 7 8



m 5 10 6 4.8 19.2 19 8.5 5.5



m 0 5 8 8 7 5.5 3.5 0



m 8 11 12 11.5 7 4.5 6 4



kN 680.0 2770.0 2088.0 1629.6 4704.0 3334.5 1402.5 374.0



deg 20 39 35 35 5 5 -22 -51



0.95 1.05 1.02 1.02 0.98 0.98 1.22 2.65



247.5 2243.1 1462.0 1141.1 411.5 291.7 -566.6 -461.9



283.7 912.0 705.1 550.3 1855.0 1314.9 639.1 290.6



115.0 501.6 301.0 240.8 963.1 953.0 426.4 126.4



Sum



4768.5



6550.6



3627.2



FS



1.44



ESA



TSA fosub



0.76



Exercise 16.13 A cross section of a levee is shown in Figure P16.13a on the next page. (a) Describe how you would determine the stability of the levee. You must provide sufficient justification for the loading conditions you would consider. (b) Use a slope stability software program such as STB2006 to analyze the stability. Vane shear test data for the medium-to-soft clay is shown in Figure P16.13b. The average water content is 40%. (c) If the upstream slope is subjected to scouring and its gradient can change, research and describe methods you would use to protect it. The subscripts p and cs on the strength parameters in Figure P16.13a denote peak and critical state condition, respectively.



Solution 16.13



Page 1



STB2006 - Slope Stability Copyright © 2006 by A. Verruijt



Students of Geotechnics, New Problem General Data Licensed User : Students of Geotechnics File Name : 16-13.stb Problem Name : New Problem Number of Slices : 1000 Relative Horizontal Force: 0.000000 Number of Nodes : 25 Number of Soil Polygons : 5



Definition Figure 2



3



1



4 13



7



16



17



2



11



12



19



8



18 14 3



20



21



4



22



23



5



24



25



5



1



6



10



9



Input Data Coordinates of Nodes Node 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23



x 17.000 17.000 23.000 23.000 17.000 0.000 0.000 58.000 58.000 0.000 0.000 6.000 33.000 58.000 58.000 38.000 30.000 57.000 48.500 0.000 58.000 0.000 58.000



y 26.000 31.000 31.000 26.000 9.000 0.000 24.000 20.000 0.000 0.000 16.000 16.000 25.000 16.000 0.000 25.000 24.000 16.000 20.000 14.000 14.000 12.000 12.000



15



Page 2



STB2006 - Slope Stability Students of Geotechnics, New Problem 24 25



0.000 58.000



10.000 10.000



Structure of Polygons Window of centers of slip circles Nodes 1 2 3 Fixed point of all slip circles Node 5 Water Polygon Nodes 6 7 17 Soil Polygons 1: Nodes 10 24 25 2: Nodes 12 13 16 3: Nodes 20 11 12 4: Nodes 22 20 21 5: Nodes 24 22 23



4



19



8



9



15 18 18 23 25



14



21



Properties of Soils Soil



Wd kN/m³ 12.700 18.200 12.700 12.700 12.700



1 2 3 4 5



Ws kN/m³ 17.800 22.050 17.800 17.800 17.800



Ko -0.500 0.500 0.500 0.500 0.500



c kN/m² 20.000 40.000 85.000 40.000 0.000



phi degrees 30.000 28.000 30.000 30.000 30.000



P/F



p=0 m



P P P P P



cap m 0.000 0.000 0.000 0.000 0.000



----------------------------------------------------------------------------



Output Data Safety Factors 2.747 2.805 2.854 2.915 2.987 3.063 3.154 3.238 3.339 3.428 3.489



2.829 2.876 2.935 2.998 3.057 3.128 3.200 3.267 3.323 3.357 3.389



2.907 2.953 2.998 3.057 3.111 3.169 3.214 3.244 3.256 3.286 3.310



2.991 3.027 3.066 3.100 3.150 3.176 3.196 3.201 3.208 3.232 3.250



3.075 3.095 3.120 3.139 3.144 3.145 3.163 3.170 3.174 3.194 3.206



3.155 3.155 3.155 3.148 3.140 3.132 3.136 3.151 3.151 3.168 3.177



3.219 3.199 3.178 3.163 3.151 3.142 3.133 3.141 3.137 3.153 3.160



3.270 3.243 3.218 3.195 3.177 3.165 3.147 3.149 3.144 3.147 3.155



Critical slip circle Center: x =



17.000, y =



31.000, Radius:



19.729, F =



2.747



3.333 3.302 3.278 3.248 3.221 3.200 3.185 3.168 3.165 3.159 3.160



3.415 3.379 3.347 3.309 3.282 3.257 3.235 3.214 3.204 3.197 3.195



3.523 3.472 3.436 3.401 3.361 3.331 3.309 3.283 3.263 3.250 3.237