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As a result of thefailure, the tension crack develops a series of little cracksalong joint sets 2 (Fig. 4).No groundwater seepage was detected through theexposed rock; therefore, it can be assumed that failureoccurred at zero cleft water pressure. Furthermore, theseismogram did not indicate any activity in the vicinity ofthe site during slope failure. The failure may be thereforeassumed to be the result of instability under static load dueto gravity, which developed along the bedding plane of theclay-filled in the limestone, and has nothing to do with thegroundwater and seismic activity.In summary, the slope face dip is 70 –80 (Figs. 3, 4),with dipping direction of 30 –45 . The sliding plane is nearthe bedding stiff intercalated clay layer in the limestoneand may be affected by the other joint set. The attitude ofthe clay is consistent with the limestone, white to gray,with thickness ranging between 0.1 and 0.3 m, in which themain mineral is montmorillonite, containing some calcium,with the unit weight, cd = 19.2 kN/m3(laboratory deter-mined). The natural water content was 8–12 %, with theliquid limit (LL) 75–84 %, plastic limit (PL) 27–32 %. Thepermeability of the clay is poor.It is obvious that this slope failure is along the existinggeological defects/discontinuities. It can be seen fromFig. 3, JS3 strikes nearly parallel to the slope face. The dipof JS3 is smaller than the dip of the slope face and veryclose to the friction angle, which is 308 empirically. Thetension cracks, which extend along the joint sets 2 andbecome the side boundary, separate the rock block from therock mass. Additionally, the JS1 and JS2, which providenegligible resistance to failure, provide release surfaces forfailure. Therefore, the above-mentioned failure of the slopewas planar type. It is obvious that the failure must be alongJS3.Methods and resultsBack analysis of the failed slopeThe limit equilibrium methods (LEM), which are routinelyadopted methods for stability calculation, are undertakento provide a factor of safety based on a comparison ofresisting forces of moments to disturbing forces ormoments. In the limit equilibrium back analysis procedure,the factor of safety is equal to 1, which introduced thethreshold of rupture. From the space geometry features ofthe exposed rock mass, the instable block includes inten-sely to moderately weathered limestone (the intenselyweathered, cd = 24.3 kN/m3; the moderately weathered, cd = 24.3 kN/m3); the vertical depth of the tension crackbehind the block is 24 m; it is assumed that there is noforce acted on the rock block along the crack (the tensionstrength of the crack is 0). The sliding plane length is 31 malong bedding weak layer plane, dipping 26 into theexcavation space; the sliding bed is the moderatelyweathered limestone.It is obvious that the failure can be determined as planartype.The factor of safety against single plane sliding in thegeneral case of a saturated slope is given by (e.g., Hoek andBray 1981):F ¼ cl þðW cos b U V sin bÞ tan uW sin b þ V cos bð1Þwhere c is cohesion across the sliding plane (kPa); l thelength of the sliding plane (m); W the overall weight of thesliding block (kN); U the resultant water force acting onthe sliding plane (kN); V the resultant water force acting onthe tension crack (kN); b the dip angle of the sliding plane( ); and / is the friction angle of the sliding plane ( ).In this case, sliding is assumed to have taken place underdry conditions and, therefore, U and V are both zero; thecalculated weight of the sliding block (W) is 14,250 kN bythe careful maps; and the average dip of the sliding plane is26 ; l is 31 m. Under the state of limit equilibrium, thefactor of stability is 1.0, the strength parameters (c and /)of the sliding plane depend on each other, and can bedescribed as: c ¼ lð2ÞTo back analysis the landslide, it is assumed that the /of the sliding plane varies between 10 and 26 , thegraphical solution for this relationship is shown in Fig. 5.From cross section II–II plotted by site measure andgeological map, the length of the sliding plane is 23.6 m,the height of the instable block is 32 m, the weight of theblock is 10,133 kN.