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The graphical solution for c and /relationship is shown in Fig. 5. Because the sizes andshapes of the sliding blocks in the two cross sections areapproximative, the continuous curves is very close, thoughthey have a intersection point, the deviation of the valuesfrom the point is easily affected by the measure and plot.For this problem, to obtain the parameters of the slidingplane, the necessary tests should be carried out.It is shown that the maximum value of back calculatedcohesion is not greater than 132 kPa. According to the siteinvestigation, the clay is stiff; from Fig. 5,if / = 10 (relatively low value), c must be 132 kPa, which indicatesthat sliding plane is in the clay. However, if / = 26 ,c must be 0. For the clay, the latter of the above solution isnot in reality, which means the sliding is not in the clay butin the contact between the stiff clay and the overlying orunderlying limestone.Direct shear tests on the clayTo reduce the disturbance to the samples, the samples oflaboratory direct shear tests were retrieved in the field. Thesamples are pided into six groups, each group includesfour samples. The field values of normal and shear stressesacted on the clay are calculated to be rn = 510 kPa,s = 243 kPa. Considering the natural stress field, fourcylindrical specimens, with height of 2 cm, cross-sectionalarea of 30 cm2were saturated. After saturation, the sam-ples were consolidated under normal stresses of 100, 200, 400, and 600 kPa. Following consolidation, the sampleswere sheared under drained condition, and at the horizontaldisplacement rate of 0.014 mm/min. The above mentionedlaboratory tests for the clay are repeated six times.The typical stress–strain curves are shown in Fig. 6.Itisassumed the clay is a Mohr–Coulomb material. When thefour tests yield, the normal and shear stress on the shearplane is as plotted in Fig. 7. From the resulting best linearfit, the effective cohesion and friction angle are 207.6 kPaand 17 , respectively (R2= 0.993). It is assumed that thesliding plane is in the clay; the factor of stability calculatedby formula (1) is 1.66 according to the above strengthparameters, whose result means the slope must be stable.Considering under high normal stress, such as400–600 kPa, which is near the natural normal stress,c = 230 kPa (Fig. 7), which is close to the natural shearstress on the shear plane, / = 15.4 ; the factor of stabilityagainst the sliding is 1.70, the slope must be stable too. Thetest results from the six groups indicate the values of c arebetween 192 and 246 kPa, and the values of / are between12 and 17 . The calculated values of F should be between1.50 and 2.20.The calculated results indicate that sliding of the failurecannot shear through the clay, which may be on the con-tacts between the clay and the limestone.In situ direct shear tests on the clay–limestone contactFor the rock, it is difficult to retrieve the undisturbed samplesin the field.Even if the samples can be retrieved in the field, thesampleswould be disturbed inevitably, the accuracy of the testresults will reduce, without any doubt. To reduce the distur-bance to the rockmass, the sampleswere artificially producedin situ, and the samples are square with the shear area of50 cm 9 50 cm (Hudson 1974–2006), the height above the shear plane is 40 cm, and the shear planes are set in the weakintercalated layer. To prevent the rock above the contact frombeing damaged or great deformation, the upper rock waspacked with cement mortar, thus, being made into a regularshape (Fig. 8). The top surfaces of the sample are all flat, andparalleling with the intercalated layer. The four sides are flatand vertical to preset shear surfaces. In test processes, theshear stresses and normal stresses acted on the shear surfacesare loaded by hydraulic jacks. At the four corners, the 6 mlong resin anchorswere preset before tests as the anti-power ofnormal loading. The four dial gauge are preset (two formonitoring the vertical displacements, the other two formonitoring shear displacements), the arithmetic mean valuesare calculated by the monitored results of both groups. Thesteel boards were laid between the samples and the jacks,which can guarantee that the stress on the contact is uniform.There are two steel boards under the normal jack; the rollerswere laid between the two boards for adapting to the sheardeformation.Before starting normal loading, the steel board, rollers, theother steel board, hydraulic jack, columns for transferringload and the anti-power equipmentswere laid in turn.Aswellas the shear loading, the similar components were laid inturn.