The problem of determining the parameters of displacement of solid rock" can be formulated as follows. In a heavy half-plane we have a cut AD (Fig. 1) which becomes filled under gravity with the material of the half-plane. We consider the density and strength characteristics of this material to be known, together with the distribution of the velocity vector v or the displacement vector u along the contour of the cut. From these data we are to determine the displacements and stresses throughout the solid rock affected by the displacement. We shall be satisfied with determinations of the mean displacements and stresses; i.e., we assume that the displacements and stresses are continuous and differentiable with respect ~o the coordinates and to time over the whole region, except perhaps for isolated lines and points. This assumption enables us to solve the problem of displacement using the equations of mechanics for a continuous medium-the equation of motion, the equation of strength (for which we use the equation expressing the Coulomb strength law), and the equation relating deformation velocity to stress (the equation of the associated-flow law).We will introduce some simplifying assumptions: we will consider only the plane case; we assume that the displacement process occurs so slowly that the influence of time can be neglected; the only bulk force is to be that of gravity; and we assume that the density is constant over the whole region.On these assumptions, the system of as follows equation of equilibrium 0,.~ +__ Ox equations representing the displacement process in the solid rock wiI1 be O'txY --O, O~xy + O:y --T;(1) OyOx dy Coulomb equation f = (~ -%)2 + 4,~ ,y --sin ,~ (% -q-% + 2k ctg .~)* = 0; equation for the associated-flow law [1, 2] ex=)" Og. ~y=), Og. ), Og O ~x O ~y ~'xy 0 ~xy (2) where k and ~ are the cohesion and the angle of internal friction of the rock, and g is the plastic potential. If, as usual [1, 2],we use a function ff for the plastic potential and bear in mind [3] the relation % / = ~ (1 _+ sin ~ cos 20) --H, %y = n sin ~ sin 20, (3) ~y l where o = (o x + Oy + 2H)/2, H = 2k cot q0, and 0 is the angle between o I and the x axis, then for the components of the deformation velocity vector we get All-Union Scientific-Research Mine Surveying Institute (VNIMI), Leningrad. Translated from Fiziko-Tekhnicheskie Problemy Razrabotki Poleznykh Iskopaemykh, No. 3, pp.

To solve the problem of plastic flow of ground we usually enlist the associated f r law [i], i.e., it is assumed that the plastic potential exactly coincides with the cot • tlon of plasticity. The associated flow law enables us to solve a wlde range of problems, but In certain cases the theoretical results conflict with the actual data. For example, from thls law it follows that any deformation of a material Is accompanied by Its expansion. Depending on its initial density, ground may either undergo expansion or compaction during deformation [2].To eliminate this discrepancy between theory and experiment, some investigators [3,4] postulate expressions for the plastic potential which differ from the plasticity conditions and contain the experimentally determined coefficient of dilatancy.In thls case, however, the characteristics for the stresses and rates of deformation do not coincide, and therefore neither do the regions of the solutions for the stresses and deformations.Other methods for taking account of the compressibility of ground have been suggested [5,6], but in these too the characteristics for the stresses and rates of deformation do not coincide.We have developed a somewhat modified version of the method proposed by Ivlev and Martynova [7] for taking account of the compressibility of ground. We have obtained equ ations representing plane deformation of compressible ground, and have examined equations for fracture; an example illustrating the qualitative side of deformation of such ground is given.I. It will be assume that the following plasticity condition holds true for the ground: where 6ij is the Kronecker delta, and ~j the rates of deformation deviator components.From Eq. (2) we haveEquations (i), (3), and (4) show that e ~ 0 only when Ell # 0. If we put o = o, = ~,-a,-0, then Ia = Is = O, whence Ell = 0 and therefore e = 0. Thus the compressibility determined by Eq. (3) depends markedly on the distortion, and material for which Eq. (2) holds true cannot be compressed by hydrostatic pressure. Thls is observed only for overdense or completely waterlogged (outflow precluded) ground.All-Unlon Sclentlflc-Research Mlne Surveying Institute, Leningrad.