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Tham khảo tài liệu 'strength analysis in geomechanics part 4', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 48 3 Some Elastic Solutions The latter equation is derived with the help of relation for vector components transformation 2.55 and sign means a derivative by z. We can check expressions above on the example of Fig. 3.1 for which the solution can be given in the following way w z Q 2n lnz Guo 3.6 The convenience of the complex variables usage consists in the opportunity of conformal transformations application when solutions for simple figures a semi-plane or a circle can be transformed to compound sections 16 . 3.1.2 Longitudinal Displacement of Strip To derive the solution of this problem by the conformal transformation of result 3.6 to the straight line l l Fig. 3.2 with the help of the Zhoukovski s relation which was deduced for an ellipse and here is used for our case as z 0.5 z z-1 z 7z2 l2 l 3.7 3.1 Longitudinal Shear 49 we put the second of these expressions into 3.6 and using 3.5 we receive w z Qln z l ỵ z l 2 1 n Guo Tị iTn Q n ựz I2 w z . 3.8 Along axis ị we compute as follows At ị l uz uo Tị 0 Tn Q rcự12 ị2 At ị l uz uo Q Gn ln ị l ự ị l 2 1 Tn 0 Tị Q nựị2 l2. 3.9 In the same manner the displacement and stresses in any point of the massif can be found. The most dangerous points are n 0 ị l and in order to investigate the fracture process there it is convenient to use decomposition z l reie which gives according to expressions 3.5 3.8 and 2.52 uz uo Q Gn ự2r ĩcos 0 2 Tg Q nV2rlx f 2 Te Q nVÕÃ. 3.10 From the third relation 3.10 we can see that the condition Te constant gives a circumference with a centre at ị l where a fracture or plastic strains should begin. 3.1.3 Deformation of Massif with Circular Hole of Unit Radius In this case Fig. 3.3 the boundary conditions are Tnlz Tz Tp p i 0. 3.11 We seek the solution in a form Fig. 3.3. Massif with circular hole 50 3 Some Elastic Solutions rc w z V A z 3.12 n 0 and the first condition 3.11 gives immediately A-1 iTz. Now with the help of 3.5 we rewrite the second 3.11 as Re ei0w eie 0 from that we have A1 iTz .
