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Men 5 F~g. 7 - Stress distribution along the boundary for an internal gear supported at the ends. Flg. 9-Stress distributionalong the boundary for an. external ge
the geom.etry factor calculationusing boundary elements .. • Enhance the program to take care of the roller boundary conditiens and to allow models to incorporate a number of t,eeth greater than three. Appendix Boundary Element Method Equations Boundary element methods use approximating functions. that satisfy the governing ,elasticity equations wi.tmn the outer periphery ,of the body being analyzed, but not the boundary ,conditions. These functions are called the fundament.al solutions. figs. 12 &; 13 showa typicatl body which is to be analyzed. The ,elasticity equations which govern the anal:ysis have been presented here in tensor notation for the sake of conciseness. for every point in the body of interest to be ina state of equilibrium, the stress field must satisfy the following equilibrium equation: equilibrium: aij.1 - 0 in 'D Here iand j vary from 1 to 2 for a two-dimensional analysis and 0 is the interior of the body. The surface loads or the tractions applied to the body are related to the stresses as These tractions are prescribed on part of the body's surface I'2' On the rest of the surface of the body, the displacements U.i have to be prescribed. Uijis the stress tensor field and 'the u, is the displacement vector field. The starting point for both the boundary element method and the finite element method is an integrated form of the equilibrium 'eq-uation: for all ut': J Uj"'uiHdO - 0 D ' where uj is an a.rbitra:ry wei,ghtin-8 function. The choice of this weighting .function is one of the differ,ences between the boundary and ~fin]teelement method. In the finite element method, these functions are chosen to be the same as the shape functions,. whereas, in the bound~ry element method, the fundamental solutions are used. Integrating by parts. (in~egration by parts in. two and ,three dimensions is carried eut using Green's identity) we get: But it can be shown lI u~ I II ..n·dI' I - J uj*ju..dO - .-1]--- - 0 r Q that Therefore, the integral becom s: J utlrijnj,dr r f u~fjidO - 0 Q or, Iut pldr - J ,uljflj
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Page 9 and 10: The Use of Boundary Elements FOlr T
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Page 35: THE USE OF BOUND.ARY ELEMENTS ...
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