By R.D. Richtmyer
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15~)since does not appear in the other equations. 16b). In fact, since 4 2 is prescribed to be 2, all we need do is substitute $2 = 2 directly into the remaining equations. 16b) in this case. 17e). 17~). 18d). 18e) by adding We now have a set of equations which can be solved directly if we take them in reverse order. 19e) we have R5 = -84/13, since $5 = 0. 19d) gives $4 = 21/13. 19~)gives 43 = 49/13. 19b) gives R, = -46113 immediately. 19a) since 42, $3, $4 are known at this stage. This gives = 179/13.
From equations i+ 1 to n. Formally this can be done by subtracting from the rth equation (i < r < n), the ith equation factored by kri("/kii(", where the 46 FINITE ELEMENTS IN PLASTICITY superscript i indicates that these coefficients have been already modified (i- 1) times prior to the elimination of the ith degree of freedom. For example, the first equation is used to eliminate & from equations 2 to n as follows : Then the second equation is used to eliminate 4 2 from equations 3 to n and so on.
2 Structural example for illustration of equation solution process. 38). , in which E(I), A(I) and L(I) are respectively the elastic modulus, crosssectional area and length of element I. The vector of applied nodal forces for each element is The vectors of the unknown nodal displacements for the elements are We also assume the following prescribed displacement values 42 = d,,. 45 = 0. 5) 44 FINITE ELEMENTS IN PLASTICITY The Theorem of Minimum Total Potential Energy will now be used to derive the stiffness equations for this problem.
4th Int'l Conference on Numerical Methods in Fluid Dynamics by R.D. Richtmyer