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Isoparametric elements and solution techniques

Isoparametric elements and solution techniques.  = ½ d 1-2 T k 1-2 d 1-2 +. + ½ d 2-4 T k 2-4 d 2-4 +….= = ½ D T KD. R=KD. gauss elimination computation time: (n order of K, b bandwith). recall: gauss elimination. rotations. isoparametric elements.

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Isoparametric elements and solution techniques

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  1. Isoparametric elements and solution techniques

  2.  = ½ d1-2Tk1-2d1-2 + + ½ d2-4Tk2-4d2-4 +….= = ½ DTKD

  3. R=KD • gauss elimination • computation time: (n order of K, b bandwith)

  4. recall: gauss elimination

  5. rotations

  6. isoparametric elements isoparametric: same shape functions for both displacements and coordinates

  7. computation of B • x = du / dX • but u=u(,  ), v=v (,  )

  8. J11* and J12* are coefficients of the first row of J-1 and

  9. gauss quadrature

  10. no strain at the Gauss points so no associated strain energy

  11. The FE would have no resistance to loads that would activate these modes Global K singular Usually such modes superposed to ‘right’ modes

  12. calculated stress =EBd are accurate at Gauss points

  13. the locations of greatest accuracy are the same Gauss points that were used for integration of the stiffness matrix

  14. Rayleigh-Ritz method Guess a displacement set that is compatible and satisfies the boundary conditions

  15. define the strain energy as function of displacement set • define the work done by external loads • write the total energy as function of the displacement set • minimize the total energy as function of the displacement and find • simulataneous equations that are solved to find displacements

  16. = (d) d  / d d1 = 0 d  / d d2 = 0 d  / d d3 = 0 d  / d d4 = 0 …… d  / d dn = 0

  17. patch tests • only for those who develops FE

  18. substructures

  19. divide the FEmodel in more parts • create a FE model of each substructure • Assemble the reduced equations KD=R • Solve equations

  20. Simmetry

  21. Constraints CD – Q =0 C is a mxn matrix m is the number of constraints n is the number of d.o.f. How to impose constraints on KD=R

  22. way 1 – Lagrange multipliers =[1 2 …. m]T T [CD-Q]=0  = 1/2DTKD – DTR + T [CD-Q]

  23. remember dAD / dD = AT dDTA/ dD = A

  24. example

  25. way 2- penalty method

  26. ½tT t = ½ [(CD-Q)T(CD-Q)]= = ½ [(CD-Q)T(CD-  Q)]= = ½ [(CD-Q)TCD- (CD-Q)T Q)]= = ½ [(DTCTCD-QTCD-DTCTQ +QTQ)]= ½[·]; d(½[·])/dD= =½[2(CTC)-(QTC)T- CTQ]= =½[2(CTC)-(C)TQ- CTQ]= =½[2(CTC)-CT  Q- CTQ]=

  27. =½[2(CTC)-CT  Q- CTQ]= = CTC-CT Q (= T)

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