1. Isoparametric elements and solution techniques

2. Advanced Design for Mechanical System - Lec 2008/10/092

3. 3  = 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

4. Advanced Design for Mechanical System - Lec 2008/10/094 R=KD gauss elimination computation time: (n order of K, b bandwith)

5. Advanced Design for Mechanical System - Lec 2008/10/095 recall: gauss elimination

6. Advanced Design for Mechanical System - Lec 2008/10/096 rotations

7. Advanced Design for Mechanical System - Lec 2008/10/097

8. 8 isoparametric elements isoparametric: same shape functions for both displacements and coordinates

9. Advanced Design for Mechanical System - Lec 2008/10/099 computation of B  x = du / dX but u=u( ,  ), v=v ( ,  )

10. Advanced Design for Mechanical System - Lec 2008/10/0910

11. Advanced Design for Mechanical System - Lec 2008/10/0911 J 11 * and J 12 * are coefficients of the first row of J -1 and

12. Advanced Design for Mechanical System - Lec 2008/10/0912

13. Advanced Design for Mechanical System - Lec 2008/10/0913 gauss quadrature

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16. Advanced Design for Mechanical System - Lec 2008/10/0916

17. Advanced Design for Mechanical System - Lec 2008/10/0917

18. Advanced Design for Mechanical System - Lec 2008/10/0918 why stiffer ? In FEM we do not have distributed loads. See beam example:

19. Advanced Design for Mechanical System - Lec 2008/10/0919

20. Advanced Design for Mechanical System - Lec 2008/10/0920 no strain at the Gauss points so no associated strain energy

21. Advanced Design for Mechanical System - Lec 2008/10/0921 The FE would have no resistance to loads that would activate these modes Global K singular Usually such modes superposed to ‘right’ modes

22. Advanced Design for Mechanical System - Lec 2008/10/0922

23. Advanced Design for Mechanical System - Lec 2008/10/0923 calculated stress  =EBd are accurate at Gauss points

24. Advanced Design for Mechanical System - Lec 2008/10/0924 the locations of greatest accuracy are the same Gauss points that were used for integration of the stiffness matrix

25. Advanced Design for Mechanical System - Lec 2008/10/0925

26. Advanced Design for Mechanical System - Lec 2008/10/0926 Rayleigh-Ritz method Guess a displacement set that is compatible and satisfies the boundary conditions

27. Advanced Design for Mechanical System - Lec 2008/10/0927 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

28. Advanced Design for Mechanical System - Lec 2008/10/0928  =  (d) d  / d d 1 = 0 d  / d d 2 = 0 d  / d d 3 = 0 d  / d d 4 = 0 …… d  / d d n = 0

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30. Advanced Design for Mechanical System - Lec 2008/10/0930

31. Advanced Design for Mechanical System - Lec 2008/10/0931 patch tests only for those who develops FE

32. Advanced Design for Mechanical System - Lec 2008/10/0932 substructures

33. Advanced Design for Mechanical System - Lec 2008/10/0933 divide the FEmodel in more parts create a FE model of each substructure Assemble the reduced equations KD=R Solve equations

34. Advanced Design for Mechanical System - Lec 2008/10/0934 Simmetry

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36. Advanced Design for Mechanical System - Lec 2008/10/0936

37. Advanced Design for Mechanical System - Lec 2008/10/0937 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

38. Advanced Design for Mechanical System - Lec 2008/10/0938 way 1 – Lagrange multipliers =[ 1 2 …. m ] T T [CD-Q]=0  = 1/2D T KD – D T R + T [CD-Q]

39. Advanced Design for Mechanical System - Lec 2008/10/0939 remember dAD / dD = A T dD T A/ dD = A

40. Advanced Design for Mechanical System - Lec 2008/10/0940 example

41. Advanced Design for Mechanical System - Lec 2008/10/0941

42. Advanced Design for Mechanical System - Lec 2008/10/0942 way 2- penalty method

43. Advanced Design for Mechanical System - Lec 2008/10/0943 ½t T  t = ½ [(CD-Q) T  (CD-Q)]= = ½ [(CD-Q) T (  CD-  Q)]= = ½ [(CD-Q) T  CD- (CD-Q) T  Q)]= = ½ [(D T C T  CD-Q T  CD-D T C T  Q +Q T  Q)]= ½[·]; d(½[·])/dD= =½[2(C T  C)-(Q T  C) T - C T  Q]= =½[2(C T  C)-(  C) T Q- C T  Q]= =½[2(C T  C)-C T  Q- C T  Q]=

44. Advanced Design for Mechanical System - Lec 2008/10/0944 =½[2(C T  C)-C T  Q- C T  Q]= = C T  C-C T  Q (  =  T )

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