1. ECIV 720 A Advanced Structural Mechanics and Analysis Lecture 12: Isoparametric CST Area Coordinates Shape Functions Strain-Displacement Matrix Rayleigh-Ritz Formulation Galerkin Formulation

2. FEM Solution: Area Triangulation Area is Discretized into Triangular Shapes

3. FEM Solution: Area Triangulation One Source of Approximation

4. FEM Solution: Nodes and Elements Points where corners of triangles meet Define NODES

5. Non-Acceptable Triangulation … Nodes should be defined on corners of ALL adjacent triangles

6. FEM Solution: Nodes and Elements Each node translates in X and Y uiui vivi X Y

7. FEM Solution: Objective Use Finite Elements to Compute Approximate Solution At Nodes 1 (x 1,y 1 ) 2 (x 2,y 2 ) 3 (x 3,y 3 ) q6q6 q5q5 q4q4 q3q3 q1q1 q2q2 v u Interpolate u and v at any point from Nodal values q 1,q 2,…q 6

8. Intrinsic Coordinate System 3 (x 3,y 3 ) 2 (x 2,y 2 ) 1 (x 1,y 1 ) 3 (0,0)1 (1,0) 2 (0,1)   Map Element Define Transformation Parent  

9. Area Coordinates A1A1 1 2 3 P X Y Location of P can be defined uniquely

10. Area Coordinates and Shape Functions 3 (x 3,y 3 ) 2 (x 2,y 2 ) 1 (x 1,y 1 )   Area Coordinates are linear functions of  and  Are equal to 1 at nodes they correspond to Are equal to 0 at all other nodes 3 (0,0)1 (1,0) 2 (0,1)   Natural Choice for Shape Functions

11. Shape Functions X Y 1 (x 1,y 1 ) 2 (x 2,y 2 ) 3 (x 3,y 3 )  

12. Geometry from Nodal Values

13. Intrinsic Coordinate System Map Element Transformation 3 (0,0)1 (1,0) 2 (0,1)   Parent 3 (x 3,y 3 ) 2 (x 2,y 2 ) 1 (x 1,y 1 )  

14. Displacement Field from Nodal Values 1 2 3 q6q6 q5q5 q4q4 q3q3 q2q2 q1q1 v u  

15. Strain Tensor from Nodal Values of Displacements Strain Tensor Need Derivatives u and v functions of  and 

16. Jacobian of Transformation J J

17. Jacobian of Transformation – Physical Significance

18. 3 (x 3,y 3 ) 2 (x 2,y 2 ) 1 (x 1,y 1 )   r1r1 r2r2

19. Jacobian of Transformation – Physical Significance 3 (x 3,y 3 ) 2 (x 2,y 2 ) 1 (x 1,y 1 )   r1r1 r2r2 k Compare to Jacobian

20. Jacobian of Transformation Solve

21. Strain Tensor from Nodal Values of Displacements

23.  = B q B q Looks Familiar?

24. Strain-Displacement Matrix Is constant within each element - CST 1 (x 1,y 1 ) 2 (x 2,y 2 ) 3 (x 3,y 3 )  

25. Stresses  = B q

26. Element Stiffness Matrix k e  = B q  = D B q e keke 1 (x 1,y 1 ) 2 (x 2,y 2 ) 3 (x 3,y 3 )  

27. x y z P Formulation of Stiffness Equations T (force/area) Tt (force/length) P Assume Plane Stress x y t

28. Total Potential Approach P Tt (force/length) Total Potential

29. Total Potential Approach P Tt (force/length)

30. Total Potential Approach

31. Work Potential of Body Forces

32. WP of Body Forces fyfy fxfx u v Element e 3 1 2

33. WP of Body Forces 1 2 3 q6q6 q5q5 q4q4 q3q3 q2q2 q1q1 v u  

34. 1 2 3 q6q6 q5q5 q4q4 q3q3 q2q2 q1q1 v u  

35. Nodal Equivalent Body Force Vector

36. Total Potential Approach Work Potential of Tractions

37. WP of Traction Tt (force/length) Distributed Load acting on EDGE of element 3 1 2

38. WP of Traction Components T x1,T y1 T x2,T y2 Known Distribution Normal Pressure p 1, p 2 Known Distribution

39. WP of Traction Normal Pressure p 1, p 2 Known Distribution T x1 T y1 T x2 T y2 Directional cos Components

40. WP of Traction 3 1 2 TxTx TyTy u v

41. T x1 T y1 T x2 T y2 WP of Traction 3 1 2

42. Nodal Equivalent Traction Vector

43. Total Potential Approach Work Potential of Concentrated Loads

44. WP of Concentrated Loads P Indicates that at location of point loads a node must be defined

45. In Summary

46. After Superposition Minimizing wrt Q

47. Galerkin Approach P Tt (force/length) Galerkin

48. Galerkin Approach P Tt (force/length)

49. Galerkin Approach Introduce Virtual Displacement Field  1 2 3 66 55 44 33 22 11 yy xx  

50. Galerkin Approach Virtual Strain Energy of element e

51. Element Stiffness Matrix k e keke 1 (x 1,y 1 ) 2 (x 2,y 2 ) 3 (x 3,y 3 )    = B  e  = D B q e

52. Galerkin Approach Virtual Work Potential of Body Forces

53.  WP of Body Forces fyfy fxfx xx yy Element e 3 1 2 As we have already seen

54.  WP of Body Forces Nodal Equivalent Body Force Vector

55. Galerkin Approach Virtual Work Potential of Traction

56.  WP of Traction 3 1 2 TxTx TyTy xx yy

57. Nodal Equivalent Traction Vector

58. Galerkin Approach Virtual Work Potential of Point Loads

59.  WP of Concentrated Loads P Indicates that at location of point loads a node must be defined

60. In Summary

61. After Superposition  is arbitrary and thus

62. Stress Calculations  e  = B e q e 1 2 3 q6q6 q5q5 q4q4 q3q3 q2q2 q1q1 v u   For Each Element BC

63. Stress Calculations 1 2 3 q6q6 q5q5 q4q4 q3q3 q2q2 q1q1 v u    e  = B e q e Constant

64. Summary of Procedure Tt (force/length) Nodes should be placed at Discretize domain in CST - start & end of distributed loads - point loads

65. Summary of Procedure For Every Element Compute Strain-Displacement Matrix B 1 2 3 q6q6 q5q5 q4q4 q3q3 q2q2 q1q1 v u  

66. Summary of Procedure Element Stiffness Matrix Node Equivalent Body Force Vector

67. Summary of Procedure Node Equivalent Traction Vector For ALL loaded sides

68. Summary of Procedure Collect ALL Point Loads in Nodal Load Vector 1 N P x1 P y1 P xN P yN

69. Summary of Procedure Form Stiffness Equations

70. Summary of Procedure Apply Boundary Conditions Solve For Every Element Compute Stress

71. Example (0,0)(3,0) (3,2) (0,2) Tt=200 lb/in f x =0 f y =60 lb/in 2

72. ANSYS Solution – Coarse Mesh

80. 2-D Constant Stress Triangle First Element for Stress Analysis Does not work very well When in Bending under-predicts displacements –Slow convergence for fine mesh For in plane strain conditions – Mesh “Locks” –No Deformations Comments

81. Element Defects

82. Constant Stress Triangles Exact Y-Deflection & X-Stress about ¼ of actual

83. Element Defects x 1 =0, y 1 =0 x 2 =a, y 2 =0 x 3 =0, y 3 =a ? Spurious Shear Strain Absorbs Energy – Larger Force Required

84. Element Defects Mesh Lock Rubber Like Material ~0.5