1. ECIV 720 A Advanced Structural Mechanics and Analysis
Lecture 15:
Quadrilateral Isoparametric Elements (cont’d)
Force Vectors
Modeling Issues
Higher Order Elements

2. Integration of Stiffness Matrix
BT(8x3)
D (3x3)
ke (8x8)
B (3x8)

3. Integration of Stiffness Matrix
Each term kij in ke is expressed as
Linear Shape Functions is each Direction
Gaussian Quadrature is accurate if
We use 2 Points in each direction

4. Integration of Stiffness Matrixh

5. Choices in Numerical Integration
Numerical Integration cannot produce exact results
Accuracy of Integration is increased by using more integration points.
Accuracy of computed FE solution DOES NOT necessarily increase by using more integration points.

6. FULL Integration
A quadrature rule of sufficient accuracy to exactly integrate all stiffness coefficients kij
e.g. 2-point Gauss rule
exact for polynomials
up to 2nd order

7. Reduced Integration, Underintegration
Use of an integration rule of less than full order
Advantages
Reduced Computation Times
May improve accuracy of FE results
Stabilization
Disadvantages
Spurious Modes
(No resistance to nodal loads that tend to activate the mode)

8. Spurious Modes
Consider the 4-node plane stress element 1
t=1
E=1
v=0.3 1
8 degrees of freedom
8 modes
Solve Eigenproblem

9. Spurious Modes
Rigid Body Mode
Rigid Body Mode

10. Spurious Modes
Rigid Body Mode

11. Spurious Modes
Flexural Mode
Flexural Mode

12. Spurious Modes
Shear Mode

13. Uniform Extension Mode
Spurious Modes
Stretching Mode
Uniform Extension Mode
(breathing)

14. Element Body Forces
Total Potential
Galerkin

15. Body Forces
Integral of the form

16. Body Forces
In both approaches
Linear Shape Functions
Use same quadrature as stiffness maitrx

17. Element Traction
Total Potential
Galerkin

18. Element Traction
Similarly to triangles, traction is applied along sides of element 3 u v Tx Ty h 4 4 x 2 1

19. Traction
For constant traction along side 2-3
Traction
components
along 2-3

20. More Accurate at Integration points Stresses h x
Stresses are calculated at any x,h

21. Modeling Issues: Nodal Forces
In view of…
A node should be
placed at the location
of nodal forces
Or virtual potential energy

22. Modeling Issues: Element Shape
Square : Optimum Shape
Not always possible to use
Rectangles:
Rule of Thumb
Ratio of sides <2
Larger ratios
may be used
with caution
Angular Distortion
Internal Angle < 180o

23. Modeling Issues: Degenerate Quadrilaterals
Coincident Corner Nodes 1 2 3 4 2 1 4 x x
Integration Bias 3
Less accurate

24. Modeling Issues: Degenerate Quadrilaterals
Three nodes collinear 1 2 3 4
Integration Bias 1 2 3 4 x x
Less accurate

25. Modeling Issues: Degenerate Quadrilaterals
2 nodes
Use only as necessary to improve representation of geometry
Do not use in place of triangular elements

26. All interior angles < 180
A NoNo Situation y 3
(7,9)
Parent x h
(6,4) 4 1 2
J singular
(3,2)
(9,2) x
All interior angles < 180

27. Another NoNo Situation
x, y
not uniquely
defined x

28. FEM at a glance
It should be clear by now that the cornerstone in FEM procedures is the interpolation of the displacement field from discrete values
Where m is the number of nodes that define the interpolation and the finite element and N is a set of Shape Functions

29. FEM at a glance
m=2 x
x1=-1
x2=1
x1=-1 1 x 3
x2=1 2
m=3

30. FEM at a glance 1 2 3 q6 q5 q4 q3 q2 q1 v u x h
m=3 x h 4 1 2 3
m=4

31. FEM at a glance
In order to derive the shape functions it was assumed that the displacement field is a polynomial of any degree, for all cases considered
1-D
2-D
Coefficients ai represent generalized coordinates

32. FEM at a glance
For the assumed displacement field to be admissible we must enforce as many boundary conditions as the number of polynomial coefficients
x1=-1 1 x 3
x2=1 2
e.g.

33. FEM at a glance
This yields a system of as many equations as the number of generalized displacements
that can be solved for ai

34. FEM at a glance
Substituting ai in the assumed displacement field
and rearranging terms…

35. FEM at a glance 1 3 2 u(x)=a0+a1 x +a2 x 2 u(-1)=a0 -a1 +a2 =u1
u(0)=a0 =u3 …

36. Let’s go through the exercise1 x2 2
Assume an incomplete form of quadratic variation

37. Incomplete form of quadratic variationx1 1 x2 2
We must satisfy

38. Incomplete form of quadratic variation
And thus,

39. Incomplete form of quadratic variation
And substituting in

40. Incomplete form of quadratic variation
Which can be cast in matrix form as

41. Isoparametric Formulation
The shape functions derived for the interpolation of the displacement field are used to interpolate geometry x1 1 x2 2

42. Intrinsic Coordinate Systems
Intrinsic coordinate systems are introduced to eliminate dependency of Shape functions from geometry
1 (-1,-1)
2 (1,-1)
4 (-1,1)
3 (1,1) x h
The price?
Jacobian of transformation
Great Advantage for the money!

43. Field Variables in Discrete Form
Geometry
Displacement
e = B un
Strain Tensor
s = DB un
Stress Tensor

44. FEM at a glance
Element Strain Energy
Work Potential of Body Force
Work Potential of Surface Traction
etc

45. Higher Order Elements
Quadrilateral Elements
Recall the 4-node
Complete Polynomial
4 Boundary Conditions for admissible displacements
4 generalized displacements ai

46. Assume Complete Quadratic Polynomial
Higher Order Elements
Quadrilateral Elements
Assume Complete Quadratic Polynomial
9 generalized displacements ai
9 BC for admissible displacements

47. BT18x3 D3x3 B3x18 ke 18x18 9-node quadrilateral
9-nodes x 2dof/node = 18 dof
BT18x3
D3x3
B3x18
ke 18x18

48. 9-node element Shape Functions
Following the standard procedure the shape functions are derived as 1 2 3 4
Corner Nodes x h 5 6 7 8
Mid-Side Nodes 9
Middle Node

49. 9-node element – Shape Functions
Can also be derived from the 3-node axial element
x1=-1 1 x
x2=1 3 2

50. Construction of Lagrange Shape Functionsx
(1,h)
(1,1)
1 (-1,-1)

51. N1,2,3,4 Graphical Representation

52. N5,6,7,8 Graphical Representation

53. N9 Graphical Representation

54. Polynomials & the Pascal Triangle
Degree 1 x y 1 2 x2 xy y2 3 x3
x2y
xy2 y3 4 x4
x3y
x2y2
xy3 y4 5 x5
x4y
x3y2
x2y3
xy4 y5
…….

55. Polynomials & the Pascal Triangle
To construct a complete polynomial Q1
4-node Quad 1 x y x2 xy y2 x3
x2y
xy2 y3 x4
x3y
x2y2
xy3 y4
……. x5
x4y
x3y2
x2y3
xy4 y5 Q2
9-node Quad
etc

56. Incomplete Polynomials1 x y x2 xy y2 x3
x2y
xy2 y3 x4
x3y
x2y2
xy3 y4
……. x5
x4y
x3y2
x2y3
xy4 y5
3-node triangular

57. Incomplete Polynomials1 x y x2 xy y2 x3
x2y
xy2 y3 x4
x3y
x2y2
xy3 y4
……. x5
x4y
x3y2
x2y3
xy4 y5

58. 8 coefficients to determine for admissible displ.
8-node quadrilateral
Assume interpolation 1 2 3 4 x h 5 6 7 8
8 coefficients to determine for admissible displ.

59. BT16x3 D3x3 B3x16 ke 16x16 8-node quadrilateral
8-nodes x 2dof/node = 16 dof
BT16x3
D3x3
B3x16
ke 16x16

60. 8-node element Shape Functions
Following the standard procedure the shape functions are derived as 1 2 3 4
Corner Nodes h 5 6 7 8
Mid-Side Nodes x

61. N1,2,3,4 Graphical Representation

62. N5,6,7,8 Graphical Representation

63. Incomplete Polynomials1 x y x2 xy y2 x3
x2y
xy2 y3 x4
x3y
x2y2
xy3 y4
……. x5
x4y
x3y2
x2y3
xy4 y5

64. 6 coefficients to determine for admissible displ.
6-node Triangular
Assume interpolation 1 2 3 4 5 6
6 coefficients to determine for admissible displ.

65. BT12x3 D3x3 B3x12 ke 12x12 6-node triangular
6-nodes x 2dof/node = 12 dof 1 2 3 4 5 6
BT12x3
D3x3
B3x12
ke 12x12

66. 6-node element Shape Functions
Following the standard procedure the shape functions are derived as
Corner Nodes 1 2 3
Mid-Side Nodes 4 5 6
Li:Area coordinates

67. Other Higher Order Elements1 x y x2 xy y2 x3
x2y
xy2 y3 x4
x3y
x2y2
xy3 y4
……. x5
x4y
x3y2
x2y3
xy4 y5
12-node quad x h 1 2 3 4

68. Other Higher Order Elements1 x y x2 xy y2 x3
x2y
xy2 y3 x4
x3y
x2y2
xy3 y4
…….
x3y2
16-node quad x h 1 2 3 4 x5
x4y
x3y2
x2y3
xy4 y5