1. ECIV 720 A Advanced Structural Mechanics and Analysis
Lecture 13 & 14:
Quadrilateral Isoparametric Elements
Stiffness Matrix
Numerical Integration
Force Vectors
Modeling Issues

2. Two Dimensional – Plane Stress
Thin Planar Bodies subjected to in plane loading

3. Stress and Strain

4. Stress Strain Relationship

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

6. Constant Strain Triangle
For Every Triangular Element 1 2 3 q6 q5 q4 q3 q2 q1 v u x h
Strain-Displacement Matrix B
Constant strain

7. FEM Solution: Quadrilateral Mesh
Area is Discretized into Quadrilateral Shapes

8. FEM Solution: Quadrilateral Elements
4 (x4,y4)
3 (x3,y3) v u
P (x,y) q1 q2 X Y q4 q3
1 (x1,y1)
2 (x2,y2)
8 Degrees of Freedom

9. FEM Solution: Objectiveq8 q7 q4 q3 q1 q2 v u q6 q5
Use Finite Elements to Compute Approximate Solution At Nodes
Interpolate u and v at any point from Nodal values q1,q2,…q8

10. Intrinsic Coordinate System Shape Functions of 4-node quadrilateral
To this end…
Intrinsic Coordinate System
Shape Functions of 4-node quadrilateral
From 1-D
Direct
Jacobian of Transformation
Strain-Displacement Matrix
Stiffness Matrix

11. Intrinsic Coordinate System
Recall 1-D x1 x x2
Map Element
Define Transformation x
x1=-1
x2=1

12. Intrinsic Coordinate Systemx h 4 1 2 3 x h
4 (-1,1)
3 (1,1)
Map Element
Define Transformation
1 (-1,-1)
2 (1,-1)
Parent

13. Lagrange Shape Functions
For 1-D
N1(x)
N1=1 x
x1=-1
x2=1

14. 2-D Lagrange Shape Functions
What is the lowest order
polynomial
f1(x,-1) along side h=-1
that satisfies
f1(-1,-1) =1 & f1(1,-1)=0? h x
(1,h)
(1,1)
1 (-1,-1)

15. 2-D Lagrange Shape Functions
What is the lowest order
polynomial
f2(-1,h) along side x1=-1
that satisfies
f2(-1,-1) =1 & f2(-1,1)=0?
1 (-1,-1) h x
(1,h)
(1,1)

16. Construction of Lagrange Shape Functionsx
(1,h)
(1,1)
1 (-1,-1)
Bi-Linear Surface
& F1(1,-1)=F1(1,1)=F1(-1,1)=0?
What is the lowest order Polynomial F1(x,h) that satisfies
F1(-1,-1) =1

17. Construction of Lagrange Shape Functions
f1(x,-1)
f2(-1,h) h x
(1,h)
(1,1) xP hP P

18. Construction of Lagrange Shape Functionsx
(1,1)
1 (-1,-1)
F1(x,h)
For Every Point (x,h)

19. Construction of Lagrange Shape Functions
F2(x,h) h x
(1,h)
(1,1)
2 (1,-1)
Bi-Linear Surface

20. Construction of Lagrange Shape Functions
F3(x,h) h x
3 (1,1)
Bi-Linear Surface

21. Construction of Lagrange Shape Functions
F4(x,h)
4 (-1,1) h x
Bi-Linear Surface

22. In Summary x h
4 (-1,1)
3 (1,1)
1 (-1,-1)
2 (1,-1)

23. Shape Functions in a Direct Way
1 (-1,-1)
2 (1,-1)
4 (-1,1)
3 (1,1) x h
Assume Bi-Linear
Variation
Complete Polynomial
i=1,2,3,4
With conditions
j=1,2,3,4

24. Shape Functions in a Direct Way
Each, i, provides 4 Equations with 4 unknowns
e.g. i=1

25. Field Variables in Discrete Formq8 q7 q4 q3 q1 q2 v u q6 q5
Geometry
Displacement

26. Field Variables in Discrete Form
Geometry
Displacement

27. Intrinsic Coordinate Systemx h 4 1 2 3 x h
4 (-1,1)
3 (1,1)
Map Element
Define Jacobian
1 (-1,-1)
2 (1,-1)
Parent

28. Transformation
The Jacobian from (x,y) to (x,h) may be obtained as:
Consider any function f(x,y)
defined on area of element
1 (-1,-1)
2 (1,-1)
4 (-1,1)
3 (1,1) x h

29. Transformation
1 (-1,-1)
2 (1,-1)
4 (-1,1)
3 (1,1) x h

30. Jacobian of Transformation

31. Jacobian of Transformation

32. Jacobian of Transformation

33. Jacobian of Transformation

34. h h x x In Summary Without Proof 4 1 2 3 1 (-1,-1) 2 (1,-1) 4 (-1,1)
3 (1,1) x h
Without Proof

35. Jacobian of Transformation

36. Jacobian of Transformation

37. In Summary
Or the inverse

38. In Summary
1 (-1,-1)
2 (1,-1)
4 (-1,1)
3 (1,1) x h

39. Strain Tensor from Nodal Values of Displacements

40. Strain Tensor from Nodal Values of Displacements
Thus we need to evaluate
and

41. Strain Tensor from Nodal Values of Displacements
Recall that
Consequently,
f=u
f=v

42. Strain Tensor from Nodal Values of Displacements

43. Strain Tensor from Nodal Values of Displacements
Next we need to evaluate

44. Strain Tensor from Nodal Values of Displacements

45. Strain Tensor from Nodal Values of Displacements

46. Strain Tensor from Nodal Values of Displacements
e = AG q
e = B q
Both A and G are linear functions of x and h

47. Stresses
e = B q

48. Element Stiffness Matrix keh 4 1 2 3
Element Stiffness Matrix ke
e = B qe ke
s = D B qe
8x8 matrix

49. Element Stiffness Matrix ke
Furthermore
and
Numerical
Integration

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

51. Integration of Stiffness Matrix
Each term kij in ke is expressed as
Linear Shape Functions is each Direction

52. Numerical Integration
Integrals
Indefinite
Definite x2

53. Numerical Integration
Definite integrals can be computed numerically
Objective:
Determine points xi
Determine coefficients wi x2

54. Numerical Integration
Depending on choice of wi and xi
Midpoint Rule
Trapezoidal Rule
Simpson's
Gaussian Quadratures
etc

55. Numerical Integration – Upper & Lower Bounds
Lower Sum
L(f;xi)
a=x1 x2 x3 x4 x5 x6
x7=b
Upper Sum
U(f;xi)
a=x1 x2 x3 x4 x5 x6
x7=b

56. Numerical Integration
It can be shown that

57. Mid-Point Rule
a=x1 x2 x3 x4 x5 x6
x7=b

58. … Mid Point Rule Simple to comprehend and implement
Large number of intervals is required for accuracy
a=x1 …
b=xn

59. Trapezoid Rule
a=x1 x2 x3 x4 x5 x6
x7=b

60. Trapezoid Rule
Simple to comprehend and implement
Exact for polynomials f(x) = ax+b
Large number of intervals is required for accuracy
(Less than midpoint rule)

61. Simpson’s Rule
Step 1 h a
a+2h

62. Simpson’s Rule
Step 2 h xi
xi+1
xi+1/2 h xi
xi+1
xi+1/2
Ii+1 Ii …

63. Quadratures
Objective
Where do such formulae come from?
Theory of Interpolation….
Let
li(x): cardinal functions
Recall Shape Functions

64. It will give correct values for the integral of
Quadratures
It will give correct values for the integral of
every polynomial of degree n-1
Mid-Point Rule is an example of a
quadrature with n=1

65. Establish coefficients of quadrature for the interval
Example
Establish coefficients of quadrature for the interval
[-2,2] and nodes –1,0,1
f(x) -2 -1 1 2
Cardinal Functions:

66. Example
Cardinal Functions – Lagrange Polynomials:

67. Example
with

68. Example

69. Example -2 -1 1 2
f(x)=x2
Quadrature
Exact

70. Quadratures
So far the placement of nodes has been arbitrary
a=x1 x2 x3 x4 x5 x6
x7=b
They should belong to the interval of integration
Quadrature is accurate for polynomials of degree n-1 (n is the number of nodes)

71. Gaussian Quadrature
Karl Friedriech Gauss discovered that by a special placement of nodes the accuracy of the numerical integration could be greatly increased

72. Gaussian Quadrature
Theorem on Gaussian nodes
Let q be a polynomial of degree n such that
Let x1,x2,…,xn be the roots of q(x). Then
with xi’s as nodes is exact for all polynomials of
degree 2n-1.

73. Gaussian Quadrature 2-point
f(x2)
W2=1
f(x)
f(x1)
W1=1 -1 1

74. Gauss Points and Weights

75. Compare Let f(x)=x2 Use 2-points MidPoint X1=-1, X2=0, X3=1 X=-0.5

76. Compare
Let f(x)=x2
Trapezoidal
Use 2-points
X1=-1, X2=0, X3=1

77. Compare
Let f(x)=x2
Gauss
Use 2-points

78. 2-Dimensional Integration
Gaussian Quadrature

79. 2-D Integration 2-point formulah x

80. 2-D Integration 2-point formulax h

81. Integration of Stiffness Matrix
1 (-1,-1)
2 (1,-1)
4 (-1,1)
3 (1,1) x h
Integration over

82. Integration of Stiffness Matrix
e = AG q
e = B q
B (3x8)

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

84. 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

85. Integration of Stiffness Matrixh

86. Element Body Forces
Total Potential
Galerkin

87. Body Forces
Integral of the form

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

89. Element Traction
Total Potential
Galerkin

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

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

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

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

94. 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

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

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

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

98. 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

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

100. Convergence Considerations
For monotonic convergence of solution
Requirements
Elements (mesh) must be compatible
Elements must be complete

101. Mesh Compatibility OK
NO NO!

102. Mesh compatibility - Refinement
Acceptable Transition
Compatibility of displacements OK
However…
Stresses?

103. Convergence Considerations
We will revisit the issue…