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Finite Element Method CHAPTER 7: FEM FOR 2D SOLIDS

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1 Finite Element Method CHAPTER 7: FEM FOR 2D SOLIDS
for readers of all backgrounds G. R. Liu and S. S. Quek CHAPTER 7: FEM FOR 2D SOLIDS

Field variable interpolation Shape functions construction Using area coordinates Strain matrix Element matrices LINEAR RECTANGULAR ELEMENTS Gauss integration Evaluation of me

Coordinate mapping Strain matrix Element matrices Remarks HIGHER ORDER ELEMENTS COMMENTS (GAUSS INTEGRATION)

4 INTRODUCTION 2D solid elements are applicable for the analysis of plane strain and plane stress problems. A 2D solid element can have a triangular, rectangular or quadrilateral shape with straight or curved edges. A 2D solid element can deform only in the plane of the 2D solid. At any point, there are two components in the x and y directions for the displacement as well as forces.

5 INTRODUCTION For plane strain problems, the thickness of the element is unit, but for plane stress problems, the actual thickness must be used. In this course, it is assumed that the element has a uniform thickness h. Formulating 2D elements with a given variation of thickness is also straightforward, as the procedure is the same as that for a uniform element.

6 2D solids – plane stress and plane strain

Less accurate than quadrilateral elements Used by most mesh generators for complex geometry A linear triangular element:

8 Field variable interpolation
where (Shape functions)

9 Shape functions construction
Assume, i= 1, 2, 3 or

10 Shape functions construction
Delta function property: Therefore, Solving,

11 Shape functions construction
Area of triangle Moment matrix Substitute a1, b1 and c1 back into N1 = a1 + b1x + c1y:

12 Shape functions construction

13 Shape functions construction
where i= 1, 2, 3 J, k determined from cyclic permutation i = 1, 2 i k j j = 2, 3 k = 3, 1

14 Using area coordinates
Alternative method of constructing shape functions  2-3-P: Similarly,  3-1-P A2  1-2-P A3

15 Using area coordinates
Partitions of unity: Delta function property: e.g. L1 = 0 at if P at nodes 2 or 3 Therefore,

16 (constant strain element)
Strain matrix where (constant strain element)

17 Element matrices Constant matrix

18 Element matrices For elements with uniform density and thickness,
Eisenberg and Malvern (1973):

19 Element matrices Uniform distributed load:

Non-constant strain matrix More accurate representation of stress and strain Regular shape makes formulation easy

21 Shape functions construction
Consider a rectangular element

22 Shape functions construction
where (Interpolation)

23 Shape functions construction
Delta function property Partition of unity

24 Strain matrix Note: No longer a constant matrix!

25 Element matrices dxdy = ab dxdh Therefore,

26 Element matrices For uniformly distributed load,

27 Gauss integration For evaluation of integrals in ke and me (in practice) In 1 direction: m gauss points gives exact solution of polynomial integrand of n = 2m - 1 In 2 directions:

28 Gauss integration m xj wj Accuracy n 1 2 -1/3, 1/3 1, 1 3
2 -1/3, 1/3 1, 1 3 -0.6, 0, 0.6 5/9, 8/9, 5/9 5 4 , , , , , , 7 , , 0, , , , , , 9 6 , , , , , , , , , , 11

29 Evaluation of me

30 Evaluation of me E.g. Note: In practice, Gauss integration is often used

Rectangular elements have limited application Quadrilateral elements with unparallel edges are more useful Irregular shape requires coordinate mapping before using Gauss integration

32 Coordinate mapping Physical coordinates Natural coordinates
(Interpolation of displacements) (Interpolation of coordinates)

33 Coordinate mapping where ,

34 Coordinate mapping Substitute x = 1 into or Eliminating ,

35 Strain matrix or where (Jacobian matrix) Since ,

36 Strain matrix Therefore,
(Relationship between differentials of shape functions w.r.t. physical coordinates and differentials w.r.t. natural coordinates) Therefore, Replace differentials of Ni w.r.t. x and y with differentials of Ni w.r.t.  and 

37 Element matrices Murnaghan (1951) : dA=det |J | dxdh

38 Remarks Shape functions used for interpolating the coordinates are the same as the shape functions used for interpolation of the displacement field. Therefore, the element is called an isoparametric element. Note that the shape functions for coordinate interpolation and displacement interpolation do not have to be the same. Using the different shape functions for coordinate interpolation and displacement interpolation, respectively, will lead to the development of so-called subparametric or superparametric elements.

39 HIGHER ORDER ELEMENTS Higher order triangular elements
nd = (p+1)(p+2)/2 Node i, Argyris, 1968 :

40 HIGHER ORDER ELEMENTS Higher order triangular elements (Cont’d)
Cubic element Quadratic element

41 HIGHER ORDER ELEMENTS Higher order rectangular elements Lagrange type:
[Zienkiewicz et al., 2000]

42 HIGHER ORDER ELEMENTS Higher order rectangular elements (Cont’d)
(nine node quadratic element)

43 HIGHER ORDER ELEMENTS Higher order rectangular elements (Cont’d)
Serendipity type: (eight node quadratic element)

44 HIGHER ORDER ELEMENTS Higher order rectangular elements (Cont’d)
(twelve node cubic element)


When the Gauss integration scheme is used, one has to decide how many Gauss points should be used. Theoretically, for a one-dimensional integral, using m points can give the exact solution for the integral of a polynomial integrand of up to an order of (2m-1). As a general rule of thumb, more points should be used for a higher order of elements.

Using a smaller number of Gauss points tends to counteract the over-stiff behaviour associated with the displacement-based method. Displacement in an element is assumed using shape functions. This implies that the deformation of the element is somehow prescribed in a fashion of the shape function. This prescription gives a constraint to the element. The so- constrained element behaves stiffer than it should. It is often observed that higher order elements are usually softer than lower order ones. This is because using higher order elements gives fewer constraint to the elements.

Two Gauss points for linear elements, and two or three points for quadratic elements in each direction should be sufficient for most cases. Most of the explicit FEM codes based on explicit formulation tend to use one-point integration to achieve the best performance in saving CPU time.

49 CASE STUDY Side drive micro-motor

50 Elastic Properties of Polysilicon
CASE STUDY 10N/m Elastic Properties of Polysilicon Young’s Modulus, E 169GPa Poisson’s ratio,  0.262 Density,  2300kgm-3 10N/m 10N/m

51 CASE STUDY Analysis no. 1: Von Mises stress distribution using 24 bilinear quadrilateral elements (41 nodes)

52 CASE STUDY Analysis no. 2: Von Mises stress distribution using 96 bilinear quadrilateral elements (129 nodes)

53 CASE STUDY Analysis no. 3: Von Mises stress distribution using 144 bilinear quadrilateral elements (185 nodes)

54 CASE STUDY Analysis no. 4: Von Mises stress distribution using 24 eight-nodal, quadratic elements (105 nodes)

55 CASE STUDY Analysis no. 5: Von Mises stress distribution using 192 three-nodal, triangular elements (129 nodes)

56 CASE STUDY Analysis no. Number / type of elements
Total number of nodes in model Maximum Von Mises Stress (GPa) 1 24 bilinear, quadrilateral 41 0.0139 2 96 bilinear, quadrilateral 129 0.0180 3 144 bilinear, quadrilateral 185 0.0197 4 24 quadratic, quadrilateral 105 0.0191 5 192 linear, triangular 0.0167

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