The Finite Element Method A Practical Course CHAPTER 5: FEM FOR 2D SOLIDS

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

CONTENTS LINEAR QUADRILATERAL ELEMENTS HIGHER ORDER ELEMENTS Coordinate mapping Strain matrix Element matrices Remarks HIGHER ORDER ELEMENTS COMMENTS (GAUSS INTEGRATION) CASE STUDY

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. 2D solid element can deform only in the plane of the 2D solid. At any point, there are two components in x and y directions for the displacement as well as forces.

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 2-D elements with given variation of thickness is also straightforward, as the procedure is the same as that for a uniform element.

2D solids – plane stress and plane strain

LINEAR TRIANGULAR ELEMENTS Less accurate than quadrilateral elements Used by most mesh generators for complex geometry Linear triangular element

Field variable interpolation where (Shape functions)

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

Shape functions construction Delta function property: Therefore, Solving,

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

Shape functions construction Similarly,

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

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

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

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

Element matrices Constant matrix 

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

Element matrices Uniform distributed load:

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

Shape functions construction Consider a rectangular element

Shape functions construction where

Shape functions construction Delta function property Partition of unity

Strain matrix Note: No longer a constant matrix!

Element matrices  dxdy = ab dxdh Therefore,

Element matrices For uniformly distributed load,

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:

Gauss integration m Gauss Point xj Gauss Weight wj Accuracy order n 1 2 -1/3, 1/3 1, 1 3 -0.6, 0, 0.6 5/9, 8/9, 5/9 5 4 -0.861136, -0.339981, 0.339981, 0.861136 0.347855, 0.652145, 0.652145, 0.347855 7 -0.906180, -0.538469, 0, 0.538469, 0.906180 0.236927, 0.478629, 0.568889, 0.478629, 0.236927 9 6 -0.932470, -0.661209, -0.238619, 0.238619, 0.661209, 0.932470 0.171324, 0.360762, 0.467914, 0.467914, 0.360762, 0.171324 11

Evaluation of me

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

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

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

Coordinate mapping where ,

Coordinate mapping Substitute x = 1 into or Eliminating ,

Strain matrix or where (Jacobian matrix) Since ,

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 

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

Remarks Shape functions used for interpolating the coordinates are the same as the shape functions used for the interpolation of the displacement field. Therefore, the element is called 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.

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

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

HIGHER ORDER ELEMENTS Higher order rectangular elements Lagrange type: (Zienkiewicz et al., 2000)

HIGHER ORDER ELEMENTS Higher order rectangular elements(Cont’d) (9 node quadratic element)

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

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

ELEMENT WITH CURVED EDGES

COMMENTS (GAUSS INTEGRATION) 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 higher order of elements.

COMMENTS (GAUSS INTEGRATION) Using 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 be. It is often observed that higher order elements are usually softer than lower order ones. This is because using higher order elements gives less constraint to the elements.

COMMENTS (GAUSS INTEGRATION) Two gauss points for linear elements, and two or three points for quadratic elements in each direction should be sufficient for many 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.

CASE STUDY Side drive micro-motor

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

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

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

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

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

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

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