FE Formulation of 2D Elasticity Problem. - Google Sites

ES 622

Finite Element Methods

Spring 2017

FE Formulation of 2D Elasticity Problem. Consider an elastic 2D plate as shown in Figure 1. Compute the displacement field using one isoparametric linear quadrilateral element.

Figure 1: Elastic plate in 2D. The governing equation for this problem is Z Z Z 1 T T Π(u) =  DdΩ − u bdΩ − uT tdΓ. 2 Ω Ω ΓN

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We start by writing picking nodes, assigning nodal coordinates and degrees-of-freedom (DOFs). Since this is a 2D elasticity problem, each node will have 2 DOFs (displacements in x and y directions), as shown in Figure 2. Let us write the table of nodal coordinates for the real and the reference elements along with

Figure 2: Node and DOF numbers. quadrature abscissas (we use 2 quadrature points in each direction). I xI yI

1 0 0

2 1 0

3 1 1

4 0 1

ξI ηI

1 -1 -1

2 1 -1

3 1 1

4 -1 1

ξIquad ηIquad

1 √ −1/ 3 √ −1/ 3

2 √ 1/ 3 √ 1/ 3

The shape functions written for the reference element read N1 = 0.25(1 − ξ)(1 − η)

N2 = 0.25(1 + ξ)(1 − η)

N3 = 0.25(1 + ξ)(1 + η)

N4 = 0.25(1 − ξ)(1 + η)

We write another table of some values that will be useful in further computations:

Gaurav ([email protected])

IIT Gandhinagar

1

ES 622

Finite Element Methods

K (I, J) (ξIquad , ηJquad ) N1 N2 N3 N4 N1,ξ N1,η N2,ξ N2,η N3,ξ N3,η N4,ξ N4,η xξ xη yξ yη j ξx ξy ηx ηy b N1,x b1,y N b2,x N b2,y N b3,x N b3,y N b N4,x b4,y N

1 (1,1) (-0.5774,-0.5774) 0.6220 0.1667 0.0447 0.1667 -0.3943 -0.3943 0.3943 -0.1057 0.1057 0.1057 -0.1057 0.3943 0.5 0.0 0.0 0.5 0.25 2 0 0 2 -0.7887 -0.7887 0.7887 -0.2113 0.2113 0.2113 -0.2113 0.7887

2 (1,2) (-0.5774,0.5774) 0.1667 0.0447 0.1667 0.6220 -0.1057 -0.3943 0.1057 -0.1057 0.3943 0.1057 -0.3943 0.3943 0.5 0.0 0.0 0.5 0.25 2 0 0 2 -0.2113 -0.7887 0.2113 -0.2113 0.7887 0.2113 -0.7887 0.7887

Spring 2017

3 (2,1) (0.5774,-0.5774) 0.1667 0.6220 0.1667 0.0447 -0.3943 -0.1057 0.3943 -0.3943 0.1057 0.3943 -0.1057 0.1057 0.5 0.0 0.0 0.5 0.25 2 0 0 2 -0.7887 -0.2113 0.7887 -0.7887 0.2113 0.7887 -0.2113 0.2113

4 (2,2) (0.5774,0.5774) 0.0447 0.1667 0.6220 0.1667 -0.1057 -0.1057 0.1057 -0.3943 0.3943 0.3943 -0.3943 0.1057 0.5 0.0 0.0 0.5 0.25 2 0 0 2 -0.2113 -0.2113 0.2113 -0.7887 0.7887 0.7887 -0.7887 0.2113

Using these values, we can compute the B matrix each Gauss point (refer to class notes for exact form of B bI . Computing BT DB at all Gauss points and adding them yields the 8 × 8 stiffness matrix as in terms of N u1 u1   1.1538 v1   0.4808 u2  −0.7692 K = v2   0.0962 u3  −0.5769 v3  −0.4808 u4  0.1923 v4 −0.0962 

v1 0.4808 1.1538 −0.0962 0.1923 −0.4808 −0.5769 0.0962 −0.7692

u2 −0.7692 −0.0962 1.1538 −0.4808 0.1923 0.0962 −0.5769 0.4808

v2 0.0962 0.1923 −0.4808 1.1538 −0.0962 −0.7692 0.4808 −0.5769

u3 −0.5769 −0.4808 0.1923 −0.0962 1.1538 0.4808 −0.7692 0.0962

v3 −0.4808 −0.5769 0.0962 −0.7692 0.4808 1.1538 −0.0962 0.1923

u4 0.1923 0.0962 −0.5769 0.4808 −0.7692 −0.0962 1.1538 −0.4808

 v4 −0.0962  −0.7692  0.4808   11 −0.5769  × 10 N/m 0.0962   0.1923   −0.4808 1.1538

(2) DOFs u1 , v1 , u4 and v4 are fixed. Applying these boundary conditions yields the following 4 × 4 stiffness

Gaurav ([email protected])

IIT Gandhinagar

2

ES 622

Finite Element Methods

Spring 2017

matrix: 

 u2 v2 u3 v3 u2  0.0962   1.1538 −0.4808 0.1923  11  K = v2 −0.4808 1.1538 −0.0962 −0.7692  × 10 N/m   u3 0.1923 −0.0962 1.1538 0.4808 v3 0.0962 −0.7692 0.4808 1.1538 Since only nodal loads are specified, the force vector can be simply written as   u2 10 v 0  × 103 N. F= 2 u3  0  v3 10

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Solving for the displacement vector d, we get   u2 0.2032 v  0.2863   × 10−6 m. d= 2 u3 −0.1435 v3 0.3204

Gaurav ([email protected])

IIT Gandhinagar

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