Week 11 MECH3361

Recap 2D elements

Displacement Shape function Strain

T3,3-node constant strain triangle (CST)

Linear: u=b1+b2 x+b3 yv=b4+b5 x+b6 y

(6 d.o.f.)

N i=1

2 A(α i+β i x+γi y )

(i=1,2,3)

A=12|1 x1 y1

1 x2 y21 x3 y3

|

Constant strain

ε xx=∂u∂ x=b2

ε yy=∂ v∂ y=b6

2 ε xy=(∂u∂ y+∂ v∂ x )

¿b3+b5

T6quadratic triangular elementlinear strain triangle (LST)

Quadratic: u=b1+b2 x+b3 y+b4 x2+b5 xy+b6 y2

v=b7+b8 x+b9 y+b10 x2+b11 xy+b12 y2

(12 d.o.f.)

N1=ξ (2ξ−1)N2=η(2η−1 )N3=(1−ξ−η )[2(1−ξ−η)−1 ]N4=4 ξηN5=4 η(1−ξ−η)N6=4(1−ξ−η )ξ

Fully-linearε xx=b2+2b4 x+b5 y

ε yy=b9+b11 x+2b12 y

2 ε xy=(b3+b5 )+(b3+2 b10 ) x+(2 b6+2 b11) y

Linear Quadrilateral Element (Q4)

Bilinear u=b1+b2 x+b3 y+b4 xyv=b5+b6 x+b7 y+b8 xy

(8 d.o.f.)

N1=14(1−ξ )(1−η )

,

N2=14(1+ξ )(1−η)

,

N3=14(1+ξ )(1+η)

,

N4=14(1−ξ )(1+η )

Half-linear

ε xx=∂u∂ x=b2+b4 y

ε yy=∂v∂ y=b7+b8 x

2 ε xy=∂ v∂ x+∂u∂ y

¿b6+b8 y+b3+b4 x

x

y

x

y

mapping

1 2

3

4

5

6

7

8

1 2

3

5

6

7

8

4

Quadratic Quadrilateral Element (Q8)

Quadraticu=b1+b2 x+b3 y+b4 x2+b5 xy+b6 y2+b7 x2 y+b8 xy 2

v=b9+b10 x+b11 y+b12 x2+b13 xy+b14 y2

+b15 x2 y+b16 xy2

(16 d.o.f.)

N1=−14(1−ξ )(1−η )(1+ξ+η)

N2=−14(1+ξ )(1−η)(1−ξ+η)

N3=−14(1+ξ )(1+η)(1−ξ−η)

N4=−14(1−ξ )(1+η)(1+ξ−η )

N5=12(1−ξ2)(1−η)

N6=12(1+ξ )(1−η2 )

N7=12(1−ξ2)(1+η )

N8=12(1−ξ )(1−η2)

Half-quadraticεxx=b2 x+2b4 x+b5 y+2 b7 xy+b8 y2

ε yy=b11 y+b13 y+2 b14 y+b15 x2+2b16 xy

εxy=b3+b5 x+2 b6 y+b7 x2+2 b8 xy+b10 x+2 b12 x+b13 y+2 b15 xy+b16 y2

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Week 11 MECH3361

8.7 Axisymmetric ProblemsEngineering problemsConditions: Both (1) Rotational structures and (2) Axisymmetric loading/boundary is applied.

Cylindrical coordinates ( x , y , z )⇒(r , θ , z )

Displacement field: u(r , z ) , w (r , z )Due to axisymmetry, there is no v – circumferential component of displacement vanishes.

Strain – displacement relation

ε rr=∂u(r , z )∂r

, εθϑ=u(r , z )

r, ε zz=

∂w (r , z )∂ z

, εrz=∂w(r , z )∂r

+∂u(r , z )∂ z

Note that ε rθ=εθz=0

Stress-strain relation {σrr

σθθ

σ zzσrz}= E(1+ν )(1−2 ν ) [1−ν ν ν 0

ν 1−ν ν 0ν ν 1−ν 0

0 0 0 1−2ν2

]{εrr

εθθ

εzzεrz}

Axisymmetric Elements:The element is in 2D but represents a ring swept around the axis.

Elemental stiffness matrix

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Week 11 MECH3361

Applications: Rotating Flywheel:

Axisymmetric Condition:(1) Structure is rotational (2) The loading of body force due to angular velocity is axisymmetric.So such a problem can be modelled as a 2D axisymmetric problem.

Examples for 2D finite element method:Example 8.9 Calculate stiffness matrix of the CST element (plane stress). Soln: To install the elemental stiffness matrix for CST element, we can use Matlab as follows:Step 1: Material matrix:

Plane stress:

[ E ]= E1−ν2 [1 ν 0

ν 1 0

0 0 1−ν2 ]

Or, plane strain:

[ E ]= E(1+ν )(1−2 ν ) [1−ν ν 0

ν 1−ν 0

0 0 1−2 ν2 ]

Step 2 Calculate [B] matrix:

[B1 ]= 12 A [ y23 0 y31 0 y12 0

0 x32 0 x13 0 x21

x32 y23 x13 y31 x21 y12]

Step 3 Calculate stiffness matrix [k]

[K e ]=∫V[B1 ]

T [ E ][B1 ]dV=∫V

dV ([B1 ]T [E ][B1 ])=tA [B1 ]

T [E ][B1 ]

¿ t4 A [

y23 0 x32

0 x32 y23

y31 0 x13

0 x13 y31y12 0 x21

0 x21 y12

](E1−ν2 [1 ν 0ν 1 0

0 0 1−ν2 ])[ y23 0 y31 0 y12 0

0 x32 0 x13 0 x21

x32 y23 x13 y31 x21 y12]

Step 4: Matlab CalculationE=1;nu=0.3;t=1;x1=0;

3

1(0,0) 2(1,0)

3(0.5,0.5)

x

y

Week 11 MECH3361

y1=0;x2=1;y2=0;x3=0.5;y3=0.5;A=(x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2))/2;x21=x2-x1;x13=x1-x3;x32=x3-x2;y12=y1-y2;y31=y3-y1;y23=y2-y3;Bmat=[y23 0 y31 0 y12 0; 0 x32 0 x13 0 x21; x32 y23 x13 y31 x21 y12];Emat=(E/(1-nu*nu))*[1 nu 0; nu 1 0; 0 0 (1-nu)/2];Ke=t/(4*A)*Bmat'*Emat*Bmat

Ke =

0.3709 0.1786 -0.1786 -0.0137 -0.1923 -0.1648 0.1786 0.3709 0.0137 0.1786 -0.1923 -0.5495 -0.1786 0.0137 0.3709 -0.1786 -0.1923 0.1648 -0.0137 0.1786 -0.1786 0.3709 0.1923 -0.5495 -0.1923 -0.1923 -0.1923 0.1923 0.3846 0 -0.1648 -0.5495 0.1648 -0.5495 0 1.0989

Example 8.10 Sketch the distribution of v-displacement and y-strain in these two connected CST elements for the global displacement vector {u} as shown.

x

y

v

D (1,0)

C (1,1)

B (0,1)

A (0,0)

x

y

D (1,0)

C (1,1)

B (0,1)

A (0,0)

1 2

132

21

401

D

D

C

C

B

B

A

A

vuvuvuvu

u1 2

Soln

Step 1: Select v-displacement in each node: v={v A v B vC v D }T= {0 −1 −2 1 }T .Step 2: Draw the displacement in each corresponding node and connect them as shown.Step 3: Calculate y-strains in each of the CST elements according to the follow equation.

{εxxε yy

ε xy}=Bd= 1

2 A [ y2− y3 0 y3− y1 0 y1− y2 00 x3−x2 0 x1−x3 0 x2−x1

x3−x2 y2− y3 x1−x3 y3− y1 x2−x1 y1− y2]{

u1

v1

u2

v2

u3

v3

}Note that node numbering of 1, 2, 3 should be in count-clockwise sequence. For element #1, we can choose: Node (1,2,3) = Node (A,D,C) (we cannot use Node (A,C,D)). For element #2, we can choose Node (1,2,3) = Node (C,B,A) (we cannot use Node (A,B,C)). For element #1: nodal coordinates: For element #2: nodal coordinates:

4

Week 11 MECH3361

x(1)={x1

y1

x2

y2

x3

y3

}={x A

y A

x D

yD

xC

yC

}={001011 } x(2)={x1

y1

x2

y2

x3

y3

}={xC

yC

x B

yB

x A

y A

}={110100 }Strain in element #1 (Node 1 = Node A, Node 2 = Node D, Node 3 = Node C)

ε yy(1)=1

2 A [ (x3−x2)v1+( x1−x3 )v2+( x2−x1) v3]¿1

2 A [(xC−x D )v A+(x A−xC) vD+( xD−x A )vC ]¿1

2×0. 5 [ (1−1 )×(0 )+(0−1 )×(1 )+(1−0 )×(−2 ) ]=−3

Strain in element #2 (Node 1 = Node C, Node 2 = Node B, Node 3 = Node A)

ε yy(2)=1

2 A [ (x3−x2)v1+ (x1−x3 )v2+( x2−x1) v3]¿1

2 A [ (x A−xB )vC+(xC−x A )v B+ (xB−xC )v A ]¿1

2×0.5 [ (0−0 )× (−2 )+ (1−0 )×(−1 )+(0−1 )×(0 ) ]=−1Step 4: plot the y-strain yy for

each constant strain element (as shown on the right). Note that the strain distributions are constant!

Example 8.11 Determine coordinate G1, G2, G3, G4 of Gaussian points in Cartesian coordinate system of iso-parametric element Q4 as shown (the nodal coordinates have been give in the figure). Compute Jacobian matrix [J] and det[J] for the Q4 element.

A(1,0) B(2,0)

C(2.25,1.5)

D(1.25,1)

x

y

A’(-1, -1)

B’(1, -1)

C’(1, 1)D’(-1, 1)

* *

**

G1G2

G3 G4

0.5774

-0.5774

Soln: Q4 and Q8 are “iso-parametric” element in which displacement and coordinate share the same “shape functions” for mapping and interpolation, as follows:

Coordinate mapping: x=∑i=1

4

N i xi , y=∑i=1

4

N i y i

Displacement interpolation: u=∑

i=1

4

N i ui , v=∑i=1

4

N i vi

Step 1: Node numbering: Node (1,2,3,4) = Node (A,B,C,D)Step 2: Coordinate in terms of Shape functionsShape functions

5

x

y

yy

D (1,0)

C (1,1)

B (0,1)

A (0,0)

1 2

-1

-3

-1-3

Week 11 MECH3361

N1=14(1−ξ )(1−η )

, N2=

14(1+ξ )(1−η)

, N3=

14(1+ξ )(1+η)

, N4=

14(1−ξ )(1+η )

Cartersian coordinate (x,y) in terms of natural coordinate (, )

x=∑i=1

4

N i x i=14(1−ξ )(1−η) x1+

14(1+ξ )(1−η) x2+

14(1+ξ )(1+η) x3+

14(1−ξ )(1+η) x4

¿14(1−ξ )(1−η )1+1

4(1+ξ )(1−η )2+1

4(1+ξ )(1+η)2.25+1

4(1−ξ )(1+η )1 .25

∴ x (ξ , η)=14(6 .5+2 ξ+0 .5 η )

y=∑i=1

4

N i y i=14(1−ξ )(1−η ) y1+

14(1+ξ )(1−η) y2+

14(1+ξ )(1+η) y3+

14(1−ξ )(1+η ) y 4

¿14(1−ξ )(1−η )0+1

4(1+ξ )(1−η)0+1

4(1+ξ )(1+η )1. 5+1

4(1−ξ )(1+η)1

y (ξ , η)=14

(2 .5+0 .5 ξ+2 .5 η+0 . 5 ξη )

Step 3 Calculate the G1 coordinate:

x (ξ=0 . 5574 , η=0 .5574 )=14(6 .5+2×0. 5574+0. 5×0 .5574 )=1 .9859

y (ξ=0 . 5574 , η=0 .5574 )=14(2 .5+0 .5×0 .5574+2. 5×0 .5574+0 .5×0 .5574×0 . 5574 )=1. 0997

Step 4: Jacobian matrix

{∂ u∂ξ∂ u∂ η}=[∂ x

∂ξ∂ y∂ξ

∂ x∂η

∂ y∂η ]⏟

[ J ]

{∂u∂ x∂u∂ y}

[J ]=[∂ x∂ ξ

∂ y∂ ξ

∂ x∂ η

∂ y∂ η ]

[J ]=[∂∂ ξ [14 (6 . 5+2 ξ+0 . 5 η)] ∂∂ ξ [14 (2 .5+0. 5 ξ+2. 5 η+0 . 5ξη )]

∂∂ η [14 (6 . 5+2 ξ+0 .5 η )] ∂

∂ η [14 (2. 5+0 .5 ξ+2 .5 η+0. 5ξη )] ][J (ξ , η)]=1

4 [(2 ) (0 .5+0. 5 η )(0 .5 ) (2. 5+0 . 5 ξ ) ]

G1:

[J (0 .5774 ,0 . 5774 ) ]=1

4 [ (2) (0 . 5+0 .5×0 .5774 )(0 . 5) (2 . 5+0 .5×0 .5774 ) ]=[ 0 .5 0. 1972

0 .125 0 . 6972 ]

Step 5 Jacobian determinant:

6

Week 11 MECH3361

det [ J (ξ , η)]= 14 [ (2) (0 .5+0 .5 η )(0 .5 ) (2 .5+0 .5 ξ ) ]=4 .75+ξ−0 .25 η

A linear function of natural

coordinate (, ). At G1:

det [ J (0 .5774 ,0 . 5774 ) ]=det[ 0 .5 0 .19720 . 125 0 .6972 ]=0.5×0 .6972−0 .125×0 .1972=0 .3239

8.6 Solid Elements for 3-D Problems3D problems

3D stress vector: σ={σxx σ yy σ zz σxy σ yz σ zx }T

3D strain vector: ε={ε xx ε yy ε zz εxy ε yz ε zx }T

Hooke’s law:

σ (6×1 )={σ xx

σ yy

σ zz

σ xy

σ yz

σ zx

}= E(1+ν )(1−2 ν ) [

1−ν ν ν 0 0 0ν 1−ν ν 0 0 0ν ν 1−ν 0 0 0

0 0 0 1−2 ν2

0 0

0 0 0 0 1−2 ν2

0

0 0 0 0 0 1−2 ν2

]{εxx

ε yy

ε zz

εxy

ε yz

εzx

}=E(6×6)ε(6×1 )

Displacement in 3D:

u={u (x , y , z )v ( x , y , z )w( x , y , z )}

Finite Element Formulation

Displacement field: u=∑

i=1

N

N i ui , v=∑i=1

N

N i v i , w=∑i=1

N

N i wi

In matrix form:

u={u (x , y , z )v ( x , y , z )w( x , y , z )}(3×1 )

=[N 1 0 0 N2 0 0 ⋯0 N1 0 0 N2 0 ⋯0 0 N1 0 0 N 2 ⋯ ](3×3 N ){

u1

v1

w1

u2

v2

w2⋮

}(3 N×1)

=Nd

where Ni are shape functions for 3D element and d= {u1 v1 w1 u2 v2 w2 ⋯}T is the nodal displacement vector.

Using strain – displacement relation, we can have the following similar formula:

7

Week 11 MECH3361

ε(6×1 )=B(6×3 N )d(3 N×1)

Similarly, the Stiffness Matrix:

k(3 N×3 N )=∫V e

BT(3 N×6)E(6×6 )B(6×3 N ) dV e

Numerical quadratures are often needed to calculate the above integration.

Typical 3D elements

Tetrahedron (Tet):

Hexahedron (brick):

Penta:

Remarks: Each node has 3 degrees of freedom (u, v, w) in these 3D elements 4-node tet element is a “constant strain element” and do not use 4-node tet elements for

capturing high stress/strain gradients.

Linear Hexahedron Element (8-node brick element)Mapping:

Coordinate transformation (from natural coordinate to Cartesian coordinate)

x=∑i=1

N

N i x i , y=∑i=1

N

N i y i , z=∑i=1

N

N i zi

Shape functions:

8

Week 11 MECH3361

N1 (ξ ,η , ζ )=18(1−ξ )(1−η )(1−ζ ) , N 2(ξ ,η , ζ )=1

8(1+ξ )(1−η )(1−ζ ) ,

N3 (ξ ,η , ζ )=18(1+ξ )(1+η)(1−ζ ) , N4 (ξ ,η , ζ )=1

8(1−ξ )(1+η )(1−ζ ) ,

N5 (ξ ,η , ζ )=18(1−ξ )(1−η)(1+ζ ) , N6 (ξ ,η , ζ )=1

8(1+ξ )(1−η )(1+ζ ),

N7 (ξ ,η , ζ )=18(1+ξ )(1+η)(1+ζ ) , N 8(ξ , η ,ζ )=1

8(1−ξ )(1+η)(1+ζ ) ,

Isoparamatric Elements: The same shape functions are used as for the displacement field.

Coordinate system:x=∑

i=1

N

N i x i , y=∑i=1

N

N i y i , z=∑i=1

N

N i zi

Displacement field:u=∑

i=1

N

N i ui , v=∑i=1

N

N i v i , w=∑i=1

N

N i wi

The name of “isoparamatric element” is applied to 2D elements as well.Jacobian Matrix: (to apply strain – displacement relation, derivatives w.r.t. different coordinate systems are needed). Take u as example, one should have:

{∂ u∂ξ∂ u∂ η∂ u∂ ζ}=[∂ x∂ξ

∂ y∂ξ

∂ z∂ξ

∂ x∂η

∂ y∂η

∂ z∂ η

∂ x∂ ζ

∂ y∂ ζ

∂ z∂ ζ]{∂ u∂ x∂ u∂ y∂ u∂ z}=J {

∂ u∂ x∂ u∂ y∂u∂ z} or

{∂ u∂ x∂ u∂ y∂u∂ z}=[∂ x∂ ξ

∂ y∂ ξ

∂ z∂ ξ

∂ x∂ η

∂ y∂ η

∂ z∂ η

∂ x∂ ζ

∂ y∂ ζ

∂ z∂ ζ]−1

{∂u∂ ξ∂u∂η∂u∂ ζ}=J−1{

∂ u∂ ξ∂ u∂ η∂u∂ζ}

where J is named Jacobian matrix. Similarly for v and w, one can extend as follows:

{∂ v∂ x∂ v∂ y∂ v∂ z}=[∂ x∂ ξ

∂ y∂ ξ

∂ z∂ ξ

∂ x∂ η

∂ y∂ η

∂ z∂ η

∂ x∂ ζ

∂ y∂ ζ

∂ z∂ ζ]−1

{∂ v∂ ξ∂ v∂η∂v∂ζ}=J−1{

∂ v∂ξ∂ v∂ η∂ v∂ ζ},

{∂w∂ x∂w∂ y∂w∂ z}=[∂ x∂ ξ

∂ y∂ ξ

∂ z∂ ξ

∂ x∂ η

∂ y∂ η

∂ z∂η

∂ x∂ζ

∂ y∂ζ

∂ z∂ζ]−1

{∂w∂ ξ∂w∂ η∂w∂ζ}=J−1{

∂w∂ ξ∂w∂η∂w∂ζ}

in which, for example, the calculation of ∂u /∂ξ can be done as:∂u∂ξ=∑

i=1

8 ∂N i

∂ ξui=

∂N1

∂ξu1+

∂N2

∂ ξu2+∂N 3

∂ ξu3+

∂N4

∂ ξu4+

∂ N5

∂ξu5+

∂N6

∂ ξu6+

∂N7

∂ ξu7+

∂ N8

∂ξu8

To calculate Jacobian matrix, the isoparamatric relations of

x=∑i=1

8

N i (ξ ,η , ζ )xi , y=∑i=1

8

N i( ξ , η , ζ ) y i , z=∑i=1

8

N i(ξ ,η ,ζ ) zi can be used as:

9

Week 11 MECH3361

J=[∂ x∂ξ

∂ y∂ξ

∂ z∂ξ

∂ x∂η

∂ y∂η

∂ z∂η

∂ x∂ζ

∂ y∂ζ

∂ z∂ ζ]=[∑i=1

8 ∂N i

∂ ξx i ∑

i=1

8 ∂N i

∂ ξy i ∑

i=1

8 ∂N i

∂ξzi

∑i=1

8 ∂N i

∂ ηx i ∑

i=1

8 ∂N i

∂ηy i ∑

i=1

8 ∂N i

∂ηzi

∑i=1

8 ∂N i

∂ ζ x i ∑i=1

8 ∂N i

∂ζ y i ∑i=1

8 ∂N i

∂ζ zi]

J=[∂ N1

∂ξ∂N2

∂ξ∂N3

∂ ξ∂N 4

∂ ξ∂N5

∂ ξ∂N6

∂ ξ∂N 7

∂ ξ∂ N8

∂ξ∂ N1

∂η∂N2

∂η∂N3

∂η∂N 4

∂η∂N5

∂η∂N6

∂ η∂N 7

∂ η∂ N8

∂η∂ N1

∂ ζ∂N2

∂ζ∂N3

∂ζ∂N 4

∂ ζ∂N5

∂ζ∂N6

∂ ζ∂N 7

∂ ζ∂ N8

∂ζ][

x1 y1 z1

x2 y2 z2

x3 y3 z3

x4 y 4 z 4

x5 y5 z5

x6 y6 z6

x7 y7 z7

x8 y8 z8

]Strain – Displacement relation:

ε(6×1 )={ε xx

ε yy

εzz

ε xy

ε yz

ε zx

}={∂ u/∂ x∂v /∂ y∂w /∂ z

(∂ v /∂ x+∂ u/∂ y )/2(∂w /∂ y+∂ v /∂ z )/2(∂ u/∂ z+∂w /∂ x )/2

}=B(6×24 )d (24×1 )

in which B(6×24 )= [B1 B2 B3 B4 B5 B6 B7 B8 ]

where:

Bi=[∂ N i/∂ x 0 0

0 ∂N i /∂ y 00 0 ∂N i/∂ z0 ∂N i/∂ z ∂N i/∂ y

∂N i /∂ z 0 ∂N i /∂ x∂N i /∂ y ∂N i/∂ x 0

](6×3)

and Jacobian transformation for chain rule of differentiation is needed here.

{∂ N i

∂ x∂ N i

∂ y∂N i

∂ z}=[∂ x∂ξ

∂ y∂ξ

∂ z∂ξ

∂ x∂η

∂ y∂η

∂ z∂ η

∂ x∂ ζ

∂ y∂ζ

∂ z∂ ζ]−1

{∂N i

∂ ξ∂N i

∂ η∂N i

∂ζ}=J−1{

∂ N i

∂ξ∂ N i

∂ η∂ N i

∂ ζ}

So Jacobian matrix needs to be positive definite.

Nodal coordinates are d(24×1 )= {u1 v1 w1 u2 v2 w2 ⋯ u8 v8 w8 }T

Strain energy

U=12∫V

σΤ ε dV=12∫V(Eε )Τ ε dV=1

2∫VεΤ Eε dV=1

2∫V(Β )Τ E(Β)dV=dT [ 12∫V ΒΤ EΒ dV ]d

10

Week 11 MECH3361

Elemental Stiffness matrix: k(24×24 )=

12∫V

ΒΤ(24×6)E(6×6 )Β(6×24 )dV

In natural coordinate system: dV=(det J )dξdηdζ

k(24×24 )=12∫−1

1

∫−1

1

∫−1

1

ΒΤ(24×6 )E(6×6)Β(6×24 ) (det J )dξdηdζ

which can be calculated using numerical integration.

Example 8.12 The 8-node iso-paramatric brick element as shown in Cartesian coordinate system (x,y,z) below is mapped to the natural coordinate (,,). Use shape function to determine (1) Cartesian coordinate of centroid (2) displacement u, v, w at the centroid if the

nodal displacement is: dT=[0 0 d3 d4 0 0 d7 d8 ]1×24

Non-zero nodal displacements: d3={1,2 ,−1 } , d4={−2,0 ,−2 }, d7= {2,2,2 } , d8= {−1,1,2 }

z

x

y

(0,0,4)

(3,0,3)

(2,0,0)

(2,1,0)

(0,3,0)

(0,4,4)

(3,3,3)

(1,1,1)

(1,1,-1)(-1,1,-1)

(-1,1,1)

(1,-1,1)

(1,-1,-1)

(-1,-1,-1)

(-1,-1,1)

mapping1

26

5

4

37

8

1 2

5 6

4 3

78

Solution:x= {0,0,0 , 2,0,0 , 2,1,0 , 0,3,0 , 0,0,4 , 3,0,3 , 3,3,3 , 0,4,4 }T

u={0,0,0 , 0,0,0 , 1,2 ,−1 , −2,0 ,−2 , 0,0,0 , 0,0,0 , 2,2,2 , −1,1,2 }T

Shape functions

N1 (ξ , η , ζ )=18(1−ξ )(1−η )(1−ζ ) , N 2(ξ ,η , ζ )=1

8(1+ξ )(1−η )(1−ζ ) ,

N3 (ξ , η , ζ )=18(1+ξ )(1+η)(1−ζ ) , N4(ξ ,η , ζ )=1

8(1−ξ )(1+η )(1−ζ ) ,

N5 (ξ , η , ζ )=18(1−ξ )(1−η)(1+ζ ) , N6 (ξ ,η , ζ )=1

8(1+ξ )(1−η )(1+ζ ),

N7 (ξ , η , ζ )=18(1+ξ )(1+η)(1+ζ ) , N 8(ξ ,η ,ζ )=1

8(1−ξ )(1+η)(1+ζ ) ,

Coordinate mapping: x=∑

i=1

8

N i x i , y=∑i=1

8

N i y i , z=∑i=1

8

N i z i

x0 (ξ=0 , η=0 ,ζ=0 )=N1 x1+N2 x2+N 3 x3+N4 x4+N 5 x5+N6 x6+N 7 x7+N8 x8

¿18(1−0 )(1−0 )(1−0 )(0)+1

8(1+0 )(1−0 )(1−0 )(2)+1

8(1+0)(1+0 )(1−0 )(2)+1

8(1−0 )(1+0)(1−0)(0 )

+18(1−0 )(1−0 )(1+0)( 0)+1

8(1+0)(1−0)(1+0 )(3)+1

8(1+0 )(1+0)(1+0 )(3)+1

8(1−0 )(1+0 )(1+0)(0 )

x0=0+28+2

8+0+0+3

8+3

8+0=10

8=1. 25

Similarly,

11

Week 11 MECH3361

The Cartesian coordinate for centroid: x0=1.25 , y 0=1 . 375 , z0=1.75Displacement at the centroid:

Displacement field: u=∑

i=1

8

N i ui , v=∑i=1

8

N i v i , w=∑i=1

8

N i wi

u0 (ξ=0 , η=0 , ζ=0)=N 1 u1+N2 u2+N3 u3+N4 u4+N5 u5+N6 u6+N7 u7+N8 u8

¿18(1−0 )(1−0 )(1−0 )(0)+1

8(1+0 )(1−0 )(1−0 )(0)+1

8(1+0 )(1+0 )(1−0)(1)+1

8(1−0 )(1+0 )(1−0 )(−2 )

+18(1−0 )(1−0 )(1+0)( 0)+1

8(1+0)(1−0)(1+0 )(0 )+1

8(1+0 )(1+0)(1+0 )(2)+1

8(1−0 )(1+0 )(1+0 )(−1)

u0=0+0+ 18−2

8+0+0+ 2

8− 1

8=0

Similarly, Displacement at the centroid: u0=0 , v0=0 . 625 , w0=0 . 125 .

Matlab code: the following simple Matlab code can be used for the calculations:xi=0; nu=0; zt=0;x1=0; x2=2; x3=2; x4=0; x5=0; x6=3; x7=3; x8=0;y1=0; y2=0; y3=1; y4=3; y5=0; y6=0; y7=3; y8=4;z1=0; z2=0; z3=0; z4=0; z5=4; z6=3; z7=3; z8=4;u1=0; u2=0; u3=1; u4=-2; u5=0; u6=0; u7=2; u8=-1;v1=0; v2=0; v3=2; v4=0; v5=0; v6=0; v7=2; v8=1;w1=0; w2=0; w3=-1;w4=-2; w5=0; w6=0; w7=2; w8=2;N1=(1-xi)*(1-nu)*(1-zt)/8; N2=(1+xi)*(1-nu)*(1-zt)/8;N3=(1+xi)*(1+nu)*(1-zt)/8; N4=(1-xi)*(1+nu)*(1-zt)/8;N5=(1-xi)*(1-nu)*(1+zt)/8; N6=(1+xi)*(1-nu)*(1+zt)/8;N7=(1+xi)*(1+nu)*(1+zt)/8; N8=(1-xi)*(1+nu)*(1+zt)/8;x=N1*x1+N2*x2+N3*x3+N4*x4+N5*x5+N6*x6+N7*x7+N8*x8y=N1*y1+N2*y2+N3*y3+N4*y4+N5*y5+N6*y6+N7*y7+N8*y8z=N1*z1+N2*z2+N3*z3+N4*z4+N5*z5+N6*z6+N7*z7+N8*z8u=N1*u1+N2*u2+N3*u3+N4*u4+N5*u5+N6*u6+N7*u7+N8*u8v=N1*v1+N2*v2+N3*v3+N4*v4+N5*v5+N6*v6+N7*v7+N8*v8w=N1*w1+N2*w2+N3*w3+N4*w4+N5*w5+N6*w6+N7*w7+N8*w8

Example 8.13 For the axisymmetric problem, if a 3-node triangular element is used, please derive the formula for B matrix (or namely strain-displacement matrix).Soln:Shape function: After replacing x by using r, y by z (in axisymmetric problem, two coordinates governing the mesh are r and z).

N1=12 A [( x2 y3−x3 y2)+( y2− y3) x+( x3−x2) y ]→1

2 A [(r2 z3−r3 z2 )+( z2−z3 )r+(r3−r2 )z ]

N2=12 A [( x3 y1−x1 y3)+( y3− y1) x+( x1−x3) y ]→1

2 A [(r3 z1−r1 z3 )+( z3−z1 )r+(r1−r3) z ]

N3=12 A [( x1 y2−x2 y1)+( y1− y2 ) x+( x2−x1) y ]→

12 A [(r 1 z2−r2 z1)+( z1−z2)r+(r2−r1)z ]

Or we can write as follows:

N1=1

2 A [α1+β1 r+γ 1 z ] , N2=1

2 A [α 2+β2 r+γ2 z ] , N3=1

2 A [α 3+β3r+γ3 z ]

12

Week 11 MECH3361

in which

α 1=r2 z3−r3 z2 , α2=r3 z1−r1 z3 , α3=r 1 z2−r2 z1 ,;β1= z2−z3 , β2=z3−z2 , β3=z1−z2 ,;γ1=r3−r2 , γ 2=r 1−r 3 , γ3=r2−r1 ,

2 A=det [1 r1 z1

1 r2 z2

1 r3 z3] (area)

Displacement field can be expressed in terms of shape function as:

u=∑i=1

3

N i ui=N 1 u1+N2 u2+N3 u3 w=∑i=1

3

N i w i=N1 w1+N 2 w2+N3 w3

or

{uw}=[N1 0 N 2 0 N 3 00 N1 0 N2 0 N3 ]{

u1

w1

u2

w2

u3

w3

}Strain – displacement relation ε rr=

∂u(r , z )∂r

, εθθ=u(r , z )

r, ε zz=

∂w(r , z )∂ z

, εrz=∂w(r , z )∂r

+∂u(r , z )∂ z

{εrr

εθθ

ε zz

εrz}={

∂ u(r , z )∂ r

u (r , z )r

∂w (r , z )∂ z

∂w(r , z )∂r

+∂ u(r , z )∂ z

}= 12 A [

∂N1

∂ r0

∂N 2

∂r0

∂ N3

∂r0

ur

0 ur

0 ur

0

0∂N 1

∂ z0

∂ N2

∂ z0

∂N3

∂ z∂N 1

∂ z∂N 1

∂r∂N 2

∂ z∂ N2

∂r∂ N3

∂ z∂N3

∂ r]{u1

w1

u2

w2

u3

w3

}{εrr

εθθ

ε zz

εrz}= 1

2 A [β1 0 β2 0 β3 0

α1

r+β1+

γ 1

rz 0

α2

r+β2+

γ2

rz 0

α3

r+β3+

γ3

rz 0

0 γ 1 0 γ 2 0 γ 3

γ1 β1 γ2 β2 γ 3 β3

]{u1

w1

u2

w2

u3

w3

}=Bd

B= 12 A [

β1 0 β2 0 β3 0α1

r+ β1+

γ1

rz 0

α2

r+β2+

γ 2

rz 0

α 3

r+β3+

γ 3

rz 0

0 γ1 0 γ2 0 γ3

γ 1 β1 γ 2 β2 γ 3 β3]

Note that B is not a constant matrix!Hooke’s law

σ={σrr

σθθ

σ zzσrz}= E(1+ν )(1−2 ν ) [1−ν ν ν 0

ν 1−ν ν 0ν ν 1−ν 0

0 0 0 1−2ν2

]{εrr

εθθ

εzzεrz}=Εε

13

Week 11 MECH3361

Strain energy

U=12∫V

σΤ ε dV= 12∫V(Eε )Τ ε dV=1

2∫VεΤ Eε dV=1

2∫V(Βd )Τ E(Βd )dV=dT [12∫V ΒΤ EΒdV ]d

Elemental Stiffness Matrix [K] for Axisymmetric element k=1

2∫VΒΤ EΒ dV

. Since B is not constant, numerical integration may be needed.

8.8 Finite Element Modelling TechniquesSymmetry Reflective (mirror, bilateral) symmetry Rotational (cyclic) symmetry Axisymmetry Translational symmetry

Nature of Finite Element Solutions FE Model – A mathematical model of the

real structure, based on many approximations.

Real Structure -- Infinite number of nodes (physical points or particles), thus infinite number of DOF’s.

FE Model – finite number of nodes, thus limited number of DOF’s.

Displacement field is controlled (or constrained) by the values at a limited number of nodes

Stiffening Effect: FE Model is stiffer than the real structure. In general, displacement results are smaller in

magnitudes than the exact values. Hence, exact solution provides an upper

bound for the FEM solution. In the convergence test, it usually provides the following diagram: The FEM solution approaches the exact solution from below. This is true for displacement based FEA only!

Numerical Error Numerical Error ( in solving FE equations) Modeling Error (approximation to beam, plate … theories) Discretization Error (finite, piecewise … )

Ill-conditionLet’s look at an example:FE equilibrium equation (after apply B.C.)

[k1 −k1

−k1 k1+k2 ]{u1

u2}={P0 }

14

Week 11 MECH3361

det[k1 −k1

−k1 k1+k2 ]=k1k2

The system will be singular if k 2 is much smaller than k1 (or vice verse)

Convergence of FE SolutionsAs the mesh in an FE model is “refined” repeatedly, the FE solution will converge to the exact solution of the mathematical model of the problem (the model based on bar, beam, plane stress/strain, plate, shell, or 3-D elasticity theories or assumptions).

Types of Refinement: h-refinement: reduce the size of the element (“h” refers to the typical size of the

elements), e.g. let smart size down from 5 to 1; p-refinement: Increase the order of the polynomials on an element, i.e. use a higher

order of elements (linear to quadratic, etc.; “h” refers to the higher order in a polynomial);

r-refinement: re-arrange the nodes in the mesh; hp-refinement: Combination of the h- and p-refinements (better results!).

Adaptivity (h-, p-, and hp-Methods)pi

Stress concentration, needs dense mesh to capture the stress gradient d) t = 0.0150 (fracture of bridge) Fracture pathFracture path

Mesh quality Know the behaviours of each type of elements:

T3 and Q4: linear displacement, constant strain and stress;T6 and Q8: quadratic displacement, linear strain and stress.

Choose the right type of elements for a given problem:When in doubt, use higher order elements or a finer mesh.

Avoid elements with large aspect ratios and corner angles:Aspect ratio = Lmax / Lminwhere Lmax and Lmin are the largest and smallest characteristic lengths of an element, respectively.

15

Week 11 MECH3361

Avoid singular elements

Example 8.14: Write elemental and global equilibrium equations of the following two CST finite element model. Go on to determine the nodal displacements and strain in both elements (assume: E=1, v=0.3, thickness = 1 for a plane stress problem).

Soln: Step 1: Connectivity of FE modelNumber of elements = 2; Number of nodes = 4; d.o.f.=8

Step 2: Elemental Stiffness matrix calculation for plane stress problem[K e ]=∫

V[B1 ]

T [ E ][B1 ]dV=∫V

dV ([B1 ]T [E ][B1 ])=tA [B1 ]

T [E ][B1 ]

¿ t4 A [

y23 0 x32

0 x32 y23

y31 0 x13

0 x13 y31y12 0 x21

0 x21 y12

](E1−ν2 [1 ν 0ν 1 0

0 0 1−ν2 ])[ y23 0 y31 0 y12 0

0 x32 0 x13 0 x21

x32 y23 x13 y31 x21 y12]

Element #1 (Counter-clockwise: Node 1 = Node A, Node 2 = Node B, Node 3 = Node C)E=1;nu=0.3;t=1;x1=0; %Node Ay1=0; %Node Ax2=1; %Node By2=0; %Node Bx3=1; %Node Cy3=1; %Node CA=(x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2))/2;x21=x2-x1;x13=x1-x3;x32=x3-x2;y12=y1-y2;

16

xA(0,0) B(1,0)

C(1,1)

D(0,1)y

1

2

Fx=1

Fy=1

Week 11 MECH3361

y31=y3-y1;y23=y2-y3;Bmat=[y23 0 y31 0 y12 0; 0 x32 0 x13 0 x21; x32 y23 x13 y31 x21 y12];Emat=(E/(1-nu*nu))*[1 nu 0; nu 1 0; 0 0 (1-nu)/2];Ke=t/(4*A)*Bmat'*Emat*Bmat

Ke= 0.5495 0 -0.5495 0.1648 0 -0.1648 0 0.1923 0.1923 -0.1923 -0.1923 0 -0.5495 0.1923 0.7418 -0.3571 -0.1923 0.1648 0.1648 -0.1923 -0.3571 0.7418 0.1923 -0.5495 0 -0.1923 -0.1923 0.1923 0.1923 0 -0.1648 0 0.1648 -0.5495 0 0.5495

Element #2 (Counter-clockwise: Node 1 = Node A, Node 2 = Node C, Node 3 = Node D)E=1;nu=0.3;t=1;x1=0; %Node Ay1=0; %Node Ax2=1; %Node Cy2=1; %Node Cx3=0; %Node Dy3=1; %Node DA=(x1*(y2-y3)+x2*(y3-y1)+x3*(y1-y2))/2;x21=x2-x1;x13=x1-x3;x32=x3-x2;y12=y1-y2;y31=y3-y1;y23=y2-y3;Bmat=[y23 0 y31 0 y12 0; 0 x32 0 x13 0 x21; x32 y23 x13 y31 x21 y12];Emat=(E/(1-nu*nu))*[1 nu 0; nu 1 0; 0 0 (1-nu)/2];Ke=t/(4*A)*Bmat'*Emat*Bmat

Ke= 0.1923 0 0 -0.1923 -0.1923 0.1923 0 0.5495 -0.1648 0 0.1648 -0.5495 0 -0.1648 0.5495 0 -0.5495 0.1648 -0.1923 0 0 0.1923 0.1923 -0.1923 -0.1923 0.1648 -0.5495 0.1923 0.7418 -0.3571 0.1923 -0.5495 0.1648 -0.1923 -0.3571 0.7418

Step 3: Elemental equilibrium equations

Element #1:

[0 .5495 0 −0.5495 0 . 1648 0 −0 .1648

0 0.1923 0 . 1923 −0 .1923 −0 .1923 0−0 .5495 0.1923 0 . 7418 −0 .3571 −0 .1923 0 .16480 .1648 −0 .1923 −0.3571 0 . 7418 0 .1923 −0 .5495

0 −0 .1923 −0.1923 0 . 1923 0 .1923 0−0 .1648 0 0 . 1648 −0 .5495 0 0 .5495

]{uA

v A

uB

vB

uC

vC

}={f xA(1 )

f yA(1 )

f xB(1 )

f yB(1 )

f xC(1 )

f yC(1)}

17

Week 11 MECH3361

Element #2:

[0 .1923 0 0 −0 .1923 −0 .1923 0 .1923

0 0.5495 −0. 1648 0 0 .1648 −0 .54950 −0 .1648 0 .5495 0 −0 .5495 0 .1648

−0 .1923 0 0 0 . 1923 0 .1923 −0 .1923−0 .1923 −0 .1648 −0. 5495 0 . 1923 0 .7418 −0 .35710 .1923 −0 .5495 0 .1648 −0 .1923 −0 .3571 0 .7418

]{uA

v A

uC

vC

uD

vD

}={f xA(2)

f yA(2)

f xC(2)

f yC(2)

f xD(2)

f yD(2 )}

Step 4: Global equilibrium equationsExpand the equilibrium equations

according global displacement vector:

d= {uA v A uB vB uC vC uD vD }T

Elemental equilibrium equation: kd=f

Element #1:

[0 .5495 0 −0 . 5495 0 . 1648 0 −0 .1648 0 0

0 0. 1923 0 . 1923 −0 . 1923 −0 .1923 0 0 0−0 .5495 0. 1923 0 . 7418 −0 . 3571 −0 .1923 0 .1648 0 00 .1648 −0. 1923 −0. 3571 0 . 7418 0 .1923 −0 .5495 0 0

0 −0. 1923 −0 . 1923 0 . 1923 0 .1923 0 0 0−0 .1648 0 0 . 1648 −0 . 5495 0 0 .5495 0 0

0 0 0 0 0 0 0 00 0 0 0 0 0 0 0

]{u A

v A

uB

v B

uC

vC

uD

v D

}={f xA(1 )

f yA(1 )

f xB(1 )

f yB(1 )

f xC(1 )

f yC(1)

00

}

Element #2:

[0 .1923 0 0 0 0 −0 .1923 −0 .1923 0 .1923

0 0.5495 0 0 −0 . 1648 0 0.1648 −0 .54950 0 0 0 0 0 0 00 0 0 0 0 0 0 00 −0. 1648 0 0 0 .5495 0 −0 .5495 0 .1648

−0 .1923 0 0 0 0 0 .1923 0.1923 −0 .1923−0 .1923 −0. 1648 0 0 −0 . 5495 0 .1923 0.7418 −0.35710 .1923 −0. 5495 0 0 0 .1648 −0 .1923 −0 .3571 0 .7418

]{u A

v A

uB

v B

uC

vC

uD

v D

}={f xA(2 )

f yA(2 )

00

f xC(2 )

f yC(2 )

f xD(2 )

f yD(2)

}Global equilibrium equationsAdd the above two expanded equations (Element #1) + (Element #2):

18

Week 11 MECH3361

[0 .7418 0 −0 . 5495 0 .1648 0 −0. 3571 −0 . 1923 0 .1923

0 . 7418 0 . 1923 −0 .1923 −0 .3571 0 0 . 1648 −0 .54950 . 7418 −0 .3571 −0 .1923 0 . 1648 0 0

0 .7418 0.1923 −0. 5495 0 00.7418 0 −0 . 5495 0 .1648

Sym 0 . 7418 0 . 1923 −0 .19230 . 7418 −0 . 3571

0 .7418]{

uA

v A

uB

vB

uC

vC

uD

v D

}={f xA(1)

f yA(1)

f xB(1)

f yB(1)

f xC(1)

f yC(1)

00

}+{f xA(2 )

f yA(2 )

00

f xC(2 )

f yC(2 )

f xD(2 )

f yD(2)

}={FxA

F yA

FxB

F yB

F xC

F yC

F xD

F yD

}Step 5: Boundary and loading conditions

[0 .7418 0 −0 .5495 0 .1648 0 −0. 3571 −0 . 1923 0 .1923

0 . 7418 0 . 1923 −0 .1923 −0 .3571 0 0 . 1648 −0 .54950 . 7418 −0 .3571 −0 .1923 0 . 1648 0 0

0 .7418 0. 1923 −0. 5495 0 00. 7418 0 −0 . 5495 0 .1648

Sym 0 . 7418 0 . 1923 −0 .19230 . 7418 −0 . 3571

0 .7418]{

0000uC

vC

00

}={FxA

F yA

FxB

F yB

11

FxD

F yD

}

yD

xD

yB

xB

yA

xA

C

C

FF

FFFF

vu

Sym 11

00

0000

7418.03571.07418.01923.01923.07418.0

1648.05495.007418.0005495.01923.07418.0001648.01923.03571.07418.05495.01648.003571.01923.01923.07418.0

1923.01923.03571.001648.05495.007418.0

Step 6: Determine and plot displacementThe equilibrium equation is reduced to 2×2 as below.

[0 .7418 00 0 . 7418 ]{uC

vC}={11 }

{uC

vC}={1/0 . 7428

1/0 .7428 }={1 .34631. 3463 }

d= {uA v A uB vB uC vC uD vD }T={0 0 0 0 1 .3463 1 . 3463 0 0 }T

x

y

u

B(1,0)

C(1,1)

D(0,1)

A(0,0)

x

y

v

B(1,0)

C(1,1)

D(0,1)

A(0,0)

19

Week 11 MECH3361

Step 7: Calculate and plot elemental strains:

{εxxε yy

ε xy}=Bd= 1

2 A [ y2− y3 0 y3− y1 0 y1− y2 00 x3−x2 0 x1−x3 0 x2−x1

x3−x2 y2− y3 x1−x3 y3− y1 x2−x1 y1− y2]{

u1

v1

u2

v2

u3

v3

}Strain in element #1 (Node 1 = Node A, Node 2 = Node B, Node 3 = Node C)

ε xx(1)=1

2 A [ ( y2− y3)u1+( y3− y1)u2+( y1− y2 )u3]=12 A [ ( y B− yC )u A+( yC− y A )uB+( y A− y B)uC ]

¿12×0.5 [ (0−1 )×(0 )+ (1−0 )× (0 )+(0−0 )× (1 .3463 ) ]=0

ε yy(1)=1

2 A [ (x3−x2)v1+( x1−x3 )v2+( x2−x1) v3]=12 A [ (xC−xB )v A+ (x A−xC )v B+( xB−x A ) vC]

¿12×0. 5 [ (1−1 )×(0 )+(0−1 )×(0 )+ (1−0 )×(1. 3463 ) ]=1 .3463

ε xy(1)=1

2 A [ (x3−x2)u1+( y2− y3 )v1+ (x1−x3 )u2+( y3− y1)v2+ (x2−x1 )u3+( y1− y2 )v3 ]¿1

2 A [ (xC−xB )uA+ ( yB− yC) v A+(x A−xC )uB+ ( yC− y A ) vB+(xB−x A )uC+ ( y A− yB )vC ]¿1

2×0.5 [ (1−1 )0+(0−1 )0+(0−1 )0+ (1−0 )0+(1−0 )1. 3463+(0−0 )1 ]=1. 3463

Strain in element #2 (Node 1 = Node A, Node 2 = Node C, Node 3 = Node D)

ε xx(2)=1

2 A [ ( y2− y3 )u1+( y3− y1 )u2+( y1− y2 )u3]=12 A [( yC− y D )uA+( yD− y A )uC+( y A− yC )uD ]

¿12×0. 5 [ (1−1 )×(0 )+(1−0 )×(1 .3463 )+ (0−1 )× (0 ) ]=1 .3463

ε yy(2)=1

2 A [ (x3−x2)v1+ (x1−x3 )v2+( x2−x1) v3]=12 A [ (x D−xC )v A+(x A−xD ) vC+( xC−xA )v D ]

¿12×0. 5 [ (0−1 )×(0 )+ (0−0 )×(1. 3463 )+(1−0 )×( 0 ) ]=0

ε xy(2)=1

2 A [ (x3−x2)u1+( y2− y3) v1+(x1−x3 )u2+( y3− y1)v2+ (x2−x1 )u3+( y1− y2 )v3 ]¿1

2 A [ (x D−xC )uA+( yC− yD ) v A+(x A−x D )uC+( y D− y A )vC+ (xC−x A )uD+( y A− yC )vD ]¿1

2×0. 5 [ (0−1 )0+ (1−1 )0+(0−0 )1. 3463+(1−0 )1 .3463+ (1−0 )0+ (0−1 )0 ]=1 .3463

x

y

xx

B(1,0)

C(1,1)

D(0,1)

A(0,0)

x

y

B(1,0)

C(1,1)

D(0,1)

A(0,0)

12

yy

1 2

1.3643

1.3643

20

Week 11 MECH3361

x

y

xy

B(1,0)

C(1,1)

D(0,1)

A(0,0)

12

1.3643

Step

8: Calculate and plot elemental stresses: σ=Εε

{σ xx

σ yy

σ xy}=Εε= E

1−ν2 [1 ν 0ν 1 0

0 0 1−ν2 ]{εxx

ε yy

ε xy}= 1

1−0.32 [ 1 0 .3 00 .3 1 0

0 0 1−0. 32 ]{ε xx

ε yy

εxy}=[ 1 .098 0 .3294 0

0 .3294 1 .098 00 0 0 .3843 ]{ε xx

ε yy

εxy}

Stress in element #1: {σ xx

σ yy

σ xy}=[ 1 .098 0 .3294 0

0 .3294 1 .098 00 0 0 .3843 ]{

01.36431 .3643}={

0.44941.49800 .5243 }

St

ress in element #2: {σ xx

σ yy

σ xy}=[ 1 .098 0 .3294 0

0 .3294 1 .098 00 0 0 .3843 ]{

1.36430

1 . 3643}={1.49800.44940 .5243 }

x

y

xx

B(1,0)

C(1,1)

D(0,1)

A(0,0)

12

1.49800.4494

x

y

yy

B(1,0)

C(1,1)D(0,1)

A(0,0)

12

1.4980

0.4494

x

y

xy

B(1,0)

C(1,1)D(0,1)

A(0,0)

12 0.5243

0.5243

21